Concept Of Physics Rotational Mechanics

H C Verma

Concept Of Physics

1.   A wheel is making revolutions about its axis with uniform angular acceleration. Starting from rest, it reaches $100$ rev/sec in $4$ seconds. Find the angular acceleration. Find the angle rotated during these four seconds.

$\omega_{0}$ = $0$;$\rho$ = $100$ $\frac{rev}{s}$ ; $\omega$ = 2$\pi;\rho$ = 200$\pi\frac{rad}{s}$$\\$ $\Rightarrow \omega$ $=$ $\omega_{0}$ $=$ $\alpha$$t$ $\\$ $\Rightarrow \omega$ $=$ $\alpha$$t$ $\\$ $\Rightarrow \alpha$ $=$ $\frac{200\pi}{4}$ $ = 50\pi\frac{rad}{s^2}$ or $25\frac{rev}{s^2}$ $\\$ $\therefore$ $\theta$ $=$ $\omega_{0}t$ $+$ $\frac{1}{2}$ $\alpha$$t^2$ $= 8 \times 50$$\pi$=400$\pi$ $rad$ $\\$ $\\$ $\therefore \alpha$ $= 50\pi\frac{rad}{s^2}$ or $25\frac{rev}{s^2}$ $\\$$\theta$ = 400 $\pi$ $rad$

2.   A wheel rotating with uniform angular acceleration covers $50$ revolutions in the first five seconds after the start. Find the angular acceleration and the angular velocity at the end of five seconds.

$\\$$\theta$ = 400 $\pi$; $t = 5 sec$ $\\$$\theta$ = $\frac{1}{2}$ $\alpha$$t^2$ $\Rightarrow 100 \pi$ = $\frac{1}{2}$ $\alpha$ 25$\\$ $\Rightarrow \alpha$ $=$ 8$\pi$ $\times$$5$ = 40$\pi\frac{rad}{s}$ = $20\frac{rev}{s}$$\\$ $\\$ $\therefore \alpha$ $= 8\pi\frac{rad}{s^2}$ = $4\frac{rev}{s^2}$$\\$ $\omega$ $= 40\pi\frac{rad}{s^2}$ = $20\frac{rev}{s^2}$

3.   A wheel starting from rest is uniformly accelerated at $4$ rad/s $2$ for $10$ seconds.It is allowed to rotate uniformly for the next $10$ seconds and is finally brought to rest in the next $10$ seconds. Find the total angle rotated by the wheel.

Area under the curve will decide the total angle rotated$\\$ $\therefore$ Maximum angular velocity = $4$ $\times$ $10$ = $40 \frac{rev}{s}$$\\$ Therefore, area under the curve = $\frac{1}{2}$ $\times$ $10$ $\times$ $40$ + $40$ $\times$ $10$ + $\frac{1}{2}$ $\times$ $40$ $\times$ $10$ = 800 rad $\\$$\therefore$ Total angle rotated = $800$ rad

4.   A body rotates about a fixed axis with an angular acceleration of one radian/second/second. Through what angle does it rotate during the time in which its angular velocity increases from $5$ rad/s to $15$ rad/s.

$ \alpha$ $= 1 \frac{rad}{s^2},$ $\omega_{0}$ $= 5 \frac{rad}{s};$ $= 15 \frac{rad}{s}$ $\\$ $\therefore W$ = $W_{0}$ + $ \alpha$$t$ $\\$$\Rightarrow t$ = $ \frac{( \omega - \omega_{0})}{ \alpha }$ = $ \frac{( 15 - 5)}{ 1}$ = $10$ sec $\\$ Also, $\theta$ $=$ $\omega_{0}t$ $+$ $\frac{1}{2}$ $\alpha$$t^2$ $\\$= $5$ $\times$ $10$ + $\frac{1}{2}$ $\times$ 1 $\times$ $100$ = $100$ rad.

5.   Find the angular velocity of a body rotating with an acceleration of $2$ rev/s $2$ as it completes the 5th revolution after the start.

$\theta$ = $5$ rev, $ \alpha = 2 \frac{rev}{s^2},$ $\omega_{0}$ = $0$ ; $\omega_{0}$ = $?$ $\\$ $\omega^2$ = ( 2 $ \alpha $ $\theta$ ) $\\$ $\Rightarrow \omega$ = $\sqrt {2 \times 2 \times 5}$ = $2\sqrt 5$ $\frac{rev}{s}$ $\\$ or $\theta$ = 10 $\pi$ $rad,$ $ \alpha = 4 \pi \frac{rad}{s^2},$ $\omega_{0}$ = $0$ ; $\omega_{0}$ = $?$ $\\$ $\omega$ = $\sqrt {2 \alpha \theta}$ = $ 2 \times 4 \pi \times 10 \pi$ $\\$ = $4 \pi \sqrt 5$ $\frac{rad}{s} = 2 \sqrt 5$ $\frac{rev}{s}.$

6.   A disc of radius $10$ cm is rotating about its axis at an angular speed of $20$ rad/s. Find the linear speed of (a) a point on the rim, (b) the middle point of a radius.

A dish of radius = $10$ cm = $0.1$ m $\\$ Angular velocity = $20 \frac {rad}{s}$ $\\$ $\therefore$ Linear velocity on the rim = $\omega$r = $20 \times 0.1 = 2 \frac {m}{s}$ $\\$ $\therefore$ Linear velocity at the middle of the radius = $\omega \frac {r}{2}$ = $20 \times \frac{(0.1)}{2}$ = 1 $\frac{m}{s}$

7.   A disc rotates about its axis with a constant angular acceleration of $4 \frac{rad}{s^2}$. Find the radial and tangential accelerations of a particle at a distance of $1$ cm from the axis at the end of the first second after the disc starts rotating.

$t = 1 sec, r = 1 cm = 0.0.1 m$ $\\$ $\alpha = 4 \frac{rd}{s^2}$ $\\$ Therefore $\omega$ = $\alpha t$ = $4 \frac{rad}{s}$ $\\$ Therefore the radial acceleration, $\\$ $A_n$ = $\omega^2r$ = 0.16 $\frac{m}{s^2}$ = 16 $\frac{cm}{s^2}$ $\\$ Therefore tangential acceleration, $a_r$ = $\alpha r$ = 0.04 $\frac{m}{s^2}$ = 4 $\frac{cm}{s^2}$

8.   A block hangs from a string wrapped on a disc of radius $20$ cm free to rotate about its axis which is fixed in a horizontal position. If the angular speed of the disc is $10 \frac{rad}{s}$ at some instant, with what speed is the block going down at that instant ?

The Block is moving the rim of the pulley $\\$ The pulley is moving at a $\omega$ = $10 \frac{rad}{s}$ $\\$ Therefore the radius of the pulley = $20 cm$ $\\$ Therefore linear velocity on the rim = tangential velocity = r$\omega$ $\\$ = $20 \times 20$ = $200 \frac{cm}{s}$ = $2 \frac{m}{s}$

9.   Three particles, each of mass $200 g$, are kept at the corners of an equilateral triangle of side $10 cm$. Find the moment of inertia of the system about an axis (a) joining two of the particles and (b) passing through one of the particles and perpendi- cular to the plane of the particles.

Therefore, the $\perp$ distance from the axis (AD) = $\frac{\sqrt3}{2}$ $\times 10$ = $5 \sqrt3 cm$ $\\$ Therefore moment of inertia about the axis BC will be $\\$ $I = mr^2$ = $200 K (5\sqrt3)^2$ = $200 \times 25 \times 3$ $\\$ $= 15000 gm - cm^2$ = $1.5 \times 10^{-3} kg - m^2$ $\\$ b) The axis of rotation let pass through A and $\perp$ to the plane of triangle $\\$ Therefore the torque will be produced by mass B and C $\\$ Therefore net moment of inertia $= I = mr^2 + mr^2$ $\\$ = $2 \times 200 \times 10^2$ = $40000 gm - cm^2$ = $4 \times 10^{-3}kg-m^2$

10.   Particles of masses $1 g, 2 g, 3 g, .........., 100 g$ are kept at the marks $1 cm, 2 cm, 3 cm, ............, 100 cm$ respectively on a metre scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the metre scale.

Masses of $1 gm, 2gm, ....... 100 gm $ are kept at the marks $1 cm, 2 cm, ...... 1000 cm$ on he \times axis respectively. A perpendicular axis is passed at the $50^{th}$ particle. $\\$ Therefore on the L.H.S. side of the axis there will be 49 particles and on the R.H.S. side there are 50 particles. $\\$ Consider the two particles at the position $49 cm$ and $51 cm.$ $\\$ Moment inertial due to these two particles will be = $49 \times 1^2 + 51 + 1^2$ = $100 gm - cm^2$ $\\$ Similarly if we consider $48^{th}$ and $52^{nd}$ term we will get $100 \times 2^2 gm - cm^2$ $\\$ Therefore we will get 49 such set and one lone particle at $100 cm.$ $\\$ Therefore total moment of inertia = $100 ${$1^2 + 2^2 + 3^2 + ......... + 49^2$} + $100(50)^2.$ $\\$ = $100 \times \frac{(50 \times 51 \times 101)}{6}$ = $4292500 gm - cm^2$ $\\$ = $0.429kg - m^2$ = $0.43 kg - m^2.$

11.   Find the moment of inertia of a pair of spheres, each having a mass m and radius r, kept in contact about the tangent passing through the point of contact.

The two bodies of mass m and radius r are moving along the common tangent. $\\$ Therefore moment of inertia of the first body about XY tangent. $\\$ = $mr^2$ + $\frac {2}{5} mr^2$ $\\$ oment of inertia of the second body about XY tangent = $mr^2$ + $\frac {2}{5} mr^2$ = $\frac {7}{5} mr^2$ $\\$ Therefore, net moment of inertia = $\frac {7}{5} mr^2$ + $\frac {7}{5} mr^2$ = $\frac {14}{5} mr^2$ units.

12.   The moment of inertia of a uniform rod of mass $0.50 kg$ and length $1 m$ is $0.10 kg-m2$ about a line perpendicular to the rod. Find the distance of this line from the middle point of the rod.

Length of the rod = $1 m,$ mass of the rod = $0.5 kg$ $\\$ Let at a distance d from the center the rod is moving $\\$ Applying parallel axis theorem : $\\$ The moment of inertia about the point $\\$ $\Rightarrow$ $\frac{(mL^2)}{12} + md^2 = 0.10$ $\\$ $\Rightarrow$ $\frac{(0.5 \times 1^2)}{12}$ + $0.5 \times d^2$ = $0.10$ $\\$ $\Rightarrow$ $d^2$ = $0.2 - 0.082$ = $0.118$ $\\$ $\Rightarrow$ $d$ = $0.342 m$ from the center.

13.   Find the radius of gyration of a circular ring of radius r about a line perpendicular to the plane of the ring and passing through one of its particles.

Moment of inertia at the center and perpendicular to the plane of the ring. $\\$ So, about a point on the rim of the ring and the axis $\perp$ to the plane of the ring, the moment of inertia $\\$ = $mR^2 + mR^2$ = $2 mR^2$ (parallel axis theorem) $\\$ $\Rightarrow$ $mK^2$ = $2 mR^2$ (K = radius of the gyration) $\\$ $\Rightarrow$ K = $\sqrt {2R^2}$ = $\sqrt 2 R.$

14.   The radius of gyration of a uniform disc about a line perpendicular to the disc equals its radius. Find the distance of the line from the centre.

The moment of inertia about the center and $\perp$ to the plane of the disc of radius r and mass m is $mr^2.$ $\\$ According to the question the radius of gyration of the disc about a pint = radius of the disc. $\\$ Therefore $mk^2$ = $\frac {1}{2} mr^2$ + $md^2$ $\\$ (K = radius of the gyration about acceleration point, d = distance of the point from the center) $\\$ $\Rightarrow$ $K^2$ = $\frac{r^2}{2}$ + $d^2$ $\\$ $\Rightarrow$ $r^2$ = $\frac{r^2}{2}$ + $d^2$ $($$\therefore$ K= r$)$ $\\$ $\Rightarrow$ = $\frac{r^2}{2}$ = $d^2$ $\Rightarrow$ $d = \frac{r}{\sqrt2}$

15.   Find the moment of inertia of a uniform square plate of mass m and edge a about one of its diagonals.

Let a small cross sectional area is at a distance x from xx axis. $\\$ Therefore mass of that small section = $\frac{m}{a^2}$ $\times$ ax dx $\\$ Therefore moment of inertia about xx axis = $I_{xx}$ = 2 $\int_0^{\frac{a}{2}} $ $(\frac{m}{a^2})$ $\times$ x^2 = $(2 \times (\frac {m}{a})(\frac{x^3}{3})$ $]_0^{\frac{a}{2}}$ $\\$ = $\frac{ma^2}{12}$ $\\$ = Therefore $I_{xx}$ = $I_{xx}$ + $I_{yy}$ $\\$ = 2 $\times$ $\frac{^*ma^2}{12}$ = $\frac{^*ma^2}{6}$ $\\$ Since the two diagonals are $\perp$ to each other $\\$ = Therefore $I_{zz}$ = $I_{x'x'}$ + $I_{y'y'}$ $\\$ $\Rightarrow$ $\frac{ma^2}{6}$ = 2 $\times$ $I_{x'x'}$ (because $I_{x'x'}$ = $I_{y'y'}$ $\Rightarrow$ $I_{x'x'}$ = $\frac{ma^2}{12}$

16.   The surface density (mass/area) of a circular disc of radius a depends on the distance from the centre as p(r) = A + Br. Find its moment of inertia about the line perpendicular to the plane of the disc through its centre.

The surface density of a circular disc of radius a depends upon the distance from the center as P(r) = A + Br $\\$ Therefore the mass of the ring of radius r will be $\\$ = $\int_o^a (A+Br)2\pi r \times dr$ = $\int_o^a 2\pi Ar^3dr$ + $\int_o^a 2\pi Br^4dr$ $\\$ = $2\pi A (\frac{r^4}{4})$ + $2\pi B(\frac{r^5}{5})]_o^a$ = $2 \pi a^4 [(\frac{A}{4}) + (\frac{Ba}{5})].$

17.   A particle of mass m is projected with a speed u at an angle $0$ with the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point.

At the highest point total force acting on the particle id its weight acting downward. $\\$ Range of the particle = $u^2 sin \frac{2\pi}{g}$ $\\$ Therefore force is at a $\perp$ distance, $\Rightarrow$ $\frac{(total range)}{2}$ = $\frac{(v^2 sin 2 \theta )}{2g}$ $\\$ (From the initial point) $\\$ Therefore $\tau$ = F $\times r (\theta = angle of projection)$ $\\$ = $ mg \times v^2 sin \frac{2\theta}{2g} (v= intial velocity)$. $\\$ = $mv^2 sin \frac{2\theta}{2}$ = $mv^2 sin \theta cos \theta.$

18.   A simple pendulum of length $1$ is pulled aside to make an angle $0$ with the vertical. Find the magnitude of the torque of the weight w of the bob about the point of suspension. When is the torque zero ?

A simple of pendulum of length l is suspended from a rigid support. A body of weight W is hanging on the other point. $\\$ When the bob is at an angle $\theta$ with the vertical, then total torque acting on the point of suspension = $i = F \times r$ $\\$ $\Rightarrow$ W r $sin \theta$ = W l $sin \theta$ $\\$ At the lowest point of suspension the torque will be zero as the force acting on the body passes through the point of suspension.

19.   When a force of $6.0$ N is exerted at $30°$ to a wrench at a distance of $8$ cm from the nut, it is just able to loosen the nut. What force F would be sufficient to loosen it if it acts perpendicularly to the wrench at $16$ cm from the nut ?

A force of $6$ N acting at an angle of $30^o$ is just able to loosen the wrench at a distance $8 cm$ from it. $\\$ Therefore total torque required at B about the point 0. $\\$ 6 $sin30^o \times \frac{8}{100}$. $\\$ Therefore total torque required at B about the point 0. $\\$ = F $\times \frac{16}{100}$ $\Rightarrow$ F $\times \frac{16}{100}$ = 6 $sin 30^o \times \frac{8}{100}$ $\\$ F = $\frac{(8 \times 3)}{16}$ = 1.5 N.

20.   Calculate the total torque acting on the body shown in figure $(10-E2)$ about the point $0$.

Torque about a point = Total force $\times$ perpendicular distance from the point to that force. $\\$ Let anticlockwise torque = + ve $\\$ And clockwise acting torque = - ve $\\$ Force acting at the point B is 15 N $\\$ Therefore torque at O due to this force $\\$ $15 \times 6 \times 10^{-2} \times sin 37^o$ $\\$ $15 \times 6 \times 10^{-2} \times \frac{3}{5}$ = 0.54 N-m (anticlock wise). $\\$ Force acting at the point C is 10 N $\\$ Therefore, torque at O due to this force $\\$ $10 \times 4 \times 10^{-2}$ = 0.4 N - m (clockwise) $\\$ Force acting at the point A is 20 N $\\$ Therefore, Torque at O due to this force = $20 \times 4 \times 10^{-2} \times sin30^o$ $\\$ = $20 \times 4 \times 10^{-2} \times \frac{1}{2}$ = 0.4 N-m (anticlockwise) $\\$ Therefore resultant torque acting at 'O' = $0.54 - 0.4 + 0.4 = 0.54 N-m.

21.   A cubical block of mass m and edge a slides down a rough inclined plane of inclination $0$ with a uniform speed. Find the torque of the normal force acting on the block about its centre.

The force mg acting on the body has two components mg $sin \theta$ and mg $cos \theta$ and the body will exert a normal reaction. $\\$ Let R = Since R and mg $cos \theta$ pass through the center of the cube, there will be no torque due to R and mg $cos \theta$. $\\$ The only torque will be produced by mg $sin \theta$. $\\$ $\therefore$ i = $F \times r(r = \frac{a}{2})$ (a = ages of cube) $\\$ $\Rightarrow$ i = mg $sin \theta \times \frac {a}{2}$. $\\$ = $\frac{1}{2}$ mg a sin $\theta$

22.   A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis through its centre. A horizontal force of constant magnitude F acts on the rod at a distance of L/4 from the centre. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after the motion starts.

A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis passing through its center. $\\$ A force F is acting perpendicular to the rod at a distance $\frac {L}{4}$ from the center. $\\$ Therefore torque about the center due to this force $\\$ $i_i$ = F $\times r$ = F $\frac{L}{4}$. $\\$ This torque will produce a angular acceleration $\alpha$. $\\$ Therefore $\tau_c$ = $l_c \times \alpha$ $\\$ $\Rightarrow$ $i_c$ = $(\frac{mL^2}{12}) \times \alpha$($l_c$ of a rod = $\frac{mL^2}{12})$ $\\$ $\Rightarrow$ F $\frac {i}{4}$ = ($\frac{mL^2}{12}$) $\times \alpha$ $\Rightarrow$ $\alpha$ = 3$\frac{F}{ml}$. $\\$ Therefore $\theta$ = $\frac{1}{2} \alpha t^2$ (initially at rest) $\\$ $\Rightarrow$ $\theta$ = $\frac{1}{2} \times \frac{3F}{ml}t^2$ = $\frac{3F}{2ml}t^2$

23.   A square plate of mass $120 g$ and edge $5.0 cm$ rotates about one of the edges. If it has a uniform angular acceleration of $0.2$ rad/s $2$ what torque acts on the plate ?

A square plate of mass 120 gm and edge 5 cm rotates about one of the edge. $\\$ Let take a small area of the square of width dx and length a which is at a distance x from the axis of rotation. $\\$ Therefore mass of the small area $\\$ $\frac{m}{a^2} \times a dx (m = mass of the square ; a = side of the plate)$ $\\$ = $\int_0^a(\frac{m}{a^2}) \times ax^2dx$ = $(\frac{m}{a})(\frac{ma^2}{3})]_0^a$ $\\$ $\frac{ma^2}{3}$ $\\$ Therefore torque produced = l $\times \alpha = (\frac{ma^2}{3}) \times \alpha$ $\\$ = ${\frac{(120 \times 10^{-3} \times 5^2 \times 10^{-4})}{3}}0.2$ $\\$ = $0.2 \times 10^{-4} = 2 \times 10^{-5} N-m.$

24.   Calculate the torque on the square plate of the previous problem if it rotates about a diagonal with the same angular acceleration.

$\\$ Moment of inertial of a square plate about its diagonal is $\frac{ma^2}{12}(m= mass of the square plate)$ $\\$ a edges of the square $\\$ Therefore torque produced = $\frac{ma^2}{12} \times \alpha$ $\\$ = ${\frac{(120 \times 10^{-3} \times 5^2 \times 10^{-4})}{12}} \times 0.2$ $\\$ = $0.5 \times 10^{-5} N-m.$

25.   A flywheel of moment of inertia 5.0 kg-m 2 is rotated at a speed of 60 rad/s. Because of the friction at the axle, it comes to rest in 5.0 minutes. Find (a) the average torque of the friction, (b) the total work done by the friction and (c) the angular momentum of the wheel 1 minute before it stops rotating.

$\\$ A flywheel of moment of inertia $5$ kg m is rotated at a speed of $60 \frac{rad}{s}.$ The flywheel comes to rest due to the friction at the axle after 5 minutes. $\\$ Therefore, the angular deceleration produced due to frictional force = $\omega = \omega_0 + \alpha t$ $\\$ $\Rightarrow$ $\omega_0$ = $-\alpha t(\omega = 0)$ $\\$ $\Rightarrow$ $\alpha$ = $-(\frac{60}{5} \times 60)$ = $\frac{-1}{5} \frac{rad}{s^2}.$ $\\$ a) Therefore total work done in stopping the wheel by frictional force $\\$ W = $\frac{1}{2} i\omega^2$ = $\frac{1}{2} \times 5 \times (60 \times 60)$ = 9000 Joule = 9 KJ. $\\$ b)Therefore torque produced by the frictional force (R) is $\\$ $l_R$ = $l \times \alpha$ = $5 \times (\frac{-1}{5})$ = lN - m opposite to the rotation of the wheel. $\\$ c) Angular velocity after 4 minutes $\\$ $\Rightarrow \omega$ = $\omega_0 + \alpha t$ 60 - $\frac{240}{5}$ = 12 $\frac{rad}{s}$ $\\$ Therefore angular momentum about the center = $1 \times \omega$ = $5 \times 12$ = $60 kg-\frac{m^2}{s}.$

26.   Because of the friction between the water in oceans with the earth's surface, the rotational kinetic energy of the earth is continuously decreasing. If the earth's angular speed decreases by 0.0016 rad/day in 100 years, find the average torque of the friction on the earth. Radius of the earth is 6400 km and its mass is 6.0 x 10 24 kg.

$\\$ The earth's angular speed decreases by $0.0016 \frac{rad}{day}$ in 1000 years. $\\$ Therefore the torque produced by the ocean water in decreasing earth's angular velocity $\\$ $\tau = l \alpha$ $\\$ = $\frac{2}{5}mr^2 \times \frac{(\omega - \omega_0)}{t}$ $\\$ = $\frac{2}{6} \times 6 \times 10^{24} \times 64^2 \times 10^{10} \times [\frac{0.0016}{(26400^2 \times 100 \times 365)}] (1 year = 365 days = 365 \times 56400 sec)$ $\\$ = $5.678 \times 10^{20} N-m.$

27.   A wheel rotating at a speed of 600 rpm (revolutions per minute) about its axis is brought to rest by applying a constant torque for 10 seconds. Find the angular deceleration and the angular velocity 5 seconds after the application of the torque.

$\\$ A wheel rotating at a speed of 600 rpm. $\\$ $\omega_0$ = 600rpm = 10 revolutions per second. $\omega$ = 0 $\\$ Therefore $\omega_0$ = -$\alpha t$ $\\$ $\Rightarrow \alpha$ = -$\frac{10}{10}$ = -1 $\frac{rev}{s^2}$ $\\$ $\Rightarrow \omega$ = $\omega_0 + \alpha t$ = $10 -1 \times 5 = 5 \frac{rev}{s}.$ $\\$ Therefore angular deceleration = 1 $\frac{rev}{s^2}$ and angular velocity of after 5 sec is 5 $\frac{rev}{s}.$n$\\$

28.   A wheel of mass 10 kg and radius 20 cm is rotating at an angular speed of 100 rev/min when the motor is turned off. Neglecting the friction at the axle, calculate the force that must be applied tangentially to the wheel to bring it to rest in 10 revolutions.

$\\$ $\omega$ = 100 $\frac{rev}{min}$ = $\frac{5}{8} \frac{rev}{s}$ = $\frac{10\pi}{3} \frac{rad}{s}$ $\\$ $\theta$ = 10 rev = 20 $\pi$ rad, r = 0.2 m $\\$ After 10 revolutions the wheel will come to rest by a tangential force $\\$ Therefore the angular deceleration produced by the force = $\alpha$ = $\frac{\omega^2}{2\theta}$ $\\$ Therefore the torque by which the wheel will come to an rest = $l_{cm}$ $\times \alpha$ $\\$ F $\times r$ = $l_{cm} \times \alpha$ $\rightarrow$ F$\times$ 0.2 = $\frac{1}{2}mr^2 \times [\frac{(\frac{10\pi}{3})^2}{(2 \times 20\pi)}]$ $\\$ $\Rightarrow$ F = $\frac{1}{2} \times 10 \times 0.2 \times \frac{100 \pi^2}{(9 \times 2 \times 20 \pi)} $ $\\$ = $\frac{5\pi}{18}$ = $\frac{15.7}{18} = 0.87 N.$

29.   A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?

$\\$ A cylinder is moving with an angular velocity 50 $\frac{rev}{s}$ brought in contact with another identical cylinder cylinder in rest. The first and second cylinder has common acceleration and deceleration as 1 $\frac{rad}{s^2}$ respectively. $\\$ Let after t sec their angular velocity will be same '$\omega$'. $\\$ For the first cylinder $\omega$ = 50 - $\alpha t$ $\\$ $\Rightarrow$ t = $\frac{(\omega - 50)}{-1}$ $\\$ And for the $2^{nd}$ cylinder $\omega$ = $\alpha_2t$ $\\$ $\Rightarrow t = \frac{\omega}{l}$ $\\$ So, $\omega = \frac{(\omega-50)}{-1}$ $\\$ $\Rightarrow 2\omega = 50 \Rightarrow \omega = 25 \frac{rev}{s}$ $\\$ $\Rightarrow t = \frac{25}{1} sec = 25 sec.$

30.   A body rotating at 20 rad/s is acted upon by a constant torque providing it a deceleration of 2 rad/s 2 At what time will the body have kinetic energy same as the initial value if the torque continues to act ?

Answer

30   None

$\\$ Initial angular velocity = 20 $\frac{rad}{s}$ $\\$ Therefore $\alpha = 2 \frac{rad}{s^2}$ $\\$ $\Rightarrow t_1 = \frac{\omega}{\alpha_1} = \frac{20}{2} = 10 sec$ $\\$ Therefore 10 sec it will come to rest. $\\$ Since the same torque is continues to act on the body it will produce same angular acceleration and since the initial kinetic energy = the kinetic energy at a instant. $\\$ So initial angular velocity = angular velocity at that instant $\\$ Therefore time require to come to that angular velocity, $\\$ $t_2 = \frac{\omega_2}{\alpha_2} = \frac{20}{2} = 10 sec$ $\\$ therefore time required = $t_1 + t_2 = 20 sec.$

31.   A light rod of length 1 m is pivoted at its center and two masses of 5 kg and 2 kg are hung from the ends as shown in figure (10-E3). Find the initial angular acceleration of the rod assuming that it was horizontal in the beginning.

$\\$ $l_{net} = l_{net} \times \alpha$ $\\$ $\Rightarrow F_1r_1 - F_2r_2$ = $(m_1r_1^2 + m_2r_2^2) \times \alpha - 2 \times 10 \times 0.5$ $\\$ $\Rightarrow 5 \times 10 \times 0.5$ = $(5 \times (\frac{1}{2})^2 + 2 \times (\frac{1}{2})^2) \times \alpha$ $\\$ $\Rightarrow 15 = \frac{7}{4} \alpha$ $\\$ $\Rightarrow \alpha = \frac{60}{7} = 8.57 \frac{rad}{s^2}.$

32.   Suppose the rod in the previous problem has a mass of 1 kg distributed uniformly over its length. (a) Find the initial angular acceleration of the rod. (b) Find the tension in the supports to the blocks of mass 2 kg and 5 kg.

$\\$ In this problem the rod has a mass 1 kg $\\$ a) $\tau_{net} = l_{net} \alpha$ $\\$ $\Rightarrow 5 \times 10 \times 10.5 - 2 \times 10 \times 0.5$ $\\$ = $(5 \times (\frac{1}{2})^2 + 2 \times (\frac{1}{2})^2) + \frac{1}{12} \times \alpha$ $\\$ $\Rightarrow 15 = (1.75 + 0.084) \alpha$ $\\$ $\Rightarrow \alpha = \frac{1500}{(175 + 8.4)} = \frac{1500}{183.4} = 8.1 \frac{rad}{s^2} (g = 10)$ $\\$ = $8.01 \frac{rad}{s^2} (if g = 9.8)$ $\\$ b) $T_1 - m_1g = m_1a$ $\\$ $\Rightarrow T_1 = m_1a + m_1g = 2(a+g)$ $\\$ = $2(\alpha r + g) = 2(8 \times 0.5 + 9.8)$ $\\$ = 27.6 N on the first body. $\\$ In the second body $\\$ $\Rightarrow m_2g - T_2 = m_2a \Rightarrow T_2 m_2g - m_2a$ $\\$ $\Rightarrow T_2= 4(g-a) = 5(9.8 - 8 \times 0.5) = 29 N.$

33.   Figure (10-E4) shows two blocks of masses m and M connected by a string passing over a pulley. The horizontal table over which the mass m slides is smooth. The pulley has a radius r and moment of inertia I about its axis and it can freely rotate about this axis. Find the acceleration of the mass M assuming that the string does not slip on the pulley.

$\\$ According to the question $\\$ $Mg - T_1 = Ma$ ...(1) $\\$ $T_2 ma$ ....(2) $\\$ $(T_1 - T_2) = 1 \frac{a}{r^2}$ ...(3) [because a = $r\alpha$]...[T.r=l$(\frac{a}{r})$] $\\$ If we add the equation 1 and 2 we will get $\\$ Mg + $(T_2 - T_1) = Ma + ma$ ...(4) $\\$ $\Rightarrow$ Mg - $\frac{la}{r^2} = Ma + ma$ $\\$ $\Rightarrow (M + m + \frac{l}{r^2})a = Mg$ $\\$ $\Rightarrow a = \frac{Mg}{M + m + \frac{l}{r^2}}$

34.   A string is wrapped on a wheel of moment of inertia 0.20 kg-m 2 and radius 10 cm and goes through a light pulley to support a block of mass 2.0 kg as shown in figure (10-E5). Find the acceleration of the block.

$\\$ $l = 0.20 kg-m^2 (Bigger pulley)$ $\\$ r = 10 cm = 0.1 m, smaller pulley is light $\\$ mass of the block, m= 2kg $\\$ therefore mg - T = ma ...(1) $\\$ $\Rightarrow T \frac{la}{r^2}$ ...(2) $\\$ $\Rightarrow mg = (m + \frac{l}{r^2})a =>\frac{(2 \times 9.8)}{[2 + (\frac{0.2}{0.01})]} = a$ $\\$ $= \frac{19.6}{22} = 0.89 \frac{m}{s^2}$ $\\$ Therefore, acceleration of the block = 0.89 $\frac{m}{s^2}$

35.   Suppose the smaller pulley of the previous problem has its radius 5.0 cm and moment of inertia 0.10 kg-m2. Find the tension in the part of the string joining the pulleys.

$\\$ $m = 2 kg, i_1 = 0.10 kg-m^2, r_1 = 5 cm = 0.05 m$ $\\$ $i_2 = 0.20 kg-m^2, r_2 = 10 cm = 0.1 m$ $\\$ Therefore $mg - T_1 = ma$ ...(1) $\\$ $(T_1 - T_2)r_1 = l_1 \alpha$ ...(2) $\\$ $T_2r_2 = l_2 \alpha$ ...(3) $\\$ Substituting the value of $T_2$ in the equation (2), we get $\\$ $\Rightarrow (t_1 - \frac{l_2 \alpha}{r_1})r_2 = l_1 \alpha$ $\\$ $\Rightarrow (T_1 - \frac{l_2 a}{r_1^2}) = \frac{l_1a}{r_2^2}$ $\\$ $\Rightarrow T_1 = [(\frac{l_1}{r_1^2}) + \frac{l_2}{r_2^2}]a$ $\\$ Substituting the value of $T_1$ in the equation (1), we get $\\$ $\Rightarrow mg - [(\frac{l_1}{r_1^2}) + \frac{l_2}{r_2^2}]a = ma$ $\\$ $\Rightarrow \frac{mg}{[(\frac{l_1}{r_1^2}) + \frac{l_2}{r_2^2}] + m} = a$ $\\$ $\Rightarrow a = \frac{2 \times 9.8}{(\frac{0.1}{0.0025}) + (\frac{0.2}{0.01}) + 2} = 0.316 \frac{m}{s^2}$ $\\$ $T_2 = \frac{l_2a}{r_2^2} = \frac{0.20 \times 0.316}{0.01} = 6.32 N.$

36.   The pulleys in figure (10-E6) are identical, each having a radius R and moment of inertia I. Find the acceleration of the block M.

37.   The descending pulley shown in figure (10-E7) has a radius 20 cm and moment of inertia 0.20 kg-m2. The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is 1.0 kg.

$\\$ A is light pulley and B is the descending pulley having $l = 0.20 kg - m^2 and r = 0.2 m$ $\\$ Mass of the block = 1 kg $\\$ According to the equation $\\$ $T_1 = m_1a$ ...(1) $\\$ $(T_2 - T_1)r = l \alpha$ ...(2) $\\$ $m_2g - m_2\frac{a}{2} = T_1 + T_2$ ...(3) $\\$ $T_2 - T_1 = \frac{la}{2R^2} = 5\frac{a}{2} and T_1 = a(because \alpha = \frac{a}{2R})$ $\\$ $T_2 = \frac{7}{2}a$ $\\$ $\Rightarrow m_2g = m_2\frac{a}{2} + \frac{7}{2}a + a$ $\\$ $\Rightarrow \frac{2l}{r^2g} = \frac{2l}{r^2} \frac{a}{2} + \frac{9}{2} a$ $(\frac{1}{2} mr^2 =l)$ $\\$ $\Rightarrow 98 = 5a + 4.5 a$ $\\$ $\Rightarrow a = \frac{98}{9.5} = 10.3 ms^2$

38.   The pulley shown in figure (10-E8) has a radius 10 cm and moment of inertia 0.5 kg-m2 about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4.0 kg block.

$\\$ $m_1g sin \theta - T_1 = m_1a$ ...(1) $\\$ $(T_1 - T_2) = \frac{la}{r^2}$ ...(2) $\\$ $T_2 - m_2g sin \theta = m_2a$ ...(3) $\\$ Adding the equation (1) and (3) we will get $\\$ $m_1g sin \theta+ (T_2 - T_1) - m_2g sin \theta = (m_1 - m_2)a$ $\\$ $\Rightarrow (m_1 - m_2)g sin \theta = (m_1 + m_2 + \frac{1}{r^2})a$ $\\$ $\Rightarrow a= \frac{(m_1 - m_2)g sin \theta}{(m_1+m_2+ \frac{1}{r^2})} = 0.248 = 0.25ms^2$

39.   Solve the previous problem if the friction coefficient between the 2.0 kg block and the plane below it is 0.5 and the plane below the 4.0 kg block is frictionless.

$\\$ $m_1 = 4kg, m_2 = kg$ $\\$ Frictional co-efficient between 2 kg block and surface = 0.5 $\\$ $R = 10 cm = 0.1 m$ $\\$ $l = 0.5kg - m^2$ $\\$ $m_1g sin \theta - T_1 = m_1a$ ...(1) $\\$ $T_2 - (m_2g sin \theta + \mu m_2g cos \theta) = m_2a$ ...(2) $\\$ $(T_1 - T_2) = \frac{la}{r^2}$ $\\$ Adding equation (1) and (2) we will get $\\$ $m_1g sin \theta - (m_2g sin \theta + \mu m_2g cos \theta) + (T_2 - T_1) = m_1a + m_2a$ $\\$ $\Rightarrow 4 \times 9.8 \times (\frac{1}{\sqrt{2}}) - {(2 \times 9.8 \times (\frac{1}{\sqrt{2}}))} = (4 + 2+ \frac{0.5}{0.01})a$ $\\$ $\Rightarrow 27.80 - (13.90 + 6.95) = 65 a \Rightarrow a = 0.125 ms^{-2}.$

40.   A uniform metre stick of mass 200 g is suspended from the ceiling through two vertical strings of equal lengths fixed at the ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tensions in the two strings.

$\\$According to the question $\\$ $m_1 = 200g, l = 1 m, m_2 = 20 g$ $\\$ Therefore, $(T_1 \times r_1) - (T_2 \times r_2) - (m_1f - r_3g) = 0$ $\\$ $\Rightarrow T_1 \times 0.7 - T_2 \times 0.3 - 2 \times 0.2 \times g = 0$ $\\$ $\Rightarrow 7T_1 - 3T_2 = 3.92$ ...(1) $\\$ $T_1 + T_2 = 0.2 \times 9.8 + 0.02 \times 9.8 = 2.156$ ...(2) $\\$ From the equation (1) and (2) we will get $\\$ $10 T_1 = 10.3$ $\\$ $T_1 = 1.038 N = 1.04 N$ $\\$ Therefore $T_2 = 2.156 - 1.038 = 1.118 = 1.12 N.$

41.   A uniform ladder of length 10'0 m and mass 16.0 kg is resting against a vertical wall making an angle of 37° with it. The vertical wall is frictionless but the ground is rough. An electrician weighing 60.0 kg climbs up the ladder. If he stays on the ladder at a point 8.00 m from the lower end, what will be the normal force and the force of friction on the ladder by the ground ? What should be the minimum coefficient of friction for the elctrician to work safely ?

$\\$ $R_1 = \mu R_2, R_2 = 16g + 60g = 745 N$ $\\$ $R_1 \times 10 cos 37^o = 16g + 5 sin 37^o + 60g \times 8 \times sin 37^o$ $\\$ $\Rightarrow 8R_1 = 48g + 288g$ $\\$ $\Rightarrow R_1 = \frac{336g}{8} = 412 N =f$ $\\$ Therefore $\mu = \frac{R_1}{R_2} = \frac{412}{745} = 0.553$

42.   Suppose the friction coefficient between the ground and the ladder of the previous problem is 0.540. Find the maximum weight of a mechanic who could go up and do the work from the same position of the ladder.

$\\$ $\mu = 0.54, R_2 = 16g + mg ; R_1 = \mu R_2$ $\\$ $\Rightarrow R_1 \times 10 cos 37^o = 16g \times 5 sin 37^o + mg \times 8 \times sin 37^o$ $\\$ $\Rightarrow 8R_1 = 48g + \frac{24}{5} mg$ $\\$ $R_2 = \frac{48g + \frac{24}{5} mg}{5 \times 8 \times 0.54} \Rightarrow 16 + m = \frac{240 + 24m}{40 \times 0.54}$ $\\$ $m = 44 kg.$

43.   A 6.5 m long ladder rests against a vertical wall reaching a height of 6.0 m. A 60 kg man stands half way up the ladder. (a) Find the torque of the force exerted by the man on the ladder about the upper end of the ladder. (b) Assuming the weight of the ladder to be negligible as compared to the man and assuming the wall to be smooth, find the force exerted by the ground on the ladder.

$\\$ m = 60kg, ladder length = 6.5m, height of the wall = 6 m $\\$ Therefore torque due to the weight of the body $\\$a) $\tau = 600 \times \frac{6.5}{2} sin \theta = i$ $\\$ $\Rightarrow \tau = 600 \times \frac{6.5}{2} \times \sqrt{[1-(\frac{6}{6.5})^2]}$ $\\$ $\Rightarrow \tau = 735 N-m.$ $\\$ b) $R_2 = mg = 60 \times 9.8$ $\\$ $R_1 = \mu R_2 \Rightarrow 6.5 R_1 cos \theta = 60 sin \theta \times \frac{6.5}{2}$ $\\$ $\Rightarrow R_1 = 60g tan \theta = 60g \times (\frac{2.5}{12})[because tan \theta = \frac{2.5}{6}]$ $\\$ $\Rightarrow R_1 = (\frac{25}{2})g = 122.5 N.$

44.   The door of an almirah is 6 ft high, 1-5 ft wide and weighs 8 kg. The door is supported by two hinges situated at a distance of 1 ft from the ends. If the magnitudes of the forces exerted by the hinges on the door are equal, find this magnitude.

$\\$ According to the question $\\$ $8g = F_1 + F_2 ; N_1 = N_2$ $\\$ $Since, R_1 = R_2$ $\\$ Therefore $F_1 = F_2$ $\\$ $\Rightarrow 2F_1 = 8g \Rightarrow F_1 = 40$ $\\$ Let us take torque about the point B. we will get $N_1 \times 4 = 8g \times 0.75.$ $\\$ $\Rightarrow N_1 = \frac{(80 \times3)}{(4 \times 4)} = 15 N$ $\\$ $Therefore \sqrt{F_1^2 + N_1^2} = R_1 = \sqrt{40^2 + 15^2} = 42.72 = 43 N.$

45.   A uniform rod of length L rests against a smooth roller as shown in figure (10-E9). Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make with the horizontal is O.

$\\$ Rod has a length = L $\\$ It makes an angle $\theta$ with the floor $\\$ The vertical wall has a height = h $\\$ $R_2 = mg - R_1 cos \theta$ ...(1) $\\$ $R_1 sin \theta = \mu R_2$ ...(2) $\\$ $R_1 cos \theta \times (\frac{h}{tan \theta}) + R_1 sin \theta \times h = mg \times \frac{1}{2} cos \theta$ $\\$ $\Rightarrow R_1 (\frac{cos^2 \theta}{sin \theta})h + R_1 sin \theta h = mg \times \frac{1}{2} cos \theta$ $\\$ $\Rightarrow R_1 = \frac{\frac{mg \times L}{2 cos \theta}}{(\frac{cos^2 \theta}{sin \theta})h + sin \theta h}$ $\\$ $\Rightarrow R_1 cos \theta = \frac{\frac{mg L}{2 cos^2 \theta sin \theta}}{(\frac{cos^2 \theta}{sin \theta})h + sin \theta h}$ $\\$ $\frac{\mu = R_1 sin \theta}{R_2} = = \frac{\frac{mg L}{2 cos \theta sin \theta}}{(\frac{cos^2 \theta}{sin \theta})h + sin \theta h) mg - mg \frac{1}{2} cos^2 \theta}$ $\\$ $\mu = \frac{\frac{L}{2 cos \theta . sin \theta \times 2 sin \theta}}{2(cos^2 \theta h + sin^2 \theta h) - L cos^2 \theta sin \theta}$ $\\$ $\mu = \frac{L cos \theta sin^2 \theta}{2h - L cos^2 \theta sin \theta}.$

46.   A uniform rod of mass 300 g and length 50 cm rotates at a uniform angular speed of 2 rad/s about an axis perpendicular to the rod through an end. Calculate (a) the angular momentum of the rod about the axis of rotation, (b) the speed of the centre of the rod and (c) its kinetic energy.

$\\$A uniform rod of mass 300 grams and length 50 cm rotates with an uniform angular velocity = 2 $\frac{rad}{s}$ about an axis perpendicular to the rod through an end. $\\$ a) $L = l_\omega$ $\\$ l at the end = $\frac{mL^2}{3} = \frac{(0.3 \times 0.5^2)}{3} = 0.025 kg-m^2$ $\\$ $= 0.025 \times 2 = 0.05 kg-\frac{m^2}{s}$ $\\$b) Speed of the center of the rod $\\$ $V = \omega r = W \times (\frac{50}{2}) = 50 \frac{cm}{s} = 0.5 \frac{m}{s}.$ $\\$ c) Its kinetic energy = $\frac{1}{2} l \omega^2 = (\frac{1}{2}) \times 0.025 \times 2^2 = 0.05 Joule.$

47.   A uniform square plate of mass 2.0 kg and edge 10 cm rotates about one of its diagonals under the action of a constant torque of 0.10 N-m. Calculate the angular momentum and the kinetic energy of the plate at the end of the fifth second after the start.

$\\$ $l= 0.10 N-m ; a = 10cm = 0.1 m ; m = 2kg$ $\\$ Therefore $(\frac{ma^2}{12}) \times \alpha = 0.10 N-m $ $\\$ $\Rightarrow \omega = 60 \times 5 = 300 \frac{rad}{s}$ $\\$ Therefore $\omega = \omega_0 + \alpha t$ $\\$ $\Rightarrow \omega = 60 \times 5 = 300 \frac{rad}{s}$ $\\$ Therefore angular momentum = $l_\omega = (\frac{0.10}{60}) \times 300 = 0.50 \frac{kg-m^2}{s}$ $\\$ And 0 kinetic energy = $\frac{1}{2} l_{\omega^2} = \frac{1}{2} \times (\frac{0.10}{60}) \times 300^2 = 75 Joules.$

48.   Calculate the ratio of the angular momentum of the earth about its axis due to its spinning motion to that about the sun due to its orbital motion. Radius of the earth = 6400 km and radius of the orbit of the earth about the sun = 1.5 x 108 km.

$\\$ Angular momentum of the earth about its axis is $\\$ $= \frac{2}{5} mr^2 \times (\frac{2 \pi}{85400})(because, l = \frac{2}{5} mr^2)$ $\\$ Angular momentum of the earth about sun's axis $\\$ = $mR^2 \times (\frac{2 \pi}{86400} \times 365) (because, l = mR^2)$ $\\$ Therefore, ratio of the angular momentum = $\frac{\frac{2}{5mr^2} \times (\frac{2 \pi}{86400})}{\frac{mR^2 \times 2 \pi}{86400 \times 365}}$ $\\$ $\Rightarrow \frac{(2r^2 \times 365)}{5R^2}$ $\\$ $\Rightarrow \frac{(2.990 \times10^{10})}{(1.125 \times 10^{17})} = 2.65 \times 10^{-7}.$

49.   Two particles of masses $m_1$, and $m_2$ and are joined by a light rigid rod of length r. The system rotates at an angular speed co about an axis through the centre of mass of the system and perpendicular to the rod. Show that the angular momentum of the system is $L = \mu r^2 \omega$. where 1.1. is the reduced mass of the system defined as $\mu = \frac{m_1m_2}{m_1+m_2}.$

$\\$ Angular momentum due to the mass $m_1$ at the center of the system is = $m_1 r^{12}.$ $\\$ $m1(\frac{m_2}{m_1m_2})^2 \omega = \frac{m_1m_2^2 r^2}{(m_1+m_2)^2} \omega$ ...(1) $\\$ Similarly the angular momentum due to the mass $m_2$ at the center of the system is $m_2 r^{112}_\omega$ $\\$ $= m_2 (\frac{m_1r}{m_1m_2})^2 \omega = \frac{m_2m_1^2}{(m_1+m_2)^2 \omega}$ ...(2) $\\$ Therefore net angular momentum = $\frac{m_1m_2^2 r^2 \omega}{(m_1+m_2)^2} + \frac{m_2m_1^2 r^2 \omega}{(m_1+m_2)^2}$ $\\$ $\Rightarrow \frac{m_1m_2(m_1+m_2) r^2 \omega}{(m_1+m_2)^2} = \frac{m_1m_2}{(m_1+m_2)} r^2 \omega = \mu r^2 \omega (proved)$

50.   A dumb-bell consists of two identical small balls of mass 1/2 kg each connected to the two ends of a 50 cm long light rod. The dumb-bell is rotating about a fixed axis through the centre of the rod and perpendicular to it at an angular speed of 10 rad/s. An impulsive force of average magnitude 5.0 N acts on one of the masses in the direction of its velocity for 0.10 s. Find the new angular velocity of the system.

$\\$ $\tau = l \alpha$ $\\$ $\Rightarrow F \times r = (mr^2+mr^2) \alpha \Rightarrow 5 \times 0.25 = 2mr^2 \times \alpha$ $\\$ $\Rightarrow \alpha = \frac{1.25}{2 \times 0.5 \times 0.025 \times 0.25} = 20$ $\\$ $\omega_0 = 10 \frac{rad}{s}, t = 0.10 sec, \omega = \omega_0+\alpha t$ $\\$ $\Rightarrow \omega = 10 + 010 \times 230 = 10 + 2 = 12 \frac{rad}{s}.$

51.   A wheel of moment of inertia 0.500 kg-$m^2$ and radius 20.0 cm is rotating about its axis at an angular speed of 20.0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel.

$\\$ A wheel has $\\$ $l = 0.500 kg-m^2, r=0.2 m, \omega = 20 \frac{rad}{s}$ $\\$ Stationary particle = 0.2 kg $\\$ Therefore $l_1 \omega_1 = l_2 \omega_2$ (since external torque = 0) $\\$ $\Rightarrow 0.5 \times 10 = (0.5 + 0.2 \times 0.2^2) \omega_2$ $\\$ $\Rightarrow \frac{10}{0.508} = \omega_2 = 19.69 = 19.7 \frac{rad}{s}$

52.   A diver having a moment of inertia of 6.0 $kg-m^2$ about an axis through its centre of mass rotates at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to decrease the moment of inertia to 5.0 $kg-m^2$, what will be the new angular speed ?

$\\$ $l_1 = 6 kg - m^2, \omega_1 = 2 \frac{rad}{s}, l_2 = 5 kg-m^2$ $\\$ Since external torque = 0 $\\$ Therefore $l_1 \omega_1 = l_2 \omega_2$ $\\$ $\Rightarrow \omega_2 = \frac{(6 \times 2)}{5} = 2.4 \frac{rad}{s}.$

53.   A boy is seated in a revolving chair revolving at an angular speed of 120 revolutions per minute. Two heavy balls form part of the revolving system and the boy can pull the balls closer to himself or may push them apart. If by pulling the balls closer, the boy decreases the moment of inertia of the system from 6 $kg-m^2$ to 2 $kg-m^2$, what will be the new angular speed ?

$\\$ $\omega_1 = 120 rpm = 120 \times (\frac{2 \pi}{60}) = 4 \pi \frac{rad}{s}$ $\\$ $l_1 = 6 kg-m^2, l_2 = 2 kgm^2$ $\\$ Since two balls are inside the system $\\$ Therefore, total external torque = 0 $\\$ Therefore, $l_1 \omega_1 = l_2 \omega_2$ $\\$ $\Rightarrow + \times 4 \pi = 2 \omega_2$ $\\$ $\Rightarrow \omega_2 = 12 \pi \frac{rad}{s} = 6 \frac{rev}{s} = 360 \frac{rev}{minute}.$

54.   A boy is standing on a platform which is free to rotate about its axis. The boy holds an open umbrella in his hand. The axis of the umbrella coincides with that of the platform. The moment of inertia of "the platform plus the boy system" is $3.0 \times 10^{ -3} kg-m^2$ and that of the umbrella is $2.0 \times 10^{-3} kg-m^2$ The boy starts spinning the umbrella about the axis at an angular speed of 2.0 rev/s with respect to himself. Find the angular velocity imparted to the platform.

$\\$ $l_1 = 2 \times 10^{-3} kg-m^2 ; l_2 = 3 \times 10^{-3} kg-m^2 ; \omega_1 = 2 \frac{rad}{s}$ $\\$ From the earth reference the umbrella has a angular velocity $(\omega_1 - \omega_2)$ $\\$ And the angular velocity of the man will be $\omega_2$ $\\$ Therefore $l_1(\omega_1 - \omega_2) = l_2 \omega_2$ $\\$ $\Rightarrow 2 \times 10^{-3}(2- \omega_2) = 3 \times 10^{-3} \times \omega_2$ $\\$ $\Rightarrow 5 \omega_2 = 4 \Rightarrow \omega_2 = 0.8 \frac{rad}{s}.$

55.   A wheel of moment of inertia 0.10 $kg-m^2$ is rotating about a shaft at an angular speed of 160 rev/minute. A second wheel is set into rotation at 300 rev/minute and is coupled to the same shaft so that both the wheels finally rotate with a common angular speed of 200 rev/minute. Find the moment of inertia of the second wheel.

$\\$ Wheel (1) has $\\$ $l_1 = 0.10 kg-m^2, \omega_1 = 160 \frac{rev}{min}$ $\\$ Wheel (2) has $\\$ $l_2 = ? ; \omega_2 = 200 \frac{rev}{min}$ $\\$ Given that after they are coupled, $\omega = 200 \frac{rev}{min}$ $\\$ Therefore if we take the two wheels to bean isolated system $\\$ Total external torque = 0 $\\$ therefore, $l_1 \omega_1 + l_2 \omega_2 = (l_1+l_1) \omega$ $\\$ $\Rightarrow 0.10 \times 160 + l_2 \times 300 = (0.10 + l_2) \times 200$ $\\$ $\Rightarrow 5l_2 = 1 - 0.8 \Rightarrow l_2 = 0.04 kg-m^2.$

56.   A kid of mass M stands at the edge of a platform of radius R which can be freely rotated about its axis. The moment of inertia of the platform is I. The system is at rest when a friend throws a ball of mass m and the kid catches it. If the velocity of the ball is v horizontally along the tangent to the edge of the platform when it was caught by the kid, find the angular speed of the platform after the event.

$\\$ A kid of mass M stands at the edge of a platform of radius R which has a moment of inertia I. $\\$ A ball of m thrown to him and horizontal velocity of the ball v when he catches it. $\\$ Therefore if we take the total bodies as a system $\\$ Therefore mvR = ${l + (M+m)R^2} \omega$ $\\$ (The moment of inertia of the kid and about the axis = $(M+m)R^2$) $\\$ $\Rightarrow \omega = \frac{mvR}{1+(M+m)R^2}.$

57.   Suppose the platform of the previous problem is brought to rest with the ball in the hand of the kid standing on the rim. The kid throws the ball horizontally to his friend in a direction tangential to the rim with a speed v as seen by his friend. Find the angular velocity with which the platform will start rotating.

$\\$ Initial angular momentum = Final angular mumentum $\\$ (the total external torque = 0) $\\$ Initial angular momentum = mvR (m= mass of the ball, v= velocity of the ball, R= radius of the platform) $\\$ Therefore angular momentum = $l_ \omega + MR^2 \omega$ $\\$ Therefore mVR = $l_ \omega + MR^2 \omega$ $\\$ $\Rightarrow \omega = \frac{mVR}{(1+MR^2)}.$

58.   Suppose the platform with the kid in the previous problem is rotating in anticlockwise direction at an angular speed co. The kid starts walking along the rim with a speed v relative to the platform also in the anticlockwise direction. Find the new angular speed of the platform.

$\\$ From a inertial frame of reference when we see the (man wheel) system, we can find that the wheel moving at a speed of $\omega$ and the man with $(\omega + \frac{V}{R})$ after the man has started walking. $\\$ ($\omega$ = angular velocity after walking, $\omega$ = angular velocity of the wheel before walking.) $\\$ Since, $\Sigma l$ = 0 $\\$ Extended torque = 0 $\\$ Therefore $(1+MR^2) \omega = l_\omega + mR^2 (\omega + \frac{V}{R})$ $\\$ $\Rightarrow (l + mR^2) \omega + l_\omega + mR^2 \omega + mVR$ $\\$ $\omega = \omega- \frac{mVR}{(1+mR^2)}.$

59.   A uniform rod of mass m and length 1 is struck at an end by a force F perpendicular to the rod for a short time interval t. Calculate (a) the speed of the centre of mass, (b) the angular speed of the rod about the centre of mass, (c) the kinetic energy of the rod and (d) the angular momentum of the rod about the centre of mass after the force has stopped to act. Assume that t is so small that the rod does not appreciably change its direction while the force acts.

$\\$ A uniform rod of mass m length l is struck at an end by a force F. $\perp$ to the rod for a short time t $\\$ a)Speed of the center of mass $\\$ mv = Ft $\Rightarrow v = \frac{Ft}{m}$ $\\$ b) The angular speed of the rod about the center of mass $l_\omega - r \times p$ $\\$ $\Rightarrow (\frac{ml^2}{12}) \times \omega = (\frac{1}{2}) \times mv$ $\\$ $\Rightarrow \frac{ml^2}{12} \times \omega = ()\frac{1}{2} l \omega^2$ $\\$ $\Rightarrow \omega = \frac{6Ft}{ml}$ $\\$c) K.E = $(\frac{1}{2}) mv^2 + (\frac{1}{2}) l \omega^2$ $\\$ = $(\frac{1}{2}) \times m(\frac{Ft}{m^2}) (\frac{1}{2}) l \omega^2$ $\\$ = $(\frac{1}{2}\times m \times (\frac{F^2t^2}{m^2})) + (\frac{1}{2}) \times (\frac{ml^2}{12}) (36 \times (\frac{F^2t^2}{m^2l^2}))$ $\\$ =$\frac{F^2t^2}{2m} + \frac{3}{2}(\frac{F^2t^2}{m}) = 2 \frac{F^2t^2}{m}$ $\\$ d) Angular momentum about the center of mass: $\\$ L= mvr = $m \times \frac{Ft}{m} \times (\frac{1}{2}) = \frac{Flt}{2}$

60.   A uniform rod of length L lies on a smooth horizontal table. A particle moving on the table strikes the rod perpendicularly at an end and stops. Find the distance travelled by the centre of the rod by the time it turns through a right angle. Show that if the mass of the rod is four times that of the particle, the collision is elastic.

$\\$ Let the mass of the particle = m & the mass of the rod = M $\\$ Let the particle strikes the rod with a velocity V. $\\$ If we take the two body to be a system, $\\$ Therefore the net external torque & net external force = 0 Therefore Applying laws of conservation of linear momentum $\\$ MV' = $mV(V' = velocity of the rod after striking)$ $\\$ $\Rightarrow \frac{V'}{V} = \frac{m}{M}$ $\\$ Again applying laws of conservation of angular momentum $\\$ $\Rightarrow \frac{mVR}{2} = l \omega$ $\\$ $\Rightarrow \frac{mVR}{2} = \frac{MR^2}{12} \times \frac{\pi}{2t} \Rightarrow t = \frac{MR\pi}{m12 \times V}$ $\\$ Therefore distance travelled : $\\$ V' t= $V' \frac{MR\pi}{m12\pi} = \frac{m}{M} \times \frac{M}{m} \times \frac{R\pi}{12} = \frac{R\pi}{12}$

61.   Suppose the particle of the previous problem has a mass m and a speed v before the collision and it sticks to the rod after the collision. The rod has a mass M. (a) Find the velocity of the centre of mass C of the system constituting "the rod plus the particle". (b) Find the velocity of the particle with respect to C before the collision. (c) Find the velocity of the rod with respect to C before the collision. (d) Find the angular momentum of the particle and of the rod about the centre of mass C before the collision. (e) Find the moment of inertia of the system about the vertical axis through the centre of mass C after the collision. (f) Find the velocity of the centre of mass C and the angular velocity of the system about the centre of mass after the collision.

$\\$a) If we take the two bodies as a system therfore total external force = 0 $\\$ Applying L.C.L.M :- $\\$ mV = (M+m)v' $\\$ $\Rightarrow v' = \frac{mv}{M+m}$ $\\$b) Let the velocity of the particle w.r.t the centre of mass = V' $\\$ $\Rightarrow v' = \frac{m \times 0 + Mv}{M+m} \Rightarrow v' = \frac{Mv}{M+m}$ $\\$ c) If the body moves towards the rod with a velocity of v, i.e, the rod is moving with a velocity - v towards the particle. $\\$ Therefore the velocity of the rod w.r.t the center of mass = $V^-$ $\\$ $\Rightarrow V^- = \frac{M\times O = m\times v}{M+m} = \frac{-mv}{M+m}$ $\\$ d) The distance of the centre of mass from the particle $\\$ = $\frac{M\times\frac{l}{2}+m\times O}{M+m} = \frac{M\times \frac{l}{2}}{M+m}$ $\\$ Therefore angular momentum of the particle before the collision $\\$ $= l \omega = Mr^2 cm \omega$ $\\$ $m(\frac{m \frac{l}{2}}{M+m})^2 \times \frac{V}{\frac{l}{2}}$ $\\$ $\frac{(mM^2vl)}{2(M+m)}$ $\\$ Distance angular momentum of the about the centre of the mass $\\$ $R_{cm}^1 = \frac{M \times 0 + m \times (\frac{1}{2})}{M+m} = \frac{(\frac{ml}{2})}{M+m}$ $\\$ Therefore angular momentum of the rod about the centre of mass $\\$ $MV_{cm} R_{cm}^1$ $\\$ $M \times {\frac{(-mv)}{M+m}}{\frac{(\frac{ml}{2})}{(M+m)}}$ $\\$ $\mid\frac{-Mm^2lv}{2(M+m)^2}\mid = \frac{Mm^2lv}{2(M+m)^2}$ (If we consider the magnitude only)

$\\$ e) Momentum of inertia of the system = M.I due to rod + M.I due to particle $\\$ $\frac{Ml^2}{12} + \frac{M(\frac{ml}{2})^2}{(M+m)^2} + \frac{m(\frac{MI}{s})^2}{(M+m)^2}$ $\\$ $\frac{Ml^2(M+4m)}{12(M+m)}.$ $\\$ f) Velocity of the centre of mass $V_m = \frac{M\times 0 + mV}{M+m} = \frac{mV}{(M+m)}$ $\\$ (Velocity of center of mass of the system before the collision = Velocity of center of mass of the system after the collision) $\\$(Because External force = 0) $\\$Angular velocity of the system about the center of mass, $\\$ $P_{cm} = l_{cm} \omega$ $\\$ $\Rightarrow MV_M \times r_m + mv_m \times r_m = l_{cm} \omega$ $\\$ $\Rightarrow M \times \frac{mv}{(M+m)} \times \frac{ml}{2(M+m)} + m \times \frac{Mv}{(M+m)} \times \frac{MI}{2(M+m)} = \frac{MI^2(M+4m)}{12(M+m)} \times\omega$ $\\$ $\Rightarrow \frac{Mm^2vl+mM^2vl}{2(M+m)^2} = \frac{Ml^2(M+4m)}{12(M+m)} \times\omega$ $\\$ $\frac{\frac{Mm}{M+m}}{2(M+m)^2} = \frac{Mi^2(M+m)}{12(M+m)} \times \omega$ $\\$ $\Rightarrow \frac{6mv}{(M+4m)l} = \omega$ $\\$

62.   Two small balls A and B, each of mass m, are joined rigidly by a light horizontal rod of length L. The rod is clamped at the centre in such a way that it can rotate freely about a vertical axis through its centre. The system is rotated with an angular speed w about the axis. A particle P of mass m kept at rest sticks to the ball A as the ball collides with it. Find the new angular speed of the rod.

$\\$ Since external torque = 0 Therefore $l_1 \omega_1 = l_2 \omega_2$ $\\$ $I_1 = \frac{ml^2}{4} + \frac{ml^2}{4} = \frac{ml^2}{2}$ $\\$ $\omega_1 = \omega$ $\\$ $I_2 = \frac{2ml^2}{4 + \frac{ml^2}{4}} = \frac{3ml^2}{4}$ $\\$ Therefore $\omega_2 = \frac{l_1\omega_1}{l_2} = \frac{(\frac{ml^2}{2}) \times \omega}{\frac{3ml^2}{4}} = \frac{2\omega}{3}$

63.   Two small balls A and B, each of mass m, are joined rigidly to the ends of a light rod of lengh L (figure 10-E10). The system translates on a frictionless horizontal surface with a velocity vo in a direction perpendicular to the rod. A particle P of mass m kept at rest on the surface sticks to the ball A as the ball collides with it. Find (a) the linear speeds of the balls A and B after the collision, (b) the velocity of the centre of mass C of the system A + B + P and (c) the angular speed of the system about C after the collision.

$\\$ Two balls A & B, each of mass m are joined rigidly to the ends of a light of rod of length L. The system moves in a velocity $v_0$ in a direction $\perp$ to the rod. $\\$ A particle P of mass m kept at rest on the surface sticks to the ball A as the ball collides with it. $\\$ a) the light rod will exert a force on the ball B only along its length. So collision will not affect its velocity. $\\$ B has a velocity = $v_0$ $\\$ If we consider the three bodies to be a system $\\$ Applying L.C.L.M. $\\$ Therefore $mv_0 = 2mv' \Rightarrow v' = \frac{v_0}{2}$ $\\$ Therefore A has velocity = $\\frac{v_0}{2}$ $\\$b) if we consider the three bodies to be a system $\\$ Therefore, net external force =0 $\\$ Therefore $V_cm = \frac{m\times v_0 + 2m(\frac{v_0}{2})}{m+2m} = \frac{mv_0+mv_0}{3m} = \frac{2v_0}{3}$ (along the initial velocity as before collision) $\\$ c) The velocity of (A+P) w.r.t the center of mass = $\frac{2v_0}{3} - \frac{v_0}{2} = \frac{v_0}{6} $ & $\\$ The velocity of B w.r.t the center of mass $v_0 - \frac{2v_0}{3} = \frac{v_0}{3}$ $\\$ [Only magnitude has been taken] $\\$ Distance of the (A+P) from center of mass = $\frac{l}{3}$ & for B it is $2 \frac{l}{3}.$ $\\$ Therefore $P_{cm} = I_{cm} \times \omega$ $\\$ $\Rightarrow 2m \times \frac{v_0}{6} \times \frac{1}{3} + m \times \frac{v_0}{3} \times \frac{2l}{3} = 2m(\frac{1}{3})^2 \times \omega$ $\\$ $\Rightarrow \frac{6mv_0l}{18} = \frac{6ml}{9} \times \omega \Rightarrow \omega = \frac{v_0}{2l}$

64.   Suppose the rod with the balls A and B of the previous problem is clamped at the centre in such a way that it can rotate freely about a horizontal axis through the clamp. The system is kept at rest in the horizontal position. A particle P of the same mass m is dropped from a height h on the ball B. The particle collides with B and sticks to it. (a) Find the angular momentum and the angular speed of the system just after the collision. (b)*What should be the minimum value of h so that the system makes a full rotation after the collision.

$\\$The system is kept rest in the horizontal position and particle P falls a height h and collides with B and sticks to it. $\\$ Therefore, the velocity of the particle $\rho$ before collision = $\sqrt{2gh}$ $\\$ If we consider the two bodies P and B to be a system. Net external torque and force = 0 Therefore, $m\sqrt{2gh} = 2m \times v$ $\\$ $\Rightarrow v' = \sqrt{\frac{(2gh)}{2}}$ $\\$ Therefore angular momentum of the rod just after the collision $\\$ $\Rightarrow 2m (v' \times r) = 2m \times \sqrt{\frac{(2gh)}{2}} \times \frac{l}{2} \Rightarrow ml \sqrt{\frac{(2gh)}{2}}$ $\\$ $\omega = \frac{L}{l} = \frac{ml \sqrt{2gh}}{2(\frac{ml^2}{4} + \frac{2ml^2}{4})} = \frac{2 \sqrt{gh}}{3l} = \frac{\sqrt{8gh}}{3l}$ $\\$ b) When the mass 2m will at the top most position and the mass m at at the lowest point, they will automatically rotate. $\\$ In this position the total gain potential energy = 2 mg $\times (\frac{l}{2}) - mg ()\frac{l}{2} = mg (\frac{l}{2})$ $\\$ Therefore $\Rightarrow mg \frac{l}{2} = \frac{l}{2} l \omega^2$ $\\$ $\Rightarrow mg \frac{l}{2} = \frac{(\frac{1}{2} \times 3ml^2)}{4} \times (\frac{8gh}{9gl^2})$ $\\$ $\Rightarrow h = \frac{3l}{2}.$

65.   Two blocks of masses 400 g and 200 g are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has a moment of inertia 1.6 x 10 -4 kg-m 2 and a radius 2.0 cm. Find (a) the kinetic energy of the system as the 400 g block falls through 50 cm, (b) the speed of the blocks at this instant.

$\\$According to the question $\\$ $0.4g - T_1 = 0.4 a$ ...(1) $\\$ $T_2 - 0.2g = 0.2 a$ ...(2) $\\$ $(T_1 - T_2)r = \frac{la}{r}$ ...(3) $\\$ From equation 1,2 and 3 $\\$ $\Rightarrow a = \frac{(0.4-0.2)g}{(0.4+0.2+\frac{1.6}{0.4})} = \frac{g}{5}$ $\\$ Therefore (b) V = $\sqrt{2ah} = \sqrt{(2 \times gl^5 \times 0.5)}$ $\\$ $\Rightarrow \sqrt{(\frac{g}{5})} = \sqrt{\frac{9.8}{5}} = 1.4 \frac{m}{s}$ $\\$ a) Total kinetic energy of the system $\\$ = $\frac{1}{2} m_1V^2 + \frac{1}{2} m_2V^2 + \frac{1}{2} 18^2$ $\\$ = $(\frac{1}{2} \times 0.4 \times 1.4^2) + (\frac{1}{2} \times 0.2 \times 1.4^2) + (\frac{1}{2} \times (\frac{1.6}{4}) \times 1.4^2) = 0.98 Joule.$

66.   The pulley shown in figure (10-E11) has a radius of 20 cm and moment of inertia 0.2 $kg-m^2$. The string going over it is attached at one end to a vertical spring of spring constant 50 N/m fixed from below, and supports a 1 kg mass at the other end. The system is released from rest with the spring at its natural length. Find the speed of the block when it has descended through 10 cm. Take g = 10 $m/s^2$

$\\$ $l = 0.2 kg-m^2, r= 0.2 m, K=50 \frac{N}{m},$ $\\$ $m= 1kg, g= 10 m^2, h=0.1m$ $\\$ Therefore applying laws of conservation of energy mgh = $\frac{1}{2}mv^2 + \frac{1}{2} kx^2$ $\\$ $\Rightarrow 1 = \frac{1}{2} \times 1 \times V^2 + \frac{1}{2} \times 0.2 \times \times \frac{V^2}{0.04} + (\frac{1}{2}) \times 50 \times 0.01$ (x=h) $\\$ $\Rightarrow 1 = 0.5 v^2 + 2.5 v^2 + \frac{1}{4}$ $\\$ $\Rightarrow 3v^2 = \frac{3}{4}$ $\\$ $\Rightarrow v = \frac{1}{2} = 0.5 \frac{m}{s}$

67.   A metre stick is held vertically with one end on a rough horizontal floor. It is gently allowed to fall on the floor. Assuming that the end at the floor does not slip, find the angular speed of the rod when it hits the floor.

$\\$ Let the mass of the rod = m $\\$ Therefore applying laws of conservation of energy $\frac{1}{2} l\omega^2 = \frac{mg l}{2}$ $\\$ $\Rightarrow \frac{1}{2} \times M \frac{I^2}{3} \times \omega^2 = mg \frac{1}{2}$ $\\$ $\omega^2 = \frac{3g}{l}$ $\\$ $\Rightarrow \omega = \sqrt{\frac{3g}{l}} = 5.42 \frac{rad}{s}.$

68.   A metre stick weighing 240 g is pivoted at its upper end in such a way that it can freely rotate in a vertical plane through this end (figure 10-E12). A particle of mass 100 g is attached to the upper end of the stick through a light string of length 1 m. Initially, the rod is kept vertical and the string horizontal when the system is released from rest. The particle collides with the lower end of the stick and sticks there. Find the maximum angle through which the stick will rise.

$\\$ $\frac{1}{2} l\omega^2 - 0 = 0.1 \times 10 \times 1$ $\\$ $\Rightarrow \omega = \sqrt{20}$ $\\$ For collision $\\$ $0.1 \times 1^2 \times \sqrt{20} + 0 = [(\frac{0.24}{3}) \times 1^2 + (0.1)^2 1^2]\omega'$ $\\$ $\Rightarrow \omega' = \frac{\sqrt{20}}{[10.(0.18)]}$ $\\$ $\Rightarrow 0 - \frac{1}{2} \omega^2 = -m_1g l(1-cos \theta) - m_2g \frac{l}{2}(1-cos \theta)$ $\\$ = $0.1 \times 10(1-cos \theta) = 0.24 \times 10 \times 0.5(1-cos \theta)$ $\\$ $\Rightarrow \frac{1}{2} \times 0.18 \times (\frac{20}{3.24}) = 2.2(1-cos \theta)$ $\\$ $\Rightarrow (1-cos \theta) = \frac{1}{(2.2 \times 1.8)}$ $\\$ $1-cos \theta = 0.252$ $\\$ $cos \theta = 1-0.252 = 0.748$ $\\$ $\omega = cos^{-1} (0.748) =41^o.$

69.   A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of 60° and then released. Find the magnitude of the force acting on a particle of mass dm at the tip of the rod when the rod makes an angle of 37° with the vertical.

$\\$Let l = length of the rod, and m = mass of the rod. $\\$ Applying energy principle $(\frac{1}{2}) l\omega^2 - O = mg(\frac{1}{2}) (cos 37^o - cos60^o)$ $\\$ $\Rightarrow \frac{1}{2} \times \frac{ml^2}{3} \omega^2 = mg \times \frac{1}{2} (\frac{4}{5}- \frac{1}{2})t$ $\\$ $\Rightarrow \omega^2 = \frac{9g}{10 l} = 0.9(\frac{g}{l})$ $\\$ Again $\frac{ml2}{3} \alpha = mg(\frac{1}{2}) sin 37^o = mgl \times \frac{3}{5}$ $\\$ $\therefore \alpha = 0.9 (\frac{g}{l})$ = angular acceleration. $\\$ So, to find out the force on the particle at the tip of the rod $\\$ $F_i$ = centrifugal force = $(dm) \omega^2 l = 0.9 (dm)g$ $\\$ $F_i$ = tangential force = $(dm) \alpha l = 0.9 (dm)g$ $\\$ So, total force F= $\sqrt{(F_1^2 + F_1^2)} = 0.9 \sqrt{2} (dm) g$

70.   A cylinder rolls on a horizontal plane surface. If the speed of the centre is 25 m/s, what is the speed of the highest point ?

$\\$ A cylinder rolls in a horizontal plane having center velocity 25 m/s. $\\$ At its age the velocity is due to its rotation as well as due to its linear motion and this two velocities are same and acts in the same direction $(v=r \omega)$ $\\$ Therefore Net velocity at A =25 $\frac{m}{s} + 25 \frac{m}{s} = 50 \frac{m}{s}$

71.   A sphere of mass m rolls on a plane surface. Find its kinetic energy at an instant when its centre moves with speed v.

$\\$A sphere having mass m rolls on a plane surface. Let its radius R. Its centre moves with a velocity v $\\$ Therefore Kinetic energy = $(\frac{1}{2}) l \omega^2 + (\frac{1}{2}) mv^2$ $\\$ $\frac{1}{2} \times \frac{2}{5}mR^2 \times \frac{v^2}{R^2} + \frac{1}{2}mv^2 = \frac{2}{10}mv^2 = \frac{2+5}{10}mv^2 = \frac{7}{10}mv^2$

72.   A string is wrapped over the edge of a uniform disc and the free end is fixed with the ceiling. The disc moves down, unwinding the string. Find the downward acceleration of the disc.

$\\$Let the radius of the disc = R $\\$ Therefore according to the question and figure $\\$ $Mg-T = ma$ $\\$And the torque about the centre $\\$ $= T \times R = l \times \alpha$ $\\$ $\Rightarrow TR = (\frac{1}{2})mR^2 \times \frac{a}{R}$ $\\$ $T= \frac{1}{2}ma$ $\\$ Putting this value in the equation (1) we get $\\$ $\Rightarrow mg-(\frac{1}{2})ma = ma$ $\\$ $mg = \frac{3}{2}ma \Rightarrow a= \frac{2g}{3}$

73.   A small spherical ball is released from a point at a height h on a rough track shown in figure (10-E13). Assuming that it does not slip anywhere, find its linear speed when it rolls on the horizontal part of the track.

$\\$A small spherical ball is released from a point at a height on a rough track & the sphere does not slip. $\\$ Therefore potential energy it has gained w.r.t the surface will be converted to angular kinetic energy about the centre & linear kinetic energy. $\\$ Therefore $mgh = (\frac{1}{2})l \omega^2 + (\frac{1}{2})mv^2$ $\\$ $\Rightarrow mgh = \frac{1}{2} \times \frac{2}{5} mR^2 \omega^2 + \frac{1}{2}mv^2$ $\\$ $\Rightarrow gh = \frac{1}{5}v^2 + \frac{1}{2}v^2$ $\\$ $\Rightarrow v^2 = \frac{10}{7}gh \Rightarrow v= \sqrt{\frac{10}{7}gh}$

74.   A small disc is set rolling with a speed v on the horizontal part of the track of the previous problem from right to left. To what height will it climb up the curved part ?

$\\$A disc is set rolling with a velocity V from right to left. Let it has attained a height h. $\\$ $Therefore \frac{1}{2}mV^2 + \frac{1}{2}l \omega^2 = mgh$ $\\$ $\Rightarrow \frac{1}{2}mV^2 + (\frac{1}{2} \times (\frac{1}{2})mR^2 \omega^2 = mgh$ $\\$ $\Rightarrow (\frac{1}{2})V^2 + \frac{1}{4}V^2 = gh \Rightarrow (\frac{3}{4})V^2 = gh$ $\\$ $\Rightarrow h= \frac{3}{4} \times \frac{V^2}{g}$

75.   A sphere starts rolling down an incline of inclination $\theta$. Find the speed of its centre when it has covered a distance l.

$\\$ A sphere is rolling in inclined plane with inclination $\theta$ $\\$ Therefore according to the principle $\\$ $Mgl sin \theta = (\frac{1}{2}) l \omega^2 + (\frac{1}{2})mv^2$ $\\$ $\Rightarrow mgl sin \theta = \frac{1}{5}mv^2 + (\frac{1}{2})mv^2$ $\\$ $Gl sin \theta = \frac{7}{10} \omega^2$ $\\$ $\Rightarrow v = \sqrt{\frac{10}{7}gl sin \theta}$

76.   A hollow sphere is released from the top of an inclined plane of inclination $\theta$. (a) What should be the minimum coefficient of friction between the sphere and the plane to prevent sliding ? (b) Find the kinetic energy of the ball as it moves down a length l on the incline if the friction coefficient is half the value calculated in part (a).

$\\$A hollow sphere is released from a top of an inclined plane of inclination $\theta$ $\\$ To prevent sliding, the body will make only perfect rolling. In this condition, $\\$ $mg sin \theta - f = ma$ ...(1) $\\$ And torque about the centre $\\$ $f \times R = \frac{2}{3}mR^2 \times \frac{a}{R}$ $\\$ $\Rightarrow f = \frac{2}{3}ma$ ...(2) $\\$ Putting this value in equation (1) we get $\\$ $\Rightarrow mg sin \theta - \frac{2}{3}ma \Rightarrow a= \frac{3}{5}g sin \theta$ $\\$ $\Rightarrow mg sin \theta - f = \frac{3}{5}mg sin \theta \Rightarrow f = \frac{2}{5}mg sin \theta$ $\\$ $\Rightarrow \mu mg cos \theta = \frac{2}{5} mg sin \theta \Rightarrow \mu = \frac{2}{5} tan \theta$ $\\$ b) $\frac{1}{5} tan \theta (mg cos \theta) R = \frac{2}{3} mR^2 \alpha$ $\\$ $\Rightarrow \alpha = \frac{3}{10} \times \frac{gsin \theta}{R}$ $\\$ $a_c = g sin \theta - \frac{g}{5} sin \theta = \frac{4}{5} sin \theta$ $\\$ $\Rightarrow t^2 = \frac{2s}{a_c} = \frac{2l}{(\frac{4g sin \theta}{5})} = \frac{5l}{2 g sin \theta}$ $\\$ Again, $\omega = \alpha t$ $\\$ K.E $= \frac{1}{2}mv^2 + (\frac{1}{2}) l \omega^2 = (\frac{1}{2})m(2as) + (\frac{1}{2}) l (\alpha^2 t^2)$ $\\$ $= \frac{1}{2}m \times \frac{4g sin \theta}{5} \times 2 \times l + \frac{1}{2} \times \frac{2}{3}mR^2 \times \frac{9}{100} \frac{g^2 sin^2 \theta}{R} \times\frac{5l}{2g sin \theta}$ $\\$ $\frac{4mgl sin \theta}{5} + \frac{3mgl sin \theta}{40} = \frac{7}{8}mgl sin \theta$

77.   A solid sphere of mass m is released from rest from the rim of a hemispherical cup so that it rolls along the surface. If the rim of the hemisphere is kept horizontal, find the normal force exerted by the cup on the ball when the ball reaches the bottom of the cup.

$\\$ Total normal force = $mg + \frac{mv^2}{R-r}$ $\\$ $\Rightarrow mg (R-r) = (\frac{1}{2}) l \omega^2 + (\frac{1}{2})mv^2$ $\\$ $\Rightarrow mg (R-r) = \frac{1}{2} \times \frac{2}{5}mv^2 + \frac{1}{2}mv^2$ $\\$ $\Rightarrow \frac{7}{10}mv^2 = mg (R-r) \Rightarrow v^2 = \frac{10}{7}g(R-r)$ $\\$ Therefore total normal force = $mg + \frac{mg + m(\frac{10}{7})g(R-r)}{R-r} = mg + mg(\frac{10}{7}) = \frac{17}{7}mg$

78.   Figure (10-E14) shows a rough track, a portion of which is in the form of a cylinder of radius R. With what minimum linear speed should a sphere of radius r be set rolling on the horizontal part so that it completely goes round the circle on the cylindrical part.

$\\$ At the top most point $\\$ $\frac{mv^2}{R-r} = mg \Rightarrow v^2 = g(R-r)$ $\\$ Let the sphere is thrown with a velocity v' $\\$ Therefore applying laws of conservation of energy. $\\$ $\Rightarrow (\frac{1}{2})mv^2 + (\frac{1}{2})l \omega^2 = mg2(R-r) + (\frac{1}{2})mv^2 + (\frac{1}{2})l \omega^2$ $\\$ $\Rightarrow \frac{7}{10}v^2 = g 2(R-r) + \frac{7}{10}v^2$ $\\$ $ \Rightarrow v^2 = \frac{20}{7}g (R-r) + g(R-r)$ $\\$ $v' = \sqrt{\frac{27}{7}g(R-r)}$

79.   Figure (10-E15) shows a small spherical ball of mass m rolling down the loop track. The ball is released on the linear portion at a vertical height H from the lowest point. The circular part shown has a radius R. (a) Find the kinetic energy of the ball when it is at a point A where the radius makes an angle 0 with the horizontal. (b) Find the radial and the tangential accelerations of the centre when the ball is at A. (c) Find the normal force and the frictional force acting on the ball if H = 60 cm, R =10 cm, 0 = 0 and m = 70 g.

$\\$ a) Total kinetic energy y = $\frac{1}{2}mv^2 + \frac{1}{2} l \omega^2$ $\\$ Therefore according to the question $\\$ $mg H = \frac{1}{2}mv^2 + \frac{1}{2}l \omega^2 + mg R(1+ cos \theta)$ $\\$ $\Rightarrow mg H - mg R(1+ cos \theta) = (\frac{1}{2})mv^2 + (\frac{1}{2})l \omega^2$ $\\$ $\Rightarrow \frac{1}{2}mv^2 + \frac{1}{2}l \omega^2 = mg(H-R-R sin \theta)$ $\\$ $\Rightarrow \frac{7}{10}mv^2 = mg(H-R-R sin \theta)$ $\\$ $\frac{v^2}{R} = \frac{10}{7}g[\frac{H}{R} - 1 - sin \theta] \rightarrow$ radical acceleration $\\$ $v^2 = \frac{10}{7}g(H-R)-R sin \theta$ $\\$ $\Rightarrow 2v \frac{dv}{dt} = -\frac{10}{7}g R cos \theta \frac{d \theta}{dt}$ $\\$ $\omega R \frac{dv}{dt} = -\frac{5}{7}g R cos \theta \frac{d \theta}{dt}$ $\\$ $\Rightarrow \frac{dv}{dt} = -\frac{5}{7}g cos \theta \rightarrow$ tangential acceleration. $\\$ c) Normal force at $\theta = 0$ $\\$ $\Rightarrow \frac{mv^2}{R} = \frac{70}{1000} \frac{10}{7} \times10(\frac{0.6-0.1}{0.1}) = 5N$ $\\$ Frictional force :- $\\$ $f= mg-ma = m(g-a) = m(10-\frac{5}{7} \times 10) = 0.07(\frac{70-50}{7}) = \frac{1}{100} \times 20 = 0.2N$

80.   A thin spherical shell of radius R lying on a rough horizontal surface is hit sharply and horizontally by a cue. Where should it be hit so that the shell does not slip on the surface ?

$\\$ Let the cue strikes at a height ‘h’ above the centre, for pure rolling, $V_c = R\omega$ $\\$Applying law of conservation of angular momentum at a point A, $\\$ $mv_ch-l \omega = O$ $\\$ $mv_ch = \frac{2}{3}mR^2 \times (\frac{v_c}{R})$ $\\$h= $\frac{2R}{3}$

81.   A uniform wheel of radius R is set into rotation about its axis at an angular speed or. This rotating wheel is now placed on a rough horizontal surface with its axis horizontal. Because of friction at the contact, the wheel accelerates forward and its rotation decelerates till the wheel starts pure rolling on the surface. Find the linear speed of the wheel after it starts pure rolling.

$\\$A uniform wheel of radius R is set into rotation about its axis (case-I) at an angular speed $\omega$ $\\$ This rotating wheel is now placed on a rough horizontal. Because of its friction at contact, the wheel accelerates forward and its rotation decelerates. As the rotation decelerates the frictional force will act backward. $\\$ If we consider the net moment at A then it is zero. Therefore the net angular momentum before pure rolling and after pure rolling remains constant Before rolling the wheel was only rotating around its axis. $\\$ Therefore Angular momentum $= l \omega = (\frac{1}{2})MR^2 \omega$ ...(1) $\\$ After pure rolling the velocity of the wheel let v $\\$ Therefore angular momentum = $l_{cm} \omega + m(V \times R)$ $\\$ $= (\frac{1}{2})mR^2 \frac{V}{R} + mVR = \frac{3}{2}mVR$ ...(2) $\\$ Because, Equation (1) and (2) are equal $\\$ Therefore, $\frac{3}{2}mVR = \frac{1}{2}mR^2 \omega$ $\\$ $\Rightarrow V = \omega \frac{R}{3}$

82.   A thin spherical shell lying on a rough horizontal surface is hit by a cue in such a way that the line of action passes through the centre of the shell. As a result, the shell starts moving with a linear speed v without any initial angular velocity. Find the linear speed of the shell after it starts pure rolling on the surface.

$\\$ The shell will move with a velocity nearly equal to v due to this motion a frictional force well act in the background direction, for which after some time the shell attains a pure rolling. $\\$ If we consider moment about A, then it will be zero. Therefore, Net angular momentum about A before pure rolling = net angular momentum after pure rolling. $\\$ Now, angular momentum before pure rolling about A = M (V × R) and angular momentum after pure rolling :- $\\$ $\frac{2}{3} MR^2 \times(\frac{V_0}{R}) = M V_0 R$ $\\$ $(V_0 = velocity after pure rolling)$ $\\$ $\Rightarrow MVR = \frac{2}{3} MV_0R + MV_0R$ $\\$ $\Rightarrow\frac{5}{3}V_0 = V$ $\\$ $\Rightarrow V_0 = \frac{3V}{5}$

83.   A hollow sphere of radius R lies on a smooth horizontal surface. It is pulled by a horizontal force acting tangentially from the highest point. Find the distance travelled by the sphere during the time it makes one full rotation.

$\\$Taking moment about the centre of hollow sphere we will get $\\$ $F \times R = \frac{2}{3}MR^2 \alpha$ $\\$ $\Rightarrow \alpha = \frac{3F}{2MR}$ $\\$ Again, $2\pi = (\frac{1}{2}) \alpha t^2(From \theta = \omega_0t + \frac{1}{2} \alpha t^2)$ $\\$ $\Rightarrow t^2 = \frac{8\pi MR}{3F}$ $\\$ $\Rightarrow a_c = \frac{F}{m}$ $\\$ $\Rightarrow$ X $= \frac{1}{2} a_ct^2 = \frac{1}{2} = \frac{4\pi R}{3}$

84.   A solid sphere of mass 0.50 kg is kept on a horizontal surface. The coefficient of static friction between the surfaces in contact is 2/7. What maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface ?

$\\$ If we take moment about the centre, then $\\$ $F \times R = l \alpha \times f \times R$ $\\$ $\Rightarrow F = \frac{2}{5}mR \alpha + \mu mg$ ...(1) $\\$ Again, $F= ma_c- \mu mg$ ...(2) $\\$ $\Rightarrow a_c = \frac{F+\mu mg}{m}$ $\\$ Putting the value $a_c$ in equation (1) we get $\\$ $\Rightarrow \frac{2}{5} \times m \times (\frac{F+\mu mg}{m}) + \mu mg$ $\\$ $\Rightarrow \frac{2}{5} (F+ \mu mg) \mu mg$ $\\$ $\Rightarrow F = \frac{2}{5}F + \frac{2}{5} \times 0.5 \times 10 + \frac{2}{7} \times 0.5 \times 10$ $\\$ $\Rightarrow \frac{3F}{5} = \frac{4}{7} + \frac{10}{7} = 2$ $\\$ $\Rightarrow F = \frac{5 \times 2}{3} = \frac{10}{3} = 3.33N$

85.   A solid sphere is set into motion on a rough horizontal surface with a linear speed v in the forward direction and an angular speed v IR in the anticlockwise direction as shown in figure (10-E16). Find the linear speed of the sphere (a) when it stops rotating and (b) when slipping finally ceases and pure rolling starts.

$\\$a) if we take moment at A then external torque will be zero $\\$ Therefore, the initial angular momentum = the angular momentum after rotation stops (i.e. only leniar velocity exits) $\\$ $MV \times R - l \omega = MV_o \times R$ $\\$ $\Rightarrow MVR-\frac{2}{5}\times \frac{MR^2V}{R} = MV_oR$ $\\$ $\Rightarrow V_o = \frac{3V}{5}$ $\\$ b) Again, after some time pure rolling starts $\\$ therefore $\Rightarrow M \times v_o \times R = \frac{2}{5}MR^2 \times \frac{V'}{R} + MV'R$ $\\$ $\Rightarrow V' = \frac{3V}{7}$

86.   A solid sphere rolling on a rough horizontal surface with a linear speed v collides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in the backward direction.

$\\$When the solid sphere collides with the wall, it rebounds with velocity ‘v’ towards left but it continues to rotate in the clockwise direction. $\\$ So, the angular momentum = $mvR-(\frac{2}{5})mR^2 \times \frac{v}{R}$ $\\$ After rebounding, when pure rolling starts let the velocity be v' $\\$ and the corresponding angular velocity is $\frac{v'}{R}$ $\\$ Therefore angular momentum = $mv'R + \frac{2}{5} mR^2(\frac{v'}{R})$ $\\$ So, $mvR \times (\frac{3}{5}) = mvR \times \frac{7}{5}$ $\\$ $v'= \frac{3v}{7}$ So, the sphere will move with velocity $\frac{3v}{7}.$