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 $T_2 = 2.156 - 1.038 = 1.118 = 1.12 N.$