 # Rotational Mechanics

## Concept Of Physics

### H C Verma

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.

##### Solution :

$\omega_{0}$ = $0$;$\rho$ = $100$ $\frac{rev}{s}$ ; $\omega$ = 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.

##### Solution :

$\\$$\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.

##### Solution :

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. ##### Solution : \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. ##### Solution : A dish of radius = 10 cm = 0.1 m \\ Angular velocity = 20 \frac {rad}{s} \\ \therefore Linear velocity on the rim = \omegar = 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. ##### Solution : 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 ? ##### Solution : 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. ##### Solution : 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. ##### Solution : 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. ##### Solution : 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. ##### Solution : 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. ##### Solution : 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. ##### Solution : 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.

##### Solution : 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.

##### Solution : 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.

##### Solution : 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 ?

##### Solution : 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 ?

##### Solution :

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 $0$.

##### Solution : 