Rotational Motion Problems and Solutions Physics Class 11 Important Question

Ans. Radial component of linear momentum.

Ans. Cause of rotational motion – Torque Out come in rotational motion – Angular momentum

Ans. Is it possible to open a pen up with one finger ? Why ?

Ans. Angular momentum gained by the bullet provides better accuracy.

Ans. No, direction of centripetal force remains constant either it is clockwise and anticlockwise rotation.

Ans. No, torque is required only for producing angular acceleration. For uniform rotation, no torque is needed.

Ans. A projectile acquires angular momentum about the point of projection during flight. Does it violate the conservation of angular momentum ?

Ans. The moment of inertia plays the same role in rotational motion as the mass does in linear motion.

Ans. Yes, moment of inertia of a body changes with the change of the axis of rotation.

Ans. The moment of inertia is another name for rotational inertia. The moment of inertia is called rotational interia for the reason that it gives the measure of inertia in rotational motion.

Ans. Why are the spokes fitted in a cycle wheel ?

Ans. Two solid spheres of same mass are made of metals of different densities. Which of them has a larger moment of inertia about a diameter ?

Ans. A ring has a larger moment of inertia because its entire mass is at a larger distance from the axis of rotation than the average distance of the mass particles of the disc.

Ans. The moment of inertia will decrease, because the material particles will, on the average come closer to the axis.

Ans. If no other external force acts on a body, friction is necessary for a body to roll. Frictional force being tangential provides torque. Any force at the centre of mass will provide no torque and so cannot roll the body.

Ans. Yes, a body in translatory motion shall have angular momentum unless the fixed point about which angular momentum is taken lies on the line of motion of the body. This follows from $|L|=r p \sin \phi \cdot L=0,$ only when $\phi=0$ or $\phi=180^{\circ}$

Ans. Yes, angular momentum of the planet is constant over the entire orbit. This is because revolution of planet around the star is under the effect of gravitational force between the star and the planet. This is a radial force whose torque is zero. Therefore, angular momentum of the planet is constant (vector), whatever be the nature of the orbit.

Ans. Let $I_{1}$ and $\omega_{1}$ be the initial moment of inertia and the angular speed of the star about its own axis of rotation, and $I_{2}$ and $\omega_{2}$ be the corresponding quantities after the star is collapsed. since the collapse is due to the internal forces (gravitational attraction of the outer layers by the inner layers of the star), the angular momentum of the star remains conserved during collapse. Thus, The collapsed star rotates with a speed $10^{4}$ times the initial angular speed.

Ans. Here, initial angular speed, $\omega_{1}=40 \mathrm{rev} / \mathrm{min} ; \omega_{2}=?$ final moment of inertia, $I_{2}=\frac{2}{5} I_{1}$ As no external torque acts in the process, therefore, L= constant i.e. K.E. of rotation increases. This is because the child spends internal energy in folding back his hands.

Ans. At the highest point of projectile motion the velocity is $u \cos \theta$ acting horizontally. The perpendicular distance between the point of projection and the highest point

Ans. The torque on the planet due the sun, $\vec{\tau}=\vec{r} \times \vec{F}$ Where $\vec{r}$ is position vector of the planet $w \cdot t .$ the sun and $\vec{F}$ is the gravitational force on the planet since the gravitational force on the planet acts along the line joining the planet to the sun, the vectors and are always parallel and hence.

Ans. Angular velocity will be same for all points. Consider a small portion of length dx at a distance x from the axis. Mass of $d x=\rho . A d x,$ where $A$ is cross sectional area. Force experienced by $d x=m r \omega^{2}=\rho A d x . x \omega^{2}$ Force at the edge $=\int_{0}^{l} \rho A \omega^{2} x d x \Rightarrow F=\rho A \omega^{2} \frac{l^{2}}{2}$

Ans. Here, Moment of inertia of grindstone, $I=6 \mathrm{kg} m^{2}$ Initial angular velocity, $\omega_{1}=0$ Final angular velocity, $\omega_{2}=2 \pi n=2 \pi \times \frac{150}{60}=5 \pi \mathrm{rad} / \mathrm{sec}$ Time for which torque acts, $t=10 \mathrm{sec} .$ $\therefore$ Angular acceleration $(\alpha)=\frac{\omega_{2}-\omega_{1}}{t}=\frac{5 \pi-0}{10}=\frac{\pi}{2} r a d / \sec ^{2}$ As $\tau=I \alpha \quad \therefore \tau=6 \times \frac{\pi}{2}=3 \pi \mathrm{Ns}$

Ans. Using the relation $v=r \omega,$ we get For point $A, v_{A}=R \omega_{0},$ along $A X$ For point $B, v_{B}=R \omega_{0},$ along $B X^{\prime}$ For point $C, v_{C}=\left(\frac{R}{2}\right) \omega_{0}$ parallel to $A X,$ The disc will not rotate, because it is placed on a perfectly frictionless table. Without friction, rolling is not possible.

Ans. For a ‘uniform’ angular speed of rotation, no torque is needed, provided there is no friction. In practice, however, a torque has to be applied to the rotor to counterbalance the frictional torque. If $\tau$ be the frictional torque, then the required power of the engine transmitting the torque is given by $P=\tau \omega$ when $\omega$ is the angular speed. Here $\tau=180 N m$ and $\omega=200 r a d s^{-1},$ therefore $P=180 N m \times 200 s^{-1}=36000 W=36 k W$

Ans. We know $\vec{\tau}=\vec{r} \times \vec{F}$. If $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$ and $\vec{F}=F_{x} \hat{i}+F_{y} \hat{j}+F_{z} \hat{k},$ we have, $\tau=\tau_{x} \hat{i}+\tau_{y} \hat{j}+\tau_{z} \hat{k}=\left|\begin{array}{ccc}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {\hat{i}} & {\hat{j}} & {\hat{k}} \\ {x} & {y} & {z} \\ {F_{x}} & {F_{y}} & {F_{z}}\end{array}\right|$ Comparing the coefficients of $\hat{i}, \hat{j}$ and $\hat{k},$ we have $\tau_{z}=x F_{y}-y F_{x}$

Ans. To roll a disc, we require a torque, which can be provided only by a tangential force. As force of friction is the only tangential force in this case, it is necessary. (a) As frictional force at B opposes the velocity of point B, which is to the left, the frictional force must be to the right. The sense of frictional torque will be perpendicular to the plane of the disc and outwards. (b) As frictional force at B decreases the velocity of the point of contact B with the surface, the perfect rollling beings only when velocity of point B becomes zero. Also, force of friction would become zero at this stage.

Ans. Parallel AxesTheorem The moment of inertia about an axis passing parallel to the axis through the centre of mass of a rigid body is the sum of the moment of inertia ( $l_{m}$ ) of the body about the axis through centre of mass and the product of its mass $M$ and square of the separation $\left(a^{2}\right)$ between the parallel axis.

Ans. Perpendicular Axes Theorem According to this theorem the sum of moment of inertia of a plane lamina about two mutually perpendicular lying in its plane is equal to its moment of inertia about an axis perpendicular axes to the plane of lamina and passing through the point of intersection of first two axes.

Ans. In a fly wheel, most of the mass is concentrated at the rim. Due to this peculiar shape, the flywheel possesses a large value of moment of inertia for the given mass and the radius. If such a wheel gains or loses some rotational energy, it brings about only a very small change in its angular speed. In this way, a flywheel helps to maintain uniform motion.

Ans. Loss in P.E. = Gain in rotational Kinetic energy

Ans. When a cylinder slides without rolling, $E=\frac{1}{2} m v_{1}^{2} \Rightarrow v_{1}=\sqrt{\frac{2 E}{m}}$ When the cylinder rolls without slipping $E=\frac{1}{2} m v_{2}^{2}+\frac{1}{2} I w^{2} \Rightarrow E=\frac{1}{2} m v_{2}^{2}+\frac{1}{2}\left(\frac{1}{2} m r^{2}\right) w^{2}$ $=\frac{1}{2} m v_{2}^{2}+\frac{1}{4} m v_{2}^{2}=\frac{3}{4} m v_{2}^{2} \quad[\because v=r w]$ $\therefore v_{2}=\sqrt{\frac{4 E}{3 m}}$ As $v_{1}>v_{2},$ therefore sliding cylinder will win the race,

Ans. Consider that a stone of mass $M$ is revolved along a circular path by tying it to a thread of length $l .$ Moment of inertia of stone, $I=M l^{2}$ Suppose that a torque $\tau$ is applied to produce an angular acceleration $\alpha .$ Then, $\tau=I \alpha=\left(M l^{2}\right) \alpha$ Or $\alpha=\frac{\tau}{M l^{2}}$ It follows that to produced the same amount of angular acceleration; the larger torque is required, when longer string is used to revolve a stone of given mass. Hence, it is more difficult to revolve a stone by tying it to a longer string.

Ans. In fig. $A B C$ is an equilateral triangle. Let $A O \perp B C$, so $A O$ is also bisector of $B C$ i.e. $A O$ is median of $\triangle A B C .$ We have to calculate moment of inertia of the system about AO.

Ans.

Ans. We are given, moment of inertia of the disc about any of its diameter $=\frac{1}{4} M R^{2}$ Using theorem of perpendicular axes, moment of inertia of the disc about an axis passing through a point on its edge and normal to the disc $=2 \times \frac{1}{4} M R^{2}=\frac{1}{2} M R^{2}$

Ans. $M g-2 T=M a \quad ; \quad 2 T R=I \alpha=\frac{M R^{2}}{2} \frac{a}{R}$

Ans. The make the disc roll, a torque is required which in turn requires tangential force on it. Here, in this problem, only frictional force can provide the tangential force. (a) Frictional force at point B opposes the velocity of point B. Therefore, frictional force at the point of contact is in the same direction as the arrow. The diretion of frictional torque is such that it opposes the angular motion. since is normal to the plane of the paper in inward direction, the direction of frictional torque is normal to paper and in outward direction. (b) Frictional force at point B decreases the velocity of the point of contact B with the surface. The perfect rolling takes place, when the velocity of point B becomes zero and once this is so, the force of friction also becomes zero.

Ans. Here, M.I. of plane lamina about an axis through its C.M. and perpendicular to its plane, $I=1.6 \mathrm{M} a^{2}$ If $I_{x}$ and $I_{y}$ are moments of inertia of the plane lamina about the axes $X X^{\prime}$ and $Y Y^{\prime}$ respectively, then according to the theorem of perpendicular axes, $I_{x}+I_{y}=I$ since the lamina is symmetrical about the axes and $I_{x}=I_{y}$ $\therefore \quad I_{x}+I_{x}=I$ or $\quad I_{x}=\frac{1}{2} I=\frac{1}{2} \times 1.6 \mathrm{Ma}^{2}=0.8 \mathrm{Ma}^{2}$ According to the theorem of parallel axes, we have $I_{A B}=I_{x}+M(O Y)^{2}=0.8 M a^{2}+M(2 a)^{2}=4.8 \mathrm{Ma}^{2}$

Ans. We know that acceleration of an object down an inclined plane is given by $a=\frac{g \sin \theta}{1+I / m r^{2}}$ ... (i) For a ring, $I=m r^{2}$ $a_{\sin g}=\frac{g \sin \theta}{1+1}=0.5 g \sin \theta$ For a disc, $I=\frac{1}{2} m r^{2}$ $\therefore \quad \frac{g \sin \theta}{1+\frac{1}{2}}=\frac{2}{3} g \sin \theta \quad=0.67 g \sin \theta$ For a sphere, $I=\frac{2}{5} m r^{2}$ $\therefore \quad a_{\text {sphere }}=\frac{g \sin \theta}{1+\frac{2}{5}}=\frac{5}{7} g \sin \theta=0.71 g \sin \theta$ As is maximum it will reach the bottom at the earliest. Again, as is minimum, it will reach the bottom at the end. The result does not depend upon the mass or the radices of the rolling body.

Ans. Angular momentum with the two masses before collision $L_{i}=m 2 v \cdot a+2 a 2 m v=6 m v a$ moment of the bar about an axis through its central axis, $I_{b}=\frac{8 m(6 a)^{2}}{12}=\frac{8 \times 36 \times m a^{2}}{12}=24 m a^{2}$ (i) Moment of inertia after collision $=24 m a^{2}+m a^{2}+2 m \cdot 4 a^{2}$ $I_{f}=33 m a^{2}$ (ii) Angular momentum associated $=L_{i}=6 \mathrm{mva}$ (iii) Angular velocity after collision $=\omega=\frac{L_{i}}{I_{f}}$ i.e., $\quad \omega=\frac{6 m v a}{33 m a^{2}}=\frac{6 v}{33 a}=\frac{2 v}{11 a}$

Ans. Let $T_{1}$ and $T_{2}$ be the periods of revolution of theearth before and after its contraction ; and $R_{1}$ and $R_{2}$ be its radii before and after the contraction. Now, according to the principle of conservation of angular momentum, $I_{1} \omega_{1}=I_{2} \omega_{2}$ where $I_{1}=\frac{2}{5} M R_{1}^{2}$ and $I_{2}=\frac{2}{5} M R_{2}^{2}$ are moments of inertia of the earth before and after its contraction. it may be noted that the mass of the earth $(M)$ remains unchanged. $\therefore \frac{2}{5} M R_{1}^{2}\left(\frac{2 \pi}{T_{1}}\right)=\frac{2}{5} M R_{2}^{2}\left(\frac{2 \pi}{T_{2}}\right)$ $\Rightarrow T_{2}=\left(\frac{R_{2}}{R_{1}}\right)^{2} \times T_{1}$ ...(i) As the earth shrinks to $\frac{1}{64}$ of its volume, we have final volume $=\frac{1}{64}$ (initial volume) substituting the value of $\frac{R_{2}}{R_{1}}$ in equation (i) we get $T_{2}=\left(\frac{1}{4}\right)^{2} T_{1}=\frac{T_{1}}{16}$ i.e. duration of the day will be of 1.5 hour.

Ans. (a) Total initial angular momentum of the two discs $=I_{1} \omega_{1}+I_{2} \omega_{2}$ since the two discs are brought into contact face to face, with their axes of rotation coincident, the M.I. of the combined system will be $I_{1}+I_{2} .$ If $\omega$ is the angular speed of the combined system, then final angular momentum of the system $=\left(I_{1}+I_{2}\right) \omega$ since no external torque acts on the system, we have $I_{1} \omega_{1}+I_{2} \omega_{2}=\left(I_{1}+I_{2}\right) \omega \Rightarrow \omega=\frac{I_{1} \omega_{1}+I_{2} \omega_{2}}{I_{1}+I_{2}}$ (b) Initial K.E. of rotation of the two discs $=\frac{1}{2} I_{1} \omega_{1}^{2}+\frac{1}{2} I_{2} \omega_{2}^{2}$ Final K.E. of the combined system $=\frac{1}{2}\left(I_{1}+I_{2}\right) \omega^{2}=\frac{1}{2}\left(I_{1}+I_{2}\right)\left(\frac{I_{1} \omega_{1}+I_{2} \omega_{2}}{I_{1}+I_{2}}\right)^{2}=\frac{1}{2} \frac{\left(I_{1} \omega_{1}+I_{2} \omega_{2}\right)^{2}}{I_{1}+I_{2}}$ Now, (initial K.E.) – (final K.E.) $=\frac{1}{2} I_{1} \omega_{1}^{2}+\frac{1}{2} I_{2} \omega_{2}^{2}-\frac{1}{2} \frac{\left(I_{1} \omega_{1}+I_{2} \omega_{2}\right)^{2}}{I_{1}+I_{2}}$ $=\frac{1}{2} \times \frac{I_{1}^{2} \omega_{1}^{2}+I_{1} I_{2} \omega_{1}^{2}+I_{1} I_{2} \omega_{2}^{2}+I_{2}^{2} \omega_{2}^{2}-I_{1}^{2} \omega_{1}^{2}-I_{2}^{2} \omega_{2}^{2}-2 I_{1} I_{2} \omega_{1} \omega_{2}}{I_{1}+I_{2}}$ $=\frac{I_{1} I_{2}}{2\left(I_{1}+I_{2}\right)}\left(\omega_{1}^{2}+\omega_{2}^{2}-2 \omega_{1} \omega_{2}\right)=\frac{I_{1} I_{2}\left(\omega_{1}-\omega_{2}\right)^{2}}{2 I_{1}+I_{2}}$ = a positive quantity Hence the rotational kinetic energy of the combined system is less than the sum of the initial energies of the two discs. The loss of energy is due to dissipation of energy in frictional contact of the two discs. It may be pointed out that the two discs rotate as a single combined system with a common angular speed, because of frictional contact only. However, the angular momentum will be conserved as the frictional torque is an internal torque.