Solve the following :

Question: A square plate of mass $120 \mathrm{~g}$ and edge $5.0 \mathrm{~cm}$ rotates about one of the edges. If it has a uniform angular acceleration of $0.2 \mathrm{rad} / S^{2}$, what torque acts on the plate? Solution: Squeeze rod along axis $A^{\prime \prime}$ so it will behave as rod $I=\frac{m a^{2}}{3}$ $\tau=I \alpha$ $=\frac{(0.120)(0.05)^{2}}{3}(0.2)$ $\tau=2 \times 10^{-5} \mathrm{~N}-\mathrm{m}$...

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Question: 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. Solution: Torque $=\bar{\tau} \times \bar{r}$ $\left.\mathrm{T}=\mathrm{F}^{\left(\frac{L}{4}\right.}\right) \sin 90^{\circ}$ $1 \propto \frac{F L}{...

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Question: A cubical block of mass $m$ and edge a slides down a rough inclined plane of inclination $\theta$ with a uniform speed. Find the torque of the normal force acting on the block about its centre. Solution: Since, acceleration $=0$ ff $=m g^{\sin \theta}$ The normal contact will shift as block is not toppling. So, $\tau_{\text {Normal }}=\tau_{f f}$ $=\mathrm{mg}^{\sin \theta\left(\frac{a}{2}\right)}$ HC VERMA Solutions for Class 11 Physics Chapter 10 Rotational Mechanics...

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Question: Calculate the total torque acting on the body shown in figure about the point $O$. Solution: Torque $=F \times d_{\perp}$ Consider anticlockwise torque as negative and clockwise torque as positive. $\tau_{N}=\tau_{10}+\tau_{15}+\tau_{5}+\tau_{20}$ $=(10)(4)-(15)\left(6^{\sin 37}\right)+5(0)-20\left(4^{\cos 60^{\circ}}\right)$ $\tau_{\text {Net }}=-0.54$ $\mathrm{T}=0.54 \mathrm{~N}-\mathrm{m}$ (anticlockwise direction)...

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Question: When a force of $6.0 \mathrm{~N}$ is exerted at $30^{\circ}$ to a wrench at a distance of $8 \mathrm{~cm}$ 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 $16 \mathrm{~cm}$ from the nut? Solution: Torque in both the cases will be equal $\tau_{1}=\tau_{2}$ $\left(6 \sin 30^{\circ}\right)(8)=(F)(16)$ $F=1.6 \mathrm{~N}$...

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Question: A simple pendulum of length $I$ is pulled aside to make an angle $\theta$ 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: rorque $=F \times d_{\perp}$ $\tau=\omega l \sin \theta$ Torque will be zero at the lowest point of suspension as $\theta$ will be zero....

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Question: A particle of mass $m$ is projected with the speed $u$ at an angle $\theta$ 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: $\begin{aligned} \tau =F \times d_{\Perp} \\ =m g\left(\frac{\operatorname{Range}}{2}\right) \\ =m g\left(\frac{u^{2} \sin 2 \theta}{2 g}\right) \\ \tau =m u^{2} \sin \theta \cos \theta \end{aligned}$ Torque acts perpendicular to plane of the motion....

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Question: The surface density (mass/area) of a circular disc of radius a depends on the distance from the centre as $\mathrm{p}(\mathrm{r})=\mathrm{A}+\mathrm{Br}$. Find its moment of inertia about the line perpendicular to the plane of the disc through its centre. Solution: Surface density of disc is varying radially so considering an element of thickness 'dr' at a distance ' $r$ ' from center of disc. $\sigma=\mathrm{A}+\mathrm{Br}=\frac{d m}{d A}$ $d m=(\mathrm{A}+\mathrm{Br})(2 \pi r d r)$ $...

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Question: Find the moment of inertia of a uniform square plate of mass $m$ and edge a about one of its diagonals. Solution: If we squeeze rod along $X X^{v}$ and $Y Y^{v}$ the square becomes rod of side 'a' So, $I_{X X^{r}}=I_{Y Y^{r}}=\frac{m a^{2}}{12}$ By perpendicular axis theorem, $I_{Z Z^{r}}=I_{X X^{l}}+I_{Y Y^{r}}=I_{p p^{l}}+I_{Q Q^{l}}$ $\frac{m a^{2}}{12}+\frac{m a^{2}}{12}=I_{p p^{l}}+I_{Q Q^{\prime}}$ $I_{p p^{l}}=\frac{m a^{2}}{12}\left[\because_{p p^{l}}=I_{Q Q^{\prime}}\right.$ A...

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Question: 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: $I_{a x i s}=I_{C O M}+m d^{2}$ $m k^{2}=\frac{m r^{2}}{2}+m d^{2}$ $\Rightarrow m r^{2}=\frac{m r^{2}}{2}+m d^{2}$ $\Rightarrow d=\sqrt{2}$...

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Question: Find the radius of gyration of a circular ring of radius $r$ about a line perpendicular to the plane of the ring passing through one of its particles. Solution: .$I_{a x i s}=I_{C O M}+m x^{2}$ $I_{A A^{r}}=m r^{2}+m r^{2}$ $I_{A A^{r}}=2 m r^{2}$ $\mathrm{~m}^{k^{2}}=2 m r^{2}$ $\therefore \mathrm{k}=\sqrt{2} r$...

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Question: The moment of inertia of a uniform rod of mass $0.50 \mathrm{~kg}$ and length $1 \mathrm{~m}$ is $0.10 \mathrm{~kg}-\mathrm{m}^{2}$ about a line perpendicular to the rod. Find the distance of this line from the middle point of the rod. Solution: By parallel axis theorem, $I_{a x i s}=I_{C O M}+m x^{2}$ $I_{a x i s}=\frac{m L^{2}}{12}+m x^{2}$ $0.1=\frac{(0.5)(1)^{2}}{12}+(0.5) x^{2}$ $X=0.34 m$...

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Question: Find the moment of inertia of a pair of spheres, each having a mass $\mathrm{m}$ and radius $\mathrm{r}$, kept in contact about the tangent passing through the point of contact. Solution: By parallel axis theorem, $I_{a x i s}=I_{C O M}+m r^{2}$ $I_{A A^{r}}=\left(\frac{2}{5} m r^{2}+m r^{2}\right) 2$ $I_{A A^{r}}=\frac{14}{5} m r^{2}$...

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Question: Particles of masses $1 \mathrm{~g}, 2 \mathrm{~g}, 3 \mathrm{~g}, \ldots . ., 100 \mathrm{~g}$ are kept at the marks $1 \mathrm{~cm}, 2 \mathrm{~cm}, 3 \mathrm{~cm}, \ldots \ldots$, $100 \mathrm{~cm}$ respectively on a metrescale. Find the moment of inertia of the system of particles about a perpendicular bisector of the metre scale. Solution: The perpendicular bisector passes from $50 \mathrm{~cm}$ mark. So, there will be 49 particles on LHS and 50 particles on RHS. $\mathrm{I}=\left[...

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Solve the following :

Question: Three particles, each of mass $200 \mathrm{~g}$, are kept at the corners of an equilateral triangle of side $10 \mathrm{~cm}$. 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 perpendicular to the plane of the particles. Solution: (a) $I_{B C}=m^{\left(\frac{\sqrt{3}}{2} a\right)^{2}+m(0)^{2}+m(0)^{2}}$ $=0.2^{\left(\frac{\sqrt{3}}{2} \times 0.1\right)^{2}}$ $=1.5^{\times 10^{-3}} \mathrm{Kq}_{-} m^...

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Solve the following :

Question: A block hangs from a string wrapped on a disc of radius $20 \mathrm{~cm}$ free to rotate about its axis which is fixed in a horizontal position. If the angular speed of the disc is $10 \mathrm{rad} / \mathrm{s}$ at some instant, with what speed is the block going down at the instant? Solution: Block speed=Rim speed of the disc $\mathrm{v}=\mathrm{R} \omega=(0.2)(10)$ $v=R \omega=(0.2)(10)$ $v=2 \mathrm{~m} / \mathrm{s}$...

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Solve the following :

Question: A disc rotates about its axis with a constant angular acceleration of $4 \mathrm{rad} /{ }^{S^{2}}$. Find the radial and tangential accelerations of a particular at a distance of $1 \mathrm{~cm}$ from the axis end of the first second after the disc starts rotating. Solution: $\omega_{0}=0, \alpha=4 \mathrm{rad} /{ }^{s^{2}} ; \mathrm{t}=1 \mathrm{sec}$ $\omega={ }^{\omega} 0_{+} \alpha_{t}$ $\omega=4 \mathrm{rad} / \mathrm{sec}$ Radial acceleration $=\mathrm{R}^{\omega^{2}}$ $=(0.01)^{...

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Solve the following :

Question: A disc of radius $10 \mathrm{~cm}$ is rotating about its axis at an angular speed of $20 \mathrm{rad} / \mathrm{s}$. Find the linear speed of (a) a point on the rim. (b) the middle point of a radius. Solution: $(a) v=r w=(0.1)(20)$ $v=2 \mathrm{~m} / \mathrm{s}$ (b) $v=r \omega=\left(\frac{0.1}{2}\right)(20)$ $v=1 \mathrm{~m} / \mathrm{s}$...

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Question: Find the angular velocity of a body rotating with an acceleration of $2 \mathrm{rev} / \mathrm{s}^{2}$ as it completes the $5^{\text {th }}$ revolution after the start. Solution: $\omega_{0}=0, \alpha=2 \frac{r e v}{s^{2}}=2 \times 2 \pi=4 \pi \frac{r a d}{s^{2}}$ $\theta=5 \mathrm{rev}=5 \times 2 \pi \mathrm{rad}=10 \pi \mathrm{rad}$ Substituting values in, $\omega^{2}=\omega_{0}^{2}+2 \alpha \theta$ $\omega=4 \pi \sqrt{5} \mathrm{rad} / \mathrm{sec}$ $\omega=2^{\sqrt{5}} \mathrm{rev}...

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Solve the following :

Question: 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 $5 \mathrm{rad} / \mathrm{s}$ to $15 \mathrm{rad} / \mathrm{s}$. Solution: $\omega_{0}=5 \mathrm{rad} / \mathrm{s} ; \omega=15 \mathrm{rad} / \mathrm{sec} ; \alpha=1 \mathrm{rad} / \mathrm{sec}^{2}$ Substituting values in, $\omega^{2}=\omega_{0}^{2}+2 \alpha \theta$ $\theta=100 \mathrm{rad}$...

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Solve the following :

Question: A wheel starting from rest is uniformly accelerated at $4 \mathrm{rad} /{ }^{S^{2}}$ for 10 seconds. It is allowed to rotated 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: Angular displacement=Area under curve $=\frac{1}{2}(30+10)(40)$ $\theta=800 \mathrm{rad}$...

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Solve the following :

Question: 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: $\omega_{0}=0 ; \mathrm{t}=4 \mathrm{sec} ; \quad \theta=50$ rev $=100 \pi \frac{r a d}{\sec }$ Using, $\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2}$ $\alpha=8 \pi^{\frac{r a d}{s e c^{2}}}$ or $\alpha=4 \pi \pi^{\frac{r e v}{s^{2}}}$ Using, $\omega=\omega_{0}+\alpha t$ $\omega=40...

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Solve the following :

Question: A wheel is making revolutions about its axis with uniform angular acceleration. Starting from rest, it reaches $100 \mathrm{rev} / \mathrm{sec}$ in 4 seconds. Find the angular acceleration. Find the angle rotated during these four seconds. Solution: $\omega_{0}=0 ; \mathrm{t}=4 \mathrm{sec} ;$ $\omega=100 \frac{r e v}{\sec }=100 \times 2 \pi^{\frac{r a d}{s e c}}$ Using, $\omega=\omega_{0}+\alpha t$ $\alpha=50 \pi^{\frac{r a d}{r e c^{2}}}$ or $\alpha=25 \pi^{\frac{r a d}{s e c^{2}}}$ ...

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Solve the following :

Question: A cart of mass $M$ is at rest on a frictionless horizontal surface and a pendulum bob of mass $m$ hangs from the roof of the cart. The string breaks, the bob falls on the floor, makes several collisions on the floor and finally lands up in a small slot made in the floor. The horizontal distance between the string and the slot is $\mathrm{L}$. Find the displacement of the cart during this process. Solution: Take ${ }^{\circ}$ as origin $x_{C O M}=\frac{x m+(x+L) M}{m+M}$ $x_{C O M}^{\pr...

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Solve the following :

Question: A small particle travelling with a velocity $v$ collides elastically with a spherical body of equal mass and of radius $r$ initially kept at rest. The center of this spherical body is located a distance $(\rhor)$ away from the direction of motion of the particle. Find the final velocities of the two particles. [Hint:- The force acts along the normal to the sphere through the contact. Treat the collision as onedimensional for this direction. In the tangential direction no force acts and...

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