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Question: 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. Solution: $\because^{\tau_{e x t}}=0$ $L_{i}=L_{f}$ ${\left.\left[M^{\le...
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Question: Suppose the particle of the previous problem has a mass $m$ and speed $v$ before the collision and sticks to the rod after the collision. The rod has a mass M. (a) Find the velocity of the centre of mass $\mathrm{C}$ 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 a...
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Question: A uniform rod of length $L$ lies on a smooth horizontal table. A particle moving on a 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. Solution: Momentum conservation $P_{i}=P_{f}$ $m u+0=m(0)+M(v)$ $\mathrm{V}=\frac{\operatorname{mu}}{M}$ Angular Momentum conservation $L_{i}=L_{f}$ $0+m...
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Question: A uniform rod of mass $\mathrm{m}$ and length $\mathrm{I}$ is struck at an end by a force $F$ perpendicular to the rod for 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...
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Question: Suppose the platform with the kid in the previous problem is rotating in anticlockwise direction at an angular speed $w$. The kid starts walking along the rim with the speed $v$ relative to the platform also in the anticlockwise direction. Find the new angular speed of the platform. Solution: Let angular velocity of platform after kid start running $\omega^{\prime}$ So, angular velocity of kid with respect to earth $=\left(\omega^{\prime}+\frac{v}{R}\right)$ $\because \tau_{e x t}=0$ $...
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Question: Suppose the platform of the previous problem is brought to rest with the ball in 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 friend. Find the angular velocity with which the platform will start rotating. Solution: Considering ball, boy and platform as system $L_{i}=L_{f}$ $0+0+0=m v R+l(-w)+\left(M^{R^{2}}\right)(-w)$ $\omega=\frac{m v R}{\left(I+M R^{2}\right)}$...
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Question: 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 $\mathrm{I}$. 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. Solution: Considering ball, boy and platform as system $L_{...
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Question: A wheel of moment of inertia $0.10 \mathrm{~kg}^{-m^{2}}$ is rotating about a shaft at an angular speed of 160 rev/minute. A second wheel is set into rotation at $300 \mathrm{rev} / \mathrm{minute}$ and is coupled to the same shaft so that both the wheels finally rotate with a common angular speed of $200 \mathrm{rev} / \mathrm{minute}$. Find the moment of inertia of the second wheel. Solution: Taking wheels as a system $\tau_{\text {ext }}=0$ $L_{i}=L_{f}$ $I_{1} \omega_{1}=I_{2} \ome...
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Question: A boy is standing on a platform which is free to rotate about its axis. The boy holds an open umbrella coincides with that of the platform. The moment of inertia of " the platform plus the boy system" is $3.0$ $\times 10^{-3} \mathrm{~kg}_{-} \mathrm{m}^{2}$ and that of the umbrella is $2.0^{\times} 10^{-3} \mathrm{~kg}^{-} \mathrm{m}^{2}$. The boy starts spinning the umbrella about the axis at an angular speed of $2.0 \mathrm{rev} / \mathrm{s}$ with respect to himself. Find the angula...
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Question: 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 by $6 \mathrm{~kg}_{-} m^{2}$ to $2 \mathrm{~kg}^{-} \mathrm{m}^{2}$, what will be the new angular speed? Solution: Since, ${ }^{\tau_{\text {ext }}}=0$ $\therefore L_{i}=L_{f}$ $6(12...
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Question: A driver having a moment of inertia of $6.0 \mathrm{~kg}_{-} \mathrm{m}^{2}$ about an axis through its centre of mass rotates at an angular speed of $2 \mathrm{rad} / \mathrm{s}$ about the axis. If he folds his hands and feet to decrease the moment of inertia to $5.0 \mathrm{~kg}_{-} \mathrm{m}^{2}$, what will be the new angular speed? Solution: Since, ${ }^{\tau_{\text {ext }}}=0$ $\quad \approx L_{i}=L_{f}$ $6(2)=5^{\omega_{f}}$ $\omega_{f}=2.4 \mathrm{rad} / \mathrm{sec}$...
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Question: A driver having a moment of inertia of $6.0 \mathrm{~kg}_{-} \mathrm{m}^{2}$ about an axis through its centre of mass rotates at an angular speed of $2 \mathrm{rad} / \mathrm{s}$ about the axis. If he folds his hands and feet to decrease the moment of inertia to $5.0 \mathrm{~kg}_{-} \mathrm{m}^{2}$, what will be the new angular speed? Solution: Since, $\tau_{\text {ext }}=0$ $\therefore L_{i}=L_{f}$ $6(2)=5^{\omega_{f}}$ $\omega_{f}=2.4 \mathrm{rad} / \mathrm{sec}$...
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Question: A wheel of moment of inertia $0.500 \mathrm{~kg}_{-} \mathrm{m}^{2}$ and radius $20.0 \mathrm{~cm}$ is rotating about its axis at an angular speed of $20.0 \mathrm{rad} / \mathrm{s}$. It picks up a stationary particle of mass $200 \mathrm{~g}$ at its edge. Find the new angular speed of the wheel. Solution: Since, $\tau_{\text {ext }}=0$ $\therefore L_{i}=L_{f}$ $I_{i} \omega_{i}=I_{f} \omega_{f}$ $(0.5)(20)=\left[0.5+(0.2)^{(0.5)^{2} \omega_{f}}\right]$ $\omega_{f}=19.7 \frac{\text { r...
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Question: A dumb-bell consists of two identical small balls of mass $1 / 2 \mathrm{~kg}$ each connected to the two ends of a $50 \mathrm{~cm}$ 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 $10 \mathrm{rad} / \mathrm{s}$. An impulsive force of average magnitude $5.0 \mathrm{~N}$ acts on one of the masses in the direction of its velocity for $0.10 \mathrm{~s}$. Find the new angular velocity of the system. S...
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Question: Two particles of masses $m_{1}$ and $m_{2}$ are joined by a light rigid rod of length $r$. The system rotates at an angular speed $\omega$ 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 $\mu$ is the reduced mass of the system defined as $\mu=\frac{m_{1} m_{2}}{m_{n}+m_{2}}$ Solution: Distance of COM from $m_{1}$ will be $r_{1}=\frac{m_{2} r}{\left(m_{1}+m_{2}\right)}$ Distan...
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Question: 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 $=6400 \mathrm{~km}$ and radius of the orbit of the earth about the sun $=1.5 \times 10^{8} \mathrm{~km}$. Solution: $L_{\text {spinning }}=\mid \omega$ $=\left(\frac{2}{5} M_{e} R_{e}^{2}\right)\left(\frac{2 \pi}{T_{e}}\right)$ $=\left(\frac{2}{5} \times M_{e} \times 6400\right)^{2}\left(\frac{2 \pi}{24 \times 60 \times...
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Question: A uniform square plate of mass $2.0 \mathrm{~kg}$ and edge $10 \mathrm{~cm}$ rotates about one of its diagonals under the action of a constant torque of $0.10 \mathrm{~N}-\mathrm{m}$. Calculate the angular momentum and kinetic energy of the plate at the end of the fifth second after the start. Solution: $\mathrm{T}=\mathrm{l \alpha}$ $0.1=\frac{\mathrm{Ma}^{2}}{12}(\alpha)$ $0.1=\frac{(2)(0.1)^{2}}{12}(\alpha)$ $\alpha=60 \mathrm{rad} / \mathrm{sec}$ $\omega_{0}=0, t=5 \mathrm{sec}$ $\...
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Question: A uniform rod of mass $300 \mathrm{~g}$ and length $50 \mathrm{~cm}$ rotates at a uniform angular speed of $2 \mathrm{rad} / \mathrm{s}$ about an axis perpendicular to the rod through an end. Calculate (a) the angular momentum of the rod about the axis of the rotation, (b) the speed of the centre of the rod and (c) its kinetic energy. Solution: (a) $L=l \omega$ $=\frac{m L^{2}}{3}(\omega)$ $=\frac{(0.3)(0.5)^{2}}{3}(2)$ $=\quad \frac{m^{2}}{s}$ $=0.05 \mathrm{~kg}-\frac{1}{3}$ (b) $\ma...
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Question: A uniform rod of length $L$ sets against a smooth roller as shown in figure. Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make with the horizontal is $\theta$. Solution: Translatory equilibrium $N_{1} \cos \theta+N_{2}=m g g_{-(i)}$ $N_{1} \sin \theta=f f$-(ii) Rotational Equilibrium about bottom $N_{1}\left(\frac{\hbar}{\sin \theta}\right)=m g\left(\frac{L}{2} \cos \theta\right)$ $f f=\mu N_{2}$-(iv) Solving (i),(ii),(iii) an...
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Question: The door of an almirah is $6 \mathrm{ft}$ high, $1.5 \mathrm{ft}$ wide and weighs $8 \mathrm{~kg}$. The door is supported by two hinges situated at a distance of $1 \mathrm{ft}$ from the ends. If the magnitudes of the force exerted by the hinges on the door are equal, find this magnitude. Solution: Magnitude of forces by hinges are equal. $\sqrt{N_{1}^{2}+N_{4}^{2}}=\sqrt{N_{2}^{2}+N_{3}^{2}}$ $N_{1}^{2}+N_{4}^{2}=N_{2}^{2}+N_{3}^{2}-(\mathrm{i})$ Rotational Equilibrium at point $O$ $N...
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Question: A $6.5 \mathrm{~m}$ long ladder rests against a vertical wall reaching a height of $6.0 \mathrm{~m}$. A $60 \mathrm{~kg}$ 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. Solution: By Pythagoras theorem, $(6.5)^{2}=(6)^{2}+B^{2}$ $B=2.5 \...
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Question: 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. Solution: Here,$\mu=0.54$ Translatory Equilibrium $N_{1}=m g+16 g{ }_{-(i)}$ $N_{2}=f f=\mu N_{1}-$ (ii) Rotational Equilibrium about point 'O' $N_{2}\left(10 \sin 53^{\circ}\right)=m g(8 \cos 53)+16 g \cos 53^{\circ}$-(iii) Solving (i), (ii) and (iii) $m=44 \mathrm{~kg}$...
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Question: A uniform ladder of length $10.0 \mathrm{~m}$ and mass $16.0 \mathrm{~kg}$ is resting against a vertical wall making an angle of $37^{\circ}$ with it. The vertical wall is frictionless but the ground is rough. An electrician weighing $60.0 \mathrm{~kg}$ climbs up the ladder. If he stays on the ladder at a point $8.00 \mathrm{~m}$ 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 fo...
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Question: A uniform metre stick of mass $200 \mathrm{~g}$ is suspended from the ceiling through two vertical strings of equal lengths fixed at the ends. A small object of mass $20 \mathrm{~g}$ is placed on the stick at a distance of 70 $\mathrm{cm}$ from the left end. Find the tensions in the two strings. Solution: Translatory Equilibrium Equation $T_{1}+T_{2}=(0.2 g)+(0.02 g)$ $T_{1}+T_{2}=0.22 g$ Rotational Equilibrium Equation at point $A$ clockwise Torque= anticlockwise Torque $(0.2 g)+(0.5)...
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Question: The pulley shown in figure has a radius $10 \mathrm{~cm}$ and moment of inertia $0.5 \mathrm{~kg}-\mathrm{m}^{2}$ about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the $4.0 \mathrm{~kg}$ block. Solution: Here, $I=0.5 \mathrm{~kg}_{-} \mathrm{m}^{2}$ $\mathrm{R}=0.1 \mathrm{~m}$ Translatory Motion Equation $4 g^{\sin 45-T_{2}=4 a}-(\mathrm{i})$ $T_{1}-2 g \sin 45=2 a$-(ii) Rotational Motion Equation $\tau=\mathrm{I} \alpha$ $T_{2} R-T_{1} R=I...
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