JEE Physics System Of Particles and Rotational Motion MCQs Set 26

Answer: (b) \( \frac{mv^3}{4\sqrt2\,g} \)

Explanation:
1. At the top, the velocity is horizontal: \( v_x=v\cos45^\circ=\frac v{\sqrt2} \), and the height is \( h=\frac{v^2\sin^245^\circ}{2g}=\frac{v^2}{4g} \).
2. \( L=mv_xh=m\cdot\frac v{\sqrt2}\cdot\frac{v^2}{4g}=\frac{mv^3}{4\sqrt2\,g} \).

Teacher's Note:
1. At the highest point the perpendicular distance of the velocity line from the point of projection is exactly h.
2. The horizontal range is not needed here.

Answer: (a) a straight line through the origin with positive slope

Explanation:
1. \( J=I\omega \) with I constant, so J is directly proportional to \( \omega \).
2. The graph is a straight line through the origin whose slope is I.

Teacher's Note:
1. Direct proportion always gives a straight line through the origin.
2. The slope of the J versus omega graph is the moment of inertia.

Answer: (c) first increase and then decrease

Explanation:
1. As the ant moves in towards the centre, the moment of inertia decreases, and when it moves out to the other edge the moment of inertia increases again.
2. Angular momentum is conserved, \( I\omega=\text{constant} \), so \( \omega \) first increases (I decreasing) and then decreases (I increasing) back to the original value.

Teacher's Note:
1. The moment of inertia is least when the ant is at the centre, so omega is largest there.
2. At the far edge the disc returns to its original angular velocity.

Answer: (d) decrease by a factor of 4

Explanation:
1. The string force is central, so \( L=mvr \) is conserved: \( v'=\frac v2 \) when \( r'=2r \).
2. \( K'=\frac12m\left(\frac v2\right)^2=\frac K4 \).

Teacher's Note:
1. Angular momentum, not kinetic energy, is conserved here.
2. The work done by the person holding the string changes the kinetic energy.

Answer: (d) Angular momentum

Explanation:
1. No external torque acts in free space, so angular momentum is conserved.
2. The moment of inertia \( \frac25MR^2 \) increases, so \( \omega=\frac LI \) decreases and \( K=\frac{L^2}{2I} \) decreases.

Teacher's Note:
1. Only angular momentum is guaranteed constant when no external torque acts.
2. Both omega and kinetic energy change when the moment of inertia changes.

Answer: (c) \( \frac23mL^2\pi^2n^2 \)

Explanation:
1. \( I=\frac{mL^2}{3} \) about the end and \( \omega=2\pi n \).
2. \( K=\frac12I\omega^2=\frac12\cdot\frac{mL^2}{3}\cdot4\pi^2n^2=\frac23mL^2\pi^2n^2 \).

Teacher's Note:
1. Use the moment of inertia about the end of the rod, not about the centre.
2. Angular velocity is 2 pi times the frequency.

Answer: (d) \( 1:5 \)

Explanation:
1. \( E_{sphere}=\frac12\cdot\frac25mR^2\omega^2=\frac15mR^2\omega^2 \).
2. \( E_{disc}=\frac12\cdot\frac12mR^2(2\omega)^2=mR^2\omega^2 \).
3. Ratio \( =1:5 \).

Teacher's Note:
1. Square the angular speed when comparing energies.
2. Moments of inertia: sphere 2/5 mR squared, disc 1/2 mR squared.

Answer: (a) \( \frac14I(\omega_1-\omega_2)^2 \)

Explanation:
1. Conservation of angular momentum: \( I\omega_1+I\omega_2=2I\omega\Rightarrow\omega=\frac{\omega_1+\omega_2}{2} \).
2. Loss \( =\frac12I\omega_1^2+\frac12I\omega_2^2-\frac12(2I)\omega^2=\frac I2(\omega_1^2+\omega_2^2)-\frac I4(\omega_1+\omega_2)^2 \).
3. \( =\frac I4(\omega_1-\omega_2)^2 \).

Teacher's Note:
1. The energy lost is converted into heat by friction between the faces.
2. If the angular velocities are equal, nothing is lost.

Answer: (a) \( W_C>W_B>W_A \)

Explanation:
1. Work needed equals the initial kinetic energy \( \frac12I\omega^2 \), so it is proportional to I.
2. \( I_A=\frac25MR^2,\ I_B=\frac12MR^2,\ I_C=MR^2 \).
3. Hence \( W_C>W_B>W_A \).

Teacher's Note:
1. The same speed and mass make the moment of inertia the deciding factor.
2. The ring has all its mass at the largest distance from the axis.

Answer: (a) \( 1:3 \)

Explanation:
1. \( K_{rot}=\frac12I\omega^2=\frac12\cdot\frac12MR^2\cdot\frac{v^2}{R^2}=\frac14Mv^2 \).
2. \( K_{total}=\frac12Mv^2+\frac14Mv^2=\frac34Mv^2 \).
3. Ratio \( =\frac{1/4}{3/4}=1:3 \).

Teacher's Note:
1. In pure rolling the translational and rotational parts are always in a fixed ratio for a given shape.
2. For a disc the ratio is 1 : 2 : 3 for rotational, translational and total.

Answer: (c) \( 5\sqrt2 \) m s\( ^{-1} \)

Explanation:
1. For rolling, the velocity of any point is the sum of the velocity of the centre (5 m/s forward) and the velocity due to rotation \( \omega R=5 \) m/s.
2. At the ends of the horizontal diameter, the rotational velocity is vertical (up at one end, down at the other), perpendicular to the forward 5 m/s.
3. Speed \( =\sqrt{5^2+5^2}=5\sqrt2 \) m/s.

Teacher's Note:
1. Add the translational and rotational velocities as vectors.
2. The top point moves at 2v and the contact point is at rest.

Answer: (d) \( \sqrt{14}:\sqrt{15} \)

Explanation:
1. Acceleration down the incline: \( a=\frac{g\sin\theta}{1+K^2/R^2} \): sphere \( \frac57g\sin\theta \), disc \( \frac23g\sin\theta \).
2. \( s=\frac12at^2\Rightarrow t\propto\frac1{\sqrt a} \).
3. \( \frac{t_s}{t_d}=\sqrt{\frac{a_d}{a_s}}=\sqrt{\frac{2/3}{5/7}}=\sqrt{\frac{14}{15}} \).

Teacher's Note:
1. The body with the larger acceleration takes less time.
2. The sphere reaches the bottom first, so its time is the smaller number.

Answer: (b) \( 2E \)

Explanation:
1. For a ring, \( K_{rot}=\frac12(MR^2)\frac{v^2}{R^2}=\frac12Mv^2=K_{trans}=E \).
2. Total \( =E+E=2E \).

Teacher's Note:
1. For a ring, rotational and translational energies are equal in pure rolling.
2. The result does not depend on mass or speed.

Answer: (a) ring

Explanation:
1. \( \frac{K_{rot}}{K_{total}}=\frac{K^2/R^2}{1+K^2/R^2}=\frac12 \Rightarrow\frac{K^2}{R^2}=1 \).
2. Radius of gyration equal to the radius (\( I=MR^2 \)) describes a ring.

Teacher's Note:
1. The fraction of rotational energy rises with the ratio K squared over R squared.
2. A ring has the highest ratio of any common shape.

Answer: (a) \( \sqrt{\frac{4gL\sin\theta}{3}} \)

Explanation:
1. Vertical drop \( h=L\sin\theta \).
2. \( mgh=\frac12mv^2+\frac14mv^2=\frac34mv^2 \).
3. \( v=\sqrt{\frac{4gL\sin\theta}{3}} \).

Teacher's Note:
1. Energy conservation gives the speed directly without finding the acceleration.
2. The factor 4/3 belongs to a disc; a sphere has 10/7.

Answer: (b) \( 1:\sqrt2 \)

Explanation:
1. Sliding (smooth): \( a_1=g\sin\theta \). Rolling ring: \( a_2=\frac{g\sin\theta}{1+1}=\frac{g\sin\theta}{2} \).
2. \( t=\sqrt{\frac{2L}{a}} \), so \( \frac{t_1}{t_2}=\sqrt{\frac{a_2}{a_1}}=\frac1{\sqrt2} \).

Teacher's Note:
1. Rolling is slower than sliding because part of the energy goes into rotation.
2. Use the time formula t = sqrt(2L/a) for both.

Answer: (b) \( g>a_s>a_d \)

Explanation:
1. \( a_s=\frac57g\sin\theta \) and \( a_d=\frac23g\sin\theta \).
2. \( \frac57>\frac23 \) and both are less than \( g\sin\theta \le g \).
3. So \( g>a_s>a_d \).

Teacher's Note:
1. Rolling bodies are always slower than a frictionless sliding block.
2. The body with the smaller moment of inertia per unit mass and radius squared accelerates more.

Answer: (b) \( \frac{3v^2}{4g} \)

Explanation:
1. Initial kinetic energy: \( \frac12Mv^2+\frac14Mv^2=\frac34Mv^2 \).
2. At the highest point everything is converted to potential energy: \( Mgh=\frac34Mv^2 \).
3. \( h=\frac{3v^2}{4g} \).

Teacher's Note:
1. Rolling without slipping conserves total mechanical energy.
2. The height is greater than v squared over 2g because of the rotational energy.

Answer: (d) 1.2 m

Explanation:
1. Total kinetic energy \( =\frac34Mv^2=\frac34\times12\times16=144 \) J.
2. \( \frac12kx^2=144\Rightarrow x^2=\frac{288}{200}=1.44 \).
3. \( x=1.2 \) m.

Teacher's Note:
1. Both translational and rotational kinetic energy go into the spring.
2. Ignoring the rotational part would give the wrong value 0.98 m.

Answer: (b) Sphere

Explanation:
1. Acceleration \( a=\frac{g\sin\theta}{1+K^2/R^2} \) does not depend on mass.
2. Sphere \( \frac{K^2}{R^2}=\frac25 \), disc \( \frac12 \), so the sphere has the larger acceleration (\( \frac57 \) against \( \frac23 \) of \( g\sin\theta \)).
3. The sphere reaches the bottom first.

Teacher's Note:
1. The smaller K squared over R squared, the faster the body rolls down.
2. Mass and radius cancel out completely.