JEE Physics System Of Particles and Rotational Motion MCQs Set 60

Answer: (c) \( \frac a3 \)

Explanation:
1. The cube does not slide since \( F \lt \mu mg=\frac{mg}{2} \), and \( N=mg \).
2. Torque about the centre of mass: \( F\cdot\frac a2+f\cdot\frac a2=N x \) with \( f=F \), so \( x=\frac{Fa}{N}=\frac a3 \).

Teacher's Note:
1. Balance the torques of friction/force and the offset normal force.

Answer: (d) A is false, but R is true.

Explanation:
1. The velocity of the centre of mass depends on the external force, not on the external torque, so A is false. For example, a pair of equal and opposite forces gives a torque but no force.
2. R is true: for an isolated system the linear momentum is conserved.

Teacher's Note:
1. Torque and force are independent: the centre of mass responds only to the net external force.

Answer: (c) \( 3:1 \)

Explanation:
1. No torque: \( I_1\omega_1=I_2\omega_2 \), so \( \frac{I_1}{I_2}=\frac{\omega_2}{\omega_1}=9 \).
2. \( I=mk^2 \), so \( \frac{k_1}{k_2}=\sqrt9=3 \).

Teacher's Note:
1. Radius of gyration varies as \( \sqrt I \).

Answer: (a) \( \frac{29\pi R^5\rho}{30} \)

Explanation:
1. Cylinder: \( \frac12(\rho\pi R^2\cdot3R)R^2=\frac32\pi\rho R^5 \).
2. Each hemisphere: mass \( \frac23\pi R^3\rho \) and \( I=\frac25mR^2=\frac{4}{15}\pi\rho R^5 \), so two give \( \frac8{15}\pi\rho R^5 \).
3. Remaining: \( \frac32\pi\rho R^5-\frac8{15}\pi\rho R^5=\frac{29}{30}\pi\rho R^5 \).

Teacher's Note:
1. Subtract the moment of inertia of the removed parts about the same axis.

Answer: (b) \( \frac{I\omega^2}{3} \)

Explanation:
1. Initial: \( \frac12I(2\omega)^2+\frac12(2I)\omega^2=3I\omega^2 \). Final: \( \omega'=\frac{4\omega}{3} \), \( K_f=\frac12(3I)\frac{16\omega^2}{9}=\frac{8I\omega^2}{3} \).
2. Loss \( =3I\omega^2-\frac83I\omega^2=\frac{I\omega^2}{3} \).

Teacher's Note:
1. Compute the kinetic energy before and after the common angular velocity is reached.

Answer: (d) \( \frac{mg}{\sqrt3} \)

Explanation:
1. It topples about the far base corner. The centre of mass is \( \frac a2 \) (horizontally) from that corner and the force acts at height \( \frac{\sqrt3a}{2} \).
2. \( F\cdot\frac{\sqrt3a}{2}=mg\cdot\frac a2 \), so \( F=\frac{mg}{\sqrt3} \).

Teacher's Note:
1. At the threshold of toppling the normal force acts at the pivot corner.

Answer: (c) the mass and the distribution of mass about the axis

Explanation:
1. \( I=\sum m_ir_i^2 \) depends on how much mass there is and how far it lies from the axis.
2. It does not depend on the angular velocity.

Teacher's Note:
1. Moment of inertia is the rotational analogue of mass.

Answer: (d) The velocities of any two parts of the wheel at equal distances from the centre are equal

Explanation:
1. Rolling is rotation about M, so \( v_P=\omega\cdot MP \) and M is at rest.
2. Points at equal distance from the centre have the same speed but different directions, so their velocities are not equal as vectors.

Teacher's Note:
1. Equal speed does not mean equal velocity.

Answer: (a) angular momentum changes in direction but not in magnitude

Explanation:
1. The angular momentum about the suspension point has a constant vertical component and a rotating horizontal component, so its magnitude is constant.
2. Its direction keeps changing, so it is not conserved as a vector (gravity and tension have a net torque).

Teacher's Note:
1. A rotating vector of constant length changes only in direction.

Answer: (d) \( I_{AD}=4I_{EF} \)

Explanation:
1. \( I_{EF}=\frac{Ma^2}{12} \) (axis through the centre in the plane).
2. Parallel axis theorem for AD, at distance \( \frac a2 \): \( I_{AD}=\frac{Ma^2}{12}+M\frac{a^2}{4}=\frac{Ma^2}{3}=4I_{EF} \).

Teacher's Note:
1. AD is parallel to EF at a distance of half the side.

Answer: (d) \( \frac{\sqrt{3g}}{2} \)

Explanation:
1. The free end hits the surface when the rod makes angle \( \theta \) below the horizontal with \( \sin\theta=\frac h\ell=\frac12 \).
2. Energy: \( \frac12\cdot\frac{m\ell^2}{3}\omega^2=mg\frac\ell2\sin\theta \) gives \( \omega^2=\frac{3g\sin\theta}{\ell}=\frac{3g}{4} \).
3. With \( e=1 \) the end rebounds with the same speed, so the angular speed is unchanged: \( \frac{\sqrt{3g}}{2} \).

Teacher's Note:
1. An elastic impact reverses the angular velocity of a rod hinged elsewhere.

Answer: (d) \( 2 \) m

Explanation:
1. \( \omega=\frac{10-5}{2}=2.5 \) rad/s.
2. The point of rest lies below B (since B is the slower point) at distance \( \frac{v_B}{\omega}=\frac{5}{2.5}=2 \) m.

Teacher's Note:
1. Use \( v=\omega r \) measured from the instantaneous axis.

Answer: (c) \( \frac12m\omega^2l \)

Explanation:
1. The centre of mass is at \( \frac l2 \) and moves in a circle of that radius.
2. \( F=m\omega^2\frac l2 \).

Teacher's Note:
1. The clamp provides the centripetal force for the centre of mass.

Answer: (a) \( \frac{3g}{2l}\sin\theta \)

Explanation:
1. Torque of the weight about the pivot: \( \tau=Mg\frac l2\sin\theta \).
2. \( \alpha=\frac{\tau}{I}=\frac{Mg\frac l2\sin\theta}{\frac{Ml^2}{3}}=\frac{3g}{2l}\sin\theta \).

Teacher's Note:
1. The lever arm of the weight is \( \frac l2\sin\theta \).

Answer: (c) \( \sqrt{\frac43} \)

Explanation:
1. Disc: \( K=\frac34mv_d^2 \). Ring: \( K=mv_r^2 \).
2. Equal energies give \( \frac{v_d^2}{v_r^2}=\frac43 \), so \( \frac{v_d}{v_r}=\sqrt{\frac43} \).

Teacher's Note:
1. Total rolling kinetic energy is \( \frac12mv^2\left(1+\frac{k^2}{R^2}\right) \).

Answer: (b) \( v \)

Explanation:
1. The foot is at \( x=L\cos\alpha \), so \( v=L\sin\alpha\,\dot\alpha \).
2. The centre moves on a circle of radius \( \frac L2 \) about the corner: speed \( =\frac L2\dot\alpha=\frac{v}{2\sin\alpha} \).
3. At \( \alpha=30^\circ \): speed \( =v \).

Teacher's Note:
1. The centre of a sliding ladder moves on a circle centred at the wall-floor corner.

Answer: (a) positive

Explanation:
1. Angular momentum is conserved, so \( \omega \) increases as \( I \) decreases.
2. The kinetic energy \( \frac{L^2}{2I} \) increases, so he does positive work (by his muscles) on the system.

Teacher's Note:
1. Kinetic energy rises when \( I \) falls at constant \( L \).

Answer: (b) \( \sqrt{\frac32} \)

Explanation:
1. \( I=m\left(\frac{R^2}{4}+\frac{l^2}{12}\right) \) with \( \pi R^2l=V \) fixed, so \( R^2=\frac{V}{\pi l} \).
2. \( I\propto\frac{V}{4\pi l}+\frac{l^2}{12} \) and \( \frac{dI}{dl}=0 \) gives \( \frac{V}{4\pi l^2}=\frac{l}{6} \), so \( l^3=\frac{3V}{2\pi} \).
3. Then \( \frac{l^2}{R^2}=\frac{\pi l^3}{V}=\frac32 \), so \( \frac lR=\sqrt{\frac32} \).

Teacher's Note:
1. Eliminate \( R \) using the fixed volume and set the derivative to zero.

Answer: (c) his moment of inertia decreases

Explanation:
1. Bringing the mass closer to the axis reduces \( I \).
2. Angular momentum is conserved, so \( \omega \) increases and the kinetic energy \( \frac{L^2}{2I} \) increases.

Teacher's Note:
1. Angular momentum is conserved; \( I \) and \( \omega \) vary inversely.

Answer: (d) disc

Explanation:
1. \( \frac12mv^2\left(1+\frac{k^2}{R^2}\right)=mgh=\frac34mv^2 \).
2. \( 1+\frac{k^2}{R^2}=\frac32 \), so \( \frac{k^2}{R^2}=\frac12 \), which is a disc.

Teacher's Note:
1. Each rolling shape has its own \( \frac{k^2}{R^2} \) value: ring \( 1 \), disc \( \frac12 \), solid sphere \( \frac25 \), hollow sphere \( \frac23 \).

Answer: (b) \( \frac{3mv}{(M+3m)L} \)

Explanation:
1. Angular momentum about O is conserved: \( mvL=\left(\frac{ML^2}{3}+mL^2\right)\omega \).
2. \( \omega=\frac{3mv}{(M+3m)L} \).

Teacher's Note:
1. The impulse at the hinge has no torque about the hinge.

Answer: (a) \( \frac{ML^2}{24} \)

Explanation:
1. For an isosceles triangle about its symmetry axis, \( I=\frac{Mb^2}{24} \) with base \( b=L \).
2. \( I=\frac{ML^2}{24} \).

Teacher's Note:
1. Each half of the triangle contributes equally.

Answer: (b) \( \frac{\omega M}{M+2m} \)

Explanation:
1. No external torque: \( MR^2\omega=(M+2m)R^2\omega' \).
2. \( \omega'=\frac{M\omega}{M+2m} \).

Teacher's Note:
1. The two masses at the rim add \( 2mR^2 \) to the moment of inertia.

Answer: (b) \( \frac14 \) kg

Explanation:
1. \( I=\frac\tau\alpha=1 \) kg m\( ^2 \).
2. \( I=mk^2 \), so \( m=\frac{1}{4} \) kg.

Teacher's Note:
1. Use \( I=mk^2 \) with \( I \) found from \( \tau=I\alpha \).

Answer: (b) \( 2 \) m/s

Explanation:
1. \( \omega=\frac{v_{cm}}{R}=\frac{3}{0.3}=10 \) rad/s about the contact point.
2. The lower end of the spool is at distance \( R-r=0.2 \) m from the contact point, so \( v=\omega(R-r)=10\times0.2=2 \) m/s.

Teacher's Note:
1. Velocity of a point on a rolling body is \( \omega \) times its distance from the contact point.

Answer: (a) \( \frac12C\theta^2 \)

Explanation:
1. \( W=\int_0^\theta\tau\,d\theta=\int_0^\theta C\theta\,d\theta=\frac12C\theta^2 \).

Teacher's Note:
1. Integrate torque over angle.