JEE Physics System Of Particles and Rotational Motion MCQs Set 14

Answer: (a) converts translational energy to rotational energy

Explanation:
1. In rolling without slipping the contact point is at rest, so static friction does no work and produces no heat.
2. Gravity reduces potential energy, and friction supplies the torque that turns part of the translational energy into rotational energy.
3. Hence friction only transfers energy from translation to rotation.

Teacher's Note:
1. Static friction in pure rolling does no work and dissipates nothing.
2. It acts as the link between linear and rotational motion.

Answer: (b) \(\frac{J}{2m}\)

Explanation:
1. Just after the impulse, \(v_0 = \frac{J}{m}\) and \(\omega = 0\).
2. Friction acts at the contact point, so angular momentum about the contact point is conserved: \(m v_0 R = m v R + (mR^2)\frac{v}{R}\).
3. This gives \(m v_0 R = 2 m v R\), so \(v = \frac{v_0}{2} = \frac{J}{2m}\).

Teacher's Note:
1. When friction is the only horizontal force, take angular momentum about the contact point.
2. Pure rolling means \(v = R\omega\).

Answer: (a) Centre of the circle

Explanation:
1. Angular momentum about a point is conserved only if the torque about it is zero.
2. The only force is centripetal, directed towards the centre, so its torque about the centre is zero.
3. About any other point the force has a non-zero lever arm, so \(\vec L\) changes in direction.

Teacher's Note:
1. Torque is \(\vec r \times \vec F\), and it vanishes when \(\vec F\) passes through the chosen point.
2. A central force gives conserved angular momentum about its centre.

Answer: (a) \(6\) hours

Explanation:
1. There is no external torque, so \(I\omega\) is conserved.
2. \(I = \frac{2}{5}MR^2\) becomes \(\frac{1}{4}\) of its value when \(R\) is halved, so \(\omega\) becomes four times.
3. The period falls to \(\frac{24}{4} = 6\) hours.

Teacher's Note:
1. \(I\omega = \text{constant}\) means \(I_1 T_2 = I_2 T_1\).
2. Here \(T \propto I \propto R^2\).

Answer: (c) \(I_A < I_B\)

Explanation:
1. Mass \(= \rho\pi R^2 t\), so \(I = \frac{1}{2}MR^2 = \frac{1}{2}\rho\pi t R^4\).
2. \(I_A \propto t r^4\) and \(I_B \propto \frac{t}{4}(4r)^4 = 64\,t r^4\).
3. Therefore \(I_B = 64\,I_A\), so \(I_A < I_B\).

Teacher's Note:
1. When the thickness and radius both change, write \(I\) in terms of \(\rho\), \(t\) and \(R\).
2. The fourth power of the radius dominates.

Answer: (a) the linear velocity of objects at the equator is greater than at other places

Explanation:
1. All points on the earth have the same angular velocity \(\omega = \frac{2\pi}{24\ \text{h}}\), so (b) and (c) are false.
2. Linear speed \(v = \omega R\cos\lambda\) is largest at the equator (\(\lambda = 0\)) and zero at the poles.
3. Linear speed differs from place to place, so (d) is false.

Teacher's Note:
1. Angular velocity is common to every point of a rotating rigid body.
2. Linear speed depends on the distance from the axis.

Answer: (a) \(\frac{3v}{4a}\)

Explanation:
1. The impulse at \(O\) has no torque about \(O\), so angular momentum about \(O\) is conserved.
2. Before: \(L = M v\frac{a}{2}\) (centre of mass is at height \(\frac{a}{2}\)).
3. After: \(I_O = \frac{Ma^2}{6} + M\left(\frac{a}{\sqrt{2}}\right)^2 = \frac{2}{3}Ma^2\). So \(\omega = \frac{Mva/2}{\frac{2}{3}Ma^2} = \frac{3v}{4a}\).

Teacher's Note:
1. Collision with a fixed ridge: conserve angular momentum about the ridge.
2. Use the parallel axis theorem with distance \(\frac{a}{\sqrt{2}}\) from the edge to the centre.

Answer: (a) \(\frac{8}{25}\hat k\ \text{rad s}^{-1}\)

Explanation:
1. \(\vec r = 3\hat i + 4\hat j\) and \(\vec v = -2\hat i\), so \(\vec r\times\vec v = 4\hat j\times(-2\hat i) = 8\hat k\).
2. \(\vec\omega = \frac{\vec r\times\vec v}{r^2} = \frac{8\hat k}{25}\).
3. The position vector turns anticlockwise, so \(\omega = \frac{8}{25}\hat k\ \text{rad s}^{-1}\).

Teacher's Note:
1. Angular velocity of a moving particle about a point is \(\frac{\vec r\times\vec v}{r^2}\).
2. Only the component of velocity perpendicular to \(\vec r\) matters.

Answer: (a) \(V_Q > V_C > V_P\)

Explanation:
1. In pure rolling every point moves instantaneously about the contact point with the same \(\omega\), so speed is \(\omega\) times its distance from the contact point.
2. \(Q\) is farther from the contact point than \(C\), and \(P\) is nearer.
3. Hence \(V_Q > V_C > V_P\).

Teacher's Note:
1. In rolling, the contact point is the instantaneous axis.
2. Higher points move faster, the top point moves at \(2V_C\).

Answer: (b) angular momentum of the sphere about the point of contact with the plane is conserved

Explanation:
1. Kinetic friction acts at the contact point and dissipates energy, so total KE is not conserved.
2. Friction changes both the linear and the angular speed, so rotational KE and angular momentum about the centre of mass change.
3. About a point on the surface, friction has zero torque, and gravity and normal reaction cancel in torque, so angular momentum about the contact point stays constant.

Teacher's Note:
1. When friction is the only unbalanced horizontal force, choose the contact point as the reference.
2. Sliding friction always dissipates mechanical energy.

Answer: (c) \(r\)

Explanation:
1. The coin just slips when \(\mu m g = m\omega^2 x\), so \(x \propto \frac{1}{\omega^2}\).
2. Doubling \(\omega\) makes \(x\) one quarter of the old value.
3. New distance \(= \frac{4r}{4} = r\).

Teacher's Note:
1. At the point of slipping, maximum friction supplies the centripetal force.
2. The slipping distance varies inversely with the square of the angular speed.

Answer: (c) \(\frac{2T}{m}\)

Explanation:
1. The string pulls the ring at \(P\) along the tangent with force \(T\). Acceleration of the centre: \(a_{cm} = \frac{T}{m}\) in the direction of the pull.
2. Torque about the centre \(= TR\), \(I = mR^2\), so \(\alpha = \frac{TR}{mR^2} = \frac{T}{mR}\). The tangential acceleration of \(P\) due to rotation is \(\alpha R = \frac{T}{m}\), in the same direction as the pull.
3. The ring is not yet rotating, so there is no centripetal term. \(a_P = \frac{T}{m} + \frac{T}{m} = \frac{2T}{m}\).

Teacher's Note:
1. For a free body add the acceleration of the centre of mass and the rotational acceleration of the point.
2. Check the direction of the torque before adding the two contributions.

Answer: (d) \(16\ \text{s}\)

Explanation:
1. \(100\ \text{rpm} = \frac{100\times 2\pi}{60} = 10.47\ \text{rad s}^{-1}\).
2. The belt does not slip, so \(\omega_A r_A = \omega_C r_C\), giving \(\omega_A = 10.47\times\frac{25}{10} = 26.2\ \text{rad s}^{-1}\).
3. \(t = \frac{\omega_A}{\alpha_A} = \frac{26.2}{1.6} \approx 16.4\ \text{s}\), so about \(16\ \text{s}\).

Teacher's Note:
1. Wheels joined by a belt have equal rim speeds, not equal angular speeds.
2. Convert rpm to rad per second first.

Answer: (d) \(\frac{1}{6}\frac{l^2\omega^2}{g}\)

Explanation:
1. The angular speed is maximum at the lowest position, where the kinetic energy is \(\frac{1}{2}I\omega^2\) with \(I = \frac{ml^2}{3}\).
2. At the highest point all this energy becomes potential energy \(mgh\).
3. \(mgh = \frac{1}{2}\cdot\frac{ml^2}{3}\omega^2\), so \(h = \frac{l^2\omega^2}{6g}\).

Teacher's Note:
1. Use energy conservation for a swinging rigid body with \(I\) about the pivot.
2. The rise is of the centre of mass.

Answer: (a) \(\frac{v\cos^2\theta}{h}\)

Explanation:
1. The rod touches the block's top edge, which is at height \(h\) and horizontal distance \(x = h\tan\theta\) from the vertical through \(O\).
2. Differentiate: \(v = \frac{dx}{dt} = h\sec^2\theta\,\frac{d\theta}{dt}\).
3. So \(\omega = \frac{d\theta}{dt} = \frac{v\cos^2\theta}{h}\).

Teacher's Note:
1. Relate the geometry with an angle, then differentiate with respect to time.
2. Check the limit \(\theta = 0\): the rod is vertical and \(\omega = \frac{v}{h}\), which is sensible.

Answer: (c) \(\frac{2F}{3mR}\)

Explanation:
1. Let friction \(f\) act backwards: \(F - f = ma\) and torque about the centre \(fR = \frac{1}{2}mR^2\alpha\).
2. Rolling gives \(a = R\alpha\), so \(f = \frac{1}{2}ma\).
3. Then \(F = \frac{3}{2}ma\), so \(a = \frac{2F}{3m}\) and \(\alpha = \frac{a}{R} = \frac{2F}{3mR}\).

Teacher's Note:
1. Write the translational and rotational equations together with the rolling condition.
2. Friction here supplies the torque, so it points backwards.

Answer: (a) unchanged

Explanation:
1. The piece leaves tangentially with its speed \(\omega R\), so it carries angular momentum \(m R^2\omega\) about the axis.
2. The remaining disc has moment of inertia \(I - mR^2\) and angular momentum \(I\omega - mR^2\omega\), so \(\omega' = \frac{(I - mR^2)\omega}{I - mR^2} = \omega\).
3. The angular speed just after detaching is unchanged.

Teacher's Note:
1. A part that leaves tangentially takes its share of angular momentum with it.
2. The ratio of angular momentum to moment of inertia of the remainder is unchanged.

Answer: (a) \(6\)

Explanation:
1. Linear momentum: \(mV_0 = Mv_{cm}\). Angular momentum about \(O\): \(mV_0 x = \frac{ML^2}{12}\omega\).
2. The end \(A\) is at rest if \(v_{cm} = \omega\frac{L}{2}\) (the rotation moves \(A\) opposite to the translation).
3. So \(\frac{mV_0}{M} = \frac{12 m V_0 x}{ML^2}\cdot\frac{L}{2}\), which gives \(1 = \frac{6x}{L}\), so \(x = \frac{L}{6}\) and \(n = 6\).

Teacher's Note:
1. Use both conservation of linear momentum and of angular momentum about the centre.
2. The end opposite to the striking point moves backwards due to rotation, which is why it can be stationary.

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

Explanation:
1. \(I = \sum m_i r_i^2\) depends on each mass and its perpendicular distance from the axis.
2. It does not depend on the angular velocity.
3. So both the mass and the way it is distributed about the axis matter.

Teacher's Note:
1. Moment of inertia is the rotational analogue of mass.
2. It changes when the axis changes.

Answer: (b) \(\frac{3F}{7}\)

Explanation:
1. Let friction \(f\) act forward. Then \(F + f = Ma\), and about the centre \((F - f)R = \frac{2}{5}MR^2\cdot\frac{a}{R}\), so \(F - f = \frac{2}{5}Ma\).
2. Adding: \(2F = \frac{7}{5}Ma\), so \(a = \frac{10F}{7M}\).
3. \(f = Ma - F = \frac{10F}{7} - F = \frac{3F}{7}\) (forward).

Teacher's Note:
1. When the force acts above the centre, friction can point forward.
2. Write both equations and solve for \(f\) by eliminating \(a\).

Answer: (c) \(I_{CC'} + 3Md^2\)

Explanation:
1. By the parallel axis theorem, \(I_{CC'} = I_O + Md^2\) and \(I_{AA'} = I_O + M(2d)^2 = I_O + 4Md^2\).
2. Subtract: \(I_{AA'} - I_{CC'} = 3Md^2\).
3. So \(I_{AA'} = I_{CC'} + 3Md^2\).

Teacher's Note:
1. The parallel axis theorem applies only with the axis through the centre of mass as the reference.
2. For two non-central axes, go through \(I_O\).

Answer: (b) \(\frac{2}{5}\tan\theta\)

Explanation:
1. For a shell, \(\frac{I}{MR^2} = \frac{2}{3}\), so \(a = \frac{g\sin\theta}{1 + 2/3} = \frac{3}{5}g\sin\theta\).
2. Friction needed: \(f = \frac{I a}{R^2} = \frac{2}{3}M\cdot\frac{3}{5}g\sin\theta = \frac{2}{5}Mg\sin\theta\).
3. Since \(f \le \mu Mg\cos\theta\), \(\mu_{min} = \frac{2}{5}\tan\theta\).

Teacher's Note:
1. Rolling needs enough static friction, \(f \le \mu N\).
2. The general result is \(\mu_{min} = \frac{\tan\theta}{1 + MR^2/I}\).

Answer: (d) \(L = \sqrt{3}R\)

Explanation:
1. About the axis, \(I = \frac{1}{2}MR^2\). About the equatorial axis, \(I = M\left(\frac{R^2}{4} + \frac{L^2}{12}\right)\).
2. Equate: \(\frac{R^2}{2} = \frac{R^2}{4} + \frac{L^2}{12}\), so \(\frac{L^2}{12} = \frac{R^2}{4}\).
3. \(L^2 = 3R^2\), so \(L = \sqrt{3}R\).

Teacher's Note:
1. Know both cylinder results: \(\frac{1}{2}MR^2\) and \(M\left(\frac{R^2}{4} + \frac{L^2}{12}\right)\).
2. Equating two expressions with the same \(M\) removes the mass.

Answer: (a) \(8\ \text{rad s}^{-2}\)

Explanation:
1. From rest, \(\theta = \frac{1}{2}\alpha t^2\).
2. \(400 = \frac{1}{2}\alpha(10)^2 = 50\alpha\).
3. \(\alpha = 8\ \text{rad s}^{-2}\).

Teacher's Note:
1. A constant torque on a fixed \(I\) gives constant \(\alpha\).
2. Use the rotational kinematic equations just as for linear motion.

Answer: (c) \(v_1 < v_2\)

Explanation:
1. Angular momentum about the contact point is conserved: \(I\omega_0 = (I + mR^2)\frac{v}{R}\), so \(v = \frac{I\omega_0 R}{I + mR^2}\).
2. Solid: \(I = \frac{2}{5}mR^2\) gives \(v_1 = \frac{2}{7}\omega_0 R\). Hollow: \(I = \frac{2}{3}mR^2\) gives \(v_2 = \frac{2}{5}\omega_0 R\).
3. \(\frac{2}{7} < \frac{2}{5}\), so \(v_1 < v_2\).

Teacher's Note:
1. The body with larger \(\frac{I}{mR^2}\) keeps more of its spin and reaches a higher rolling speed.
2. Use angular momentum about the contact point.