JEE Physics System Of Particles and Rotational Motion MCQs Set 49

Answer: (c) initially up the incline and then down the incline, periodically

Explanation:
1. Let a be the acceleration of the centre down the slope and f the friction up the slope. Then \( Ma=Mg\sin\theta-kx-f \) and, for rolling, \( fR=\tfrac12MR^2\cdot\frac aR\Rightarrow f=\tfrac12Ma \).
2. Eliminating f gives \( f=\tfrac13(Mg\sin\theta-kx) \), where x is the spring extension.
3. As the cylinder oscillates, the quantity \( Mg\sin\theta-kx \) changes sign, so friction acts up the slope for part of the motion and down the slope for the rest, periodically.

Teacher's Note:
1. Friction on a rolling body is not fixed in direction; it is set by the sign of the net non-frictional force along the slope.
2. Always find f from the two equations (force and torque) together.

Answer: (c) \( \frac{1}{\sqrt3} \)

Explanation:
1. Toppling occurs about the lower base corner on the side away from the force. The apex height is \( h=\frac{\sqrt3}{2}a \) and the weight acts a/2 from that corner.
2. Just before toppling: \( F\,h=mg\,\frac a2\Rightarrow F=\frac{mg}{\sqrt3} \).
3. Before toppling there is no sliding, so friction equals F and must not exceed \( \mu mg \). Hence \( \mu\ge\frac1{\sqrt3} \).

Teacher's Note:
1. Take torques about the pivot corner, not about the centre of mass.
2. The height of an equilateral triangle of side a is \( \frac{\sqrt3}{2}a \).

Answer: (c) \( \sqrt2 \)

Explanation:
1. Each bob moves on a circle with the same starting condition (horizontal) and the same g, so the motion differs only through the length.
2. Energy: \( \tfrac12mv^2=mg\ell\Rightarrow v=\sqrt{2g\ell} \). Every part of the path scales with the length and the speed scales with \( \sqrt{\ell} \), so the time scales as \( \sqrt{\ell} \).
3. \( \frac{t_A}{t_B}=\sqrt{\frac{\ell}{\ell/2}}=\sqrt2 \)

Teacher's Note:
1. For a pendulum released from any fixed angle, the time to reach the bottom is proportional to the square root of the length.
2. This works even for large swings.

Answer: (c) \( 0.025\ \text{kg m}^2/\text{s} \)

Explanation:
1. \( \omega=\frac vR=\frac{0.50}{0.10}=5\ \text{rad/s} \).
2. \( I_{cm}=mR^2=0.50\times0.01=0.005\ \text{kg m}^2 \).
3. \( L_{spin}=I\omega=0.005\times5=0.025\ \text{kg m}^2/\text{s} \)

Teacher's Note:
1. Spin angular momentum uses only the rotation about the centre; the orbital part mvr is separate.
2. Rolling without slipping links v and ω by v = ωR.

Answer: (c) \( \frac{\tan\theta}{3} \)

Explanation:
1. For a solid cylinder, \( a=\tfrac23g\sin\theta \) and the friction required is \( f=Mg\sin\theta-Ma=\tfrac13Mg\sin\theta \).
2. Normal reaction: \( N=Mg\cos\theta \).
3. \( \mu\ge\frac fN=\frac{\tan\theta}{3} \)

Teacher's Note:
1. Friction needed for rolling is the static value, so compare it with μN.
2. For a solid sphere the corresponding value is 2 tanθ / 7.

Answer: (c) \( \frac34\sqrt{gl} \)

Explanation:
1. Moment of inertia about the hinge: \( I=\tfrac13ml^2+ml^2=\tfrac43ml^2 \).
2. Loss of potential energy: rod centre falls l/2 and the point mass falls l, so \( mg\frac l2+mgl=\tfrac32mgl \).
3. Energy: \( \tfrac12\cdot\tfrac43ml^2\omega^2=\tfrac32mgl\Rightarrow\omega^2=\frac{9g}{4l}\Rightarrow\omega=\frac32\sqrt{\frac gl} \).
4. Speed of the rod's centre: \( v=\omega\frac l2=\frac34\sqrt{gl} \)

Teacher's Note:
1. Use the hinge as the axis for I, then conserve mechanical energy.
2. The centre of the rod is at half the radius of its free end, so it moves at half the end speed.

Answer: (c) \( 2v\cos\left(\frac{vt}{2r}\right) \)

Explanation:
1. In time t the hoop turns through \( \varphi=\omega t=\frac{vt}{r} \).
2. Velocity of the marked point = velocity of centre (horizontal, v) + velocity due to rotation (magnitude v, rotated by φ from the horizontal).
3. Magnitude of the sum of two equal vectors with angle φ between them: \( 2v\cos\frac\varphi2=2v\cos\frac{vt}{2r} \).

Teacher's Note:
1. At t = 0 the two contributions point the same way, giving the top speed 2v, which matches the formula.
2. Half the angle appears because the resultant of two equal vectors bisects them.

Answer: (a) \( r=\frac{mg-\mu Mg}{M\omega^2} \)

Explanation:
1. The cord tension equals mg and pulls the block inward. The block needs the net inward force \( M\omega^2r \).
2. At the smallest r the required inward force is least, so friction must act outward at its limit: \( mg-\mu Mg=M\omega^2r \).
3. \( r_{min}=\frac{mg-\mu Mg}{M\omega^2} \)

Teacher's Note:
1. Decide the direction of friction first: smaller r needs less inward force than the tension supplies, so friction points outward.
2. At the limit use the maximum static friction μMg.

Answer: (c) \( 6I \)

Explanation:
1. About a diameter \( I=\tfrac14MR^2\Rightarrow MR^2=4I \).
2. About the perpendicular axis through the centre \( I_{cm}=\tfrac12MR^2=2I \).
3. Parallel-axis theorem for an axis through the rim: \( I_{rim}=2I+MR^2=2I+4I=6I \)

Teacher's Note:
1. Perpendicular-axis theorem gives I_z = 2 I_diameter for a disc.
2. Add MR² when moving the axis to the rim.

Answer: (c) \( v^2=\frac{mgh}{\frac m2+\frac{I}{2r^2}+\frac M3} \)

Explanation:
1. Cord speed equals v everywhere, so the shell turns with \( \omega_s=\frac vR \) and the pulley with \( \omega_p=\frac vr \).
2. Shell: \( I_s=\tfrac23MR^2\Rightarrow KE=\tfrac12\cdot\tfrac23MR^2\frac{v^2}{R^2}=\tfrac13Mv^2 \). Pulley: \( \tfrac12I\frac{v^2}{r^2} \). Object: \( \tfrac12mv^2 \).
3. Energy: \( mgh=\tfrac12mv^2+\frac{Iv^2}{2r^2}+\tfrac13Mv^2 \).
4. \( v^2=\frac{mgh}{\frac m2+\frac{I}{2r^2}+\frac M3} \)

Teacher's Note:
1. Convert every rotational energy to a multiple of v² using the no-slip condition.
2. A thin spherical shell has I = 2MR²/3.

Answer: (c) \( 2b\sqrt{\frac ab} \)

Explanation:
1. Torque \( \vec\tau=\frac{d\vec J}{dt}=2t\,\vec b \), so it points along \( \vec b \) with magnitude 2bt.
2. \( \vec J \) has components a (perpendicular to b) and \( bt^2 \) along b. The angle between τ and J is 45°, so \( \tan45^\circ=\frac{a}{bt^2}=1\Rightarrow t=\sqrt{\frac ab} \).
3. \( \tau=2bt=2b\sqrt{\frac ab} \)

Teacher's Note:
1. Torque is the time derivative of angular momentum.
2. Use the components of J along and across b to apply the 45° condition.

Answer: (d) dependent on the length of the rod, on θ and on the speed of end A at that instant

Explanation:
1. Height of A: \( y=L\cos\theta \). Differentiating, \( v_A=-L\sin\theta\,\dot\theta \).
2. So \( \omega=\dot\theta=\frac{v_A}{L\sin\theta} \) in magnitude.
3. It depends on L (inversely), on θ and on v_A.

Teacher's Note:
1. Write the constraint equation for the end points and differentiate it.
2. The rod's angular velocity is not simply proportional to its length.

Answer: (d) \( \frac{13}{24}MR^2 \)

Explanation:
1. Areas are in the ratio 4 : 1 (full plate : hole), so the full plate has mass \( \tfrac43M \) and the hole has mass \( \tfrac M3 \).
2. Full plate: \( I=\tfrac12\cdot\tfrac43MR^2=\tfrac23MR^2 \).
3. Hole about O (parallel-axis): \( \tfrac12\cdot\tfrac M3\left(\tfrac R2\right)^2+\tfrac M3\left(\tfrac R2\right)^2=\tfrac{MR^2}{24}+\tfrac{MR^2}{12}=\tfrac{MR^2}{8} \).
4. \( I=\tfrac23MR^2-\tfrac18MR^2=\tfrac{13}{24}MR^2 \)

Teacher's Note:
1. Treat the hole as a body with negative mass and subtract its moment of inertia.
2. Use the parallel-axis theorem for the hole's contribution.

Answer: (b) greater in process A than in B

Explanation:
1. Process A: \( \alpha_A=\frac{FR}{I} \), so \( a_A=\frac{FR^2}{I} \) with F = 5 N.
2. Process B: the hanging mass m = F/g has tension T less than mg. \( mg-T=ma,\ TR=I\alpha,\ a=\alpha R\Rightarrow a_B=\frac{mgR^2}{I+mR^2} \).
3. Since \( \frac{mgR^2}{I+mR^2}<\frac{mgR^2}{I} \), we get \( a_A>a_B \).

Teacher's Note:
1. A hanging mass also has to be accelerated, which reduces the tension below its weight.
2. A pulling force that equals the weight gives the larger angular acceleration.

Answer: (c) \( \frac{F}{2M}t^2,\ \frac{F}{MR}t^2 \)

Explanation:
1. Linear: \( a=\frac FM\Rightarrow x=\tfrac12at^2=\frac{F}{2M}t^2 \).
2. Angular: torque \( FR \), \( I=\tfrac12MR^2\Rightarrow\alpha=\frac{2F}{MR} \).
3. \( \theta=\tfrac12\alpha t^2=\frac{F}{MR}t^2 \)

Teacher's Note:
1. The surface is frictionless, so the disc slides and spins independently.
2. Linear and rotational motions are solved separately.

Answer: (b) 1.9 J

Explanation:
1. Speed gained: \( v=\frac Jm=\frac{7.5}{15}=0.5\ \text{m/s} \).
2. The impulse acts at the centre, so there is no rotation.
3. \( KE=\frac{J^2}{2m}=\frac{56.25}{30}=1.875\approx1.9\ \text{J} \)

Teacher's Note:
1. An impulse through the centre of mass produces no rotation.
2. KE = J²/2m is a handy shortcut.

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

Explanation:
1. Torque about the hinge: \( \tau=mg\frac L2 \).
2. Moment of inertia about the end: \( I=\tfrac13mL^2 \).
3. \( \alpha=\frac\tau I=\frac{mgL/2}{mL^2/3}=\frac{3g}{2L} \)

Teacher's Note:
1. Use the hinge as the axis.
2. The same result gives the initial acceleration of the free end as 3g/2.

Answer: (a) \( \frac vL \)

Explanation:
1. Linear momentum: \( 2Mv_{cm}=Mv\Rightarrow v_{cm}=\frac v2 \).
2. About the centre of mass the impulse has moment arm L/2: \( J\frac L2=I_{cm}\omega \) with \( I_{cm}=2M\left(\frac L2\right)^2=\frac{ML^2}{2} \).
3. \( \omega=\frac{Mv\,L/2}{ML^2/2}=\frac vL \)

Teacher's Note:
1. Take angular momentum about the centre of mass of the pair, which is the midpoint.
2. Check by seeing that the struck ball moves at v/2 + (v/L)(L/2) = v.

Answer: (d) \( 4.8\,Ma^2 \)

Explanation:
1. By perpendicular-axis theorem \( I_x+I_y=I_z=1.6Ma^2 \); by symmetry \( I_x=I_y=0.8Ma^2 \) about a line through the centre in the plane.
2. The tangent AB is parallel to such a line at a distance a (half the side) plus a (radius) = 2a.
3. Parallel-axis theorem: \( I_{AB}=0.8Ma^2+M(2a)^2=4.8Ma^2 \)

Teacher's Note:
1. The four-fold symmetry makes the two in-plane moments equal.
2. Distance from the centre to the tangent is the sum of half the side and the semicircle radius.

Answer: (b) \( 4\,mR^2 \)

Explanation:
1. Area of the hole is 1/9 of the disc, so the hole has mass m.
2. Full disc: \( \tfrac12(9m)R^2=\tfrac92mR^2 \).
3. Hole about O: \( \tfrac12m\left(\frac R3\right)^2+m\left(\frac{2R}{3}\right)^2=\tfrac1{18}mR^2+\tfrac49mR^2=\tfrac12mR^2 \).
4. \( I=\tfrac92mR^2-\tfrac12mR^2=4mR^2 \)

Teacher's Note:
1. Mass is proportional to area for a uniform disc.
2. Subtract the hole's moment of inertia taken about the original axis.

Answer: (d) 64

Explanation:
1. Mass is proportional to volume: \( m_y=\rho\pi(4R)^2\frac t4=4\rho\pi R^2t=4m_x \).
2. Moment of inertia: \( I=\tfrac12mr^2 \).
3. \( \frac{I_y}{I_x}=\frac{4m_x(4R)^2}{m_xR^2}=64 \)

Teacher's Note:
1. Both the mass and the radius factor change, so multiply the effects.
2. I scales as thickness times radius to the fourth power.

Answer: (d) \( \frac{3\rho L^3}{8\pi^2} \)

Explanation:
1. Radius: \( 2\pi R=L\Rightarrow R=\frac L{2\pi} \); mass \( m=\rho L \).
2. About a diameter \( I=\tfrac12mR^2 \); the tangent is parallel at distance R.
3. Parallel-axis theorem: \( I=\tfrac12mR^2+mR^2=\tfrac32mR^2=\frac32\rho L\cdot\frac{L^2}{4\pi^2}=\frac{3\rho L^3}{8\pi^2} \)

Teacher's Note:
1. A ring has I = ½mR² about any diameter.
2. Express R through L first, then substitute.

Answer: (c) \( \frac{Mg}{2} \)

Explanation:
1. The block tips about the bottom edge on the far side from the push.
2. At the point of toppling the normal reaction acts at that edge, so torques about it balance: \( F\cdot L=Mg\cdot\frac L2 \).
3. \( F=\frac{Mg}{2} \)

Teacher's Note:
1. Choose the pivot edge so that the normal and friction forces have no torque.
2. Height of the force is L, the weight's arm is L/2.

Answer: (b) \( \frac m2\left[r^2-\frac{lb}{6\pi r^2}(l^2+b^2)\right] \)

Explanation:
1. Mass of the cut piece: \( m'=m\frac{lb}{\pi r^2} \).
2. Its moment of inertia about the central perpendicular axis: \( \frac{m'}{12}(l^2+b^2) \).
3. Remaining: \( I=\tfrac12mr^2-\frac{m\,lb}{12\pi r^2}(l^2+b^2)=\frac m2\left[r^2-\frac{lb(l^2+b^2)}{6\pi r^2}\right] \)

Teacher's Note:
1. The mass of the cut piece comes from the area ratio.
2. The rectangle's moment about its centre is m'(l² + b²)/12.