JEE Physics System Of Particles and Rotational Motion MCQs Set 33

Answer: (b) total angular momentum and total energy

Explanation:
1. The forces on the system are gravity (parallel to the axis) and the axis reaction, so there is no external torque about the axis; total angular momentum is conserved.
2. There is no friction, so no mechanical energy is lost; total energy is conserved.
3. As the beads move outwards the moment of inertia increases, so the angular velocity changes.

Teacher's Note:
1. Conserved quantities: angular momentum (no external torque) and energy (no friction).
2. I increases as the beads move away from the axis, so ω falls.

Answer: (b) \( \frac{ML^2}{6} \)

Explanation:
1. Each rod is pivoted at its mid-point (the intersection is symmetrical), so each has \( I=\frac{ML^2}{12} \) about a perpendicular axis through its centre.
2. The two rods are in the same plane and the axis is perpendicular to it, so the moments add: \( \frac{ML^2}{12}+\frac{ML^2}{12}=\frac{ML^2}{6} \).
3. The result does not depend on the angle θ.

Teacher's Note:
1. For an axis perpendicular to the plane, only the distance of each element from the axis matters, not the orientation of the rod.
2. Moments of inertia about the same axis simply add.

Answer: (c) \( \frac32MR^2\omega \)

Explanation:
1. Angular momentum about O = spin part + orbital part of the centre of mass: \( L=I_{cm}\omega+Mv_{cm}\,R \).
2. With \( I_{cm}=\frac12MR^2 \) and \( v_{cm}=\omega R \): \( L=\frac12MR^2\omega+MR^2\omega=\frac32MR^2\omega \).

Teacher's Note:
1. The centre of mass is at height R above the plane, so its perpendicular distance from the line of motion through O is R.
2. Both the spin and orbital parts have the same sense for a rolling disc.

Answer: (c) \( \frac12MR^2 \)

Explanation:
1. Two such semicircular discs placed together make a full disc of mass 2M and radius R, with \( I=\frac12(2M)R^2=MR^2 \).
2. By symmetry each semicircle has half of this: \( \frac12MR^2 \).

Teacher's Note:
1. Use symmetry: the full disc's moment of inertia is shared equally by the two halves when the axis is through the centre.
2. The result is the same as for a complete disc of mass M.

Answer: (a) \( \frac{2}{\sqrt{15}}R \)

Explanation:
1. Mass is the same: \( I=\frac25MR^2 \).
2. Disc about an axis through its edge, perpendicular to the plane: \( I=\frac12Mr^2+Mr^2=\frac32Mr^2 \).
3. Equating: \( \frac32r^2=\frac25R^2\Rightarrow r^2=\frac{4}{15}R^2 \), so \( r=\frac{2R}{\sqrt{15}} \).

Teacher's Note:
1. Melting keeps the mass unchanged.
2. Use the parallel-axis theorem for the edge axis of the disc.

Answer: (c) \( 3ml^2 \)

Explanation:
1. The mass at A is on the axis: contribution 0.
2. B and D lie on the diagonal BD, which is parallel to the axis at a distance equal to half the other diagonal, \( \frac{l}{\sqrt2} \): each gives \( m\frac{l^2}{2} \).
3. The mass at C is at the distance of the full diagonal AC, \( \sqrt2l \): it gives \( 2ml^2 \).
4. Total: \( 0+\frac{ml^2}{2}+\frac{ml^2}{2}+2ml^2=3ml^2 \).

Teacher's Note:
1. The diagonals of a square are perpendicular, so AC is perpendicular to the axis.
2. Measure each distance perpendicular to the axis, not to the point A.

Answer: (b) \( \sqrt{2gh} \)

Explanation:
1. On the frictionless raised surface there is no torque about the centre, so the spin of the sphere does not change.
2. Thus only the translational kinetic energy is available for climbing: \( \frac12mv^2=mgh \).
3. \( v=\sqrt{2gh} \).

Teacher's Note:
1. With no friction the rotational kinetic energy cannot be converted to potential energy.
2. Compare this with the rough incline, where all the kinetic energy is available and v = √(10gh/7).

Answer: (d) velocity is zero, acceleration is \( v^2/R \)

Explanation:
1. In pure rolling the contact point is instantaneously at rest: velocity \( v-\omega R=0 \).
2. Because the centre moves with constant velocity, the contact point has only the centripetal acceleration towards the centre in the frame of the centre: \( \omega^2R=\frac{v^2}{R} \).

Teacher's Note:
1. The contact point of a rolling body has zero velocity but non-zero acceleration directed towards the centre.
2. Acceleration of any point relative to the centre is ω²R when ω is constant.

Answer: (c) is directed up the plane throughout the journey

Explanation:
1. Gravity gives a net acceleration down the plane throughout (up while the speed is decreasing, down while it is increasing).
2. The angular acceleration needed for rolling is such that the angular velocity decreases while going up and increases while coming back, and this needs a torque from friction acting up the plane.
3. Hence friction acts up the plane throughout, as the velocity itself reverses.

Teacher's Note:
1. Friction up the plane gives a torque that reduces the spin on the way up and increases it on the way down.
2. Friction is therefore opposite to the velocity on the way up and along the velocity on the way down.

Answer: (d) \( \text{zero},\ m\omega^2L \)

Explanation:
1. In the frame rotating with the rod the mass m is at rest, so its acceleration relative to P is zero.
2. The pseudo (centrifugal) force on m depends on its distance from the axis O, not on the observer's position: \( F=m\omega^2L \) directed away from O.
3. The Coriolis force is zero because m is at rest in this frame.

Teacher's Note:
1. The centrifugal force is computed from the distance of the particle from the axis of rotation.
2. A body at rest in a rotating frame has zero acceleration there; the pseudo force balances the real tension.

Answer: (c) \( \sqrt2 \)

Explanation:
1. For a rod of length x at angle φ below the horizontal, energy conservation gives \( \frac12mx^2\omega^2=mgx\sin\varphi \), so \( \omega^2=\frac{2g\sin\varphi}{x} \).
2. The angular speed at every angle scales as \( 1/\sqrt x \), so the time to turn through the same angle (90 degrees) scales as \( \sqrt x \).
3. So \( \frac{t_A}{t_B}=\sqrt{\frac{l}{l/2}}=\sqrt2 \).

Teacher's Note:
1. Dimensional reasoning also works: the time can depend only on g and the length, so t is proportional to √(l/g).
2. Halving the length reduces the time by a factor 1/√2.

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

Explanation:
1. The spin angular momentum is \( I\omega \) with \( I=mR^2 \) and \( \omega=\frac vR \).
2. \( L_{spin}=mR^2\cdot\frac vR=mRv=0.50\times0.10\times0.50=0.025\ \text{kg m}^2/\text{s} \).

Teacher's Note:
1. Spin angular momentum depends only on the rotation about the centre of mass, not on the position of the origin.
2. For a ring, I ω = m R v.

Answer: (b) \( 80\ \text{N} \)

Explanation:
1. Forces on the bar: the force P at A, the normal reaction N at A (upwards), and the hinge reaction at B.
2. Take torques about B. P acts at A, whose perpendicular distance from the line of P is 4 m (the vertical height of B above the floor): torque \( 60\times4=240\ \text{N m} \) in one sense.
3. N acts at a horizontal distance 3 m from B: torque \( N\times3 \) in the opposite sense. Equilibrium gives \( 3N=240\Rightarrow N=80\ \text{N} \).
4. Vertical equilibrium of the bar means the hinge pulls it down with 80 N, so the bar exerts 80 N on the hinge.

Teacher's Note:
1. Take moments about the hinge so that the unknown hinge force drops out.
2. The weight is neglected, so the vertical hinge force simply equals the floor's normal reaction.

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

Explanation:
1. For rolling: \( mg\sin\theta-f=ma \) and \( fR=\frac12mR^2\cdot\frac aR\Rightarrow f=\frac{ma}{2} \).
2. So \( a=\frac23g\sin\theta \) and \( f=\frac13mg\sin\theta \).
3. Limiting friction must be at least this: \( \mu mg\cos\theta\ge\frac13mg\sin\theta\Rightarrow\mu\ge\frac{\tan\theta}{3} \).

Teacher's Note:
1. Friction needed for pure rolling is f = mg sinθ × (I/(mR²+I)), here 1/3 of mg sinθ.
2. The normal force is mg cosθ.

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

Explanation:
1. The hoop turns with \( \omega=\frac vr \), so in time t the point has moved through an angle \( \theta=\frac{vt}{r} \) from the top.
2. The velocity of the point is the vector sum of v (translation) and \( \omega r=v \) (rotation), with angle θ between these two vectors: \( |v_P|=\sqrt{v^2+v^2+2v^2\cos\theta}=2v\cos\frac\theta2 \).
3. So \( |v_P|=2v\cos\frac{vt}{2r} \).

Teacher's Note:
1. At the top the two velocities are parallel (sum 2v); at the contact point they are opposite (sum 0).
2. The magnitude of the sum of two equal vectors with angle θ between them is 2v cos(θ/2).

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

Explanation:
1. The tension in the cord equals \( mg \) and acts towards the centre. Friction can act either way, with a maximum of \( \mu Mg \).
2. The required centripetal force is \( M\omega^2r \). For the smallest r, friction acts outwards at its maximum: \( M\omega^2r=mg-\mu Mg \).
3. \( r_{min}=\frac{mg-\mu Mg}{M\omega^2} \).

Teacher's Note:
1. The smaller the radius, the smaller the centripetal force needed, so friction has to act outwards.
2. The larger limit with friction acting inwards is r = (mg + μMg)/(Mω²).

Answer: (c) \( 6I \)

Explanation:
1. Diameter: \( I=\frac14MR^2\Rightarrow MR^2=4I \).
2. Perpendicular axis through the centre: \( \frac12MR^2=2I \).
3. Parallel-axis theorem for a rim point: \( 2I+MR^2=2I+4I=6I \).

Teacher's Note:
1. Use I(diameter) = MR²/4 to express MR² in terms of I.
2. Then apply the parallel axis theorem for the edge.

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

Explanation:
1. The shell's rim moves at speed v, and the pulley's rim at speed v, since the cord does not slip.
2. Energy conservation: \( mgh=\frac12mv^2+\frac12I\frac{v^2}{r^2}+\frac12\left(\frac23MR^2\right)\frac{v^2}{R^2} \).
3. \( mgh=v^2\left[\frac m2+\frac{I}{2r^2}+\frac M3\right] \).

Teacher's Note:
1. For a thin spherical shell, I = (2/3)MR².
2. The pulley and shell have the same rim speed v as the object, so R and r cancel.

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

Explanation:
1. Torque: \( \vec\tau=\frac{d\vec J}{dt}=2\vec bt \), along \( \vec b \).
2. The vector \( \vec J \) has a component \( a \) perpendicular to \( \vec b \) and a component \( bt^2 \) along \( \vec b \). The angle is 45° when \( bt^2=a\Rightarrow t=\sqrt{a/b} \).
3. So \( \tau=2bt=2b\sqrt{\frac ab} \).

Teacher's Note:
1. The angle between τ (along b) and J is found from the ratio of the components of J.
2. 45° means the two components of J are equal in magnitude.

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

Explanation:
1. Mass is proportional to area, so the cut-out piece is 1/4 of the full disc. Remaining = 3/4 of full mass, so the full disc has mass \( \frac{4M}{3} \) and the hole has mass \( \frac M3 \).
2. Full disc: \( \frac12\cdot\frac{4M}{3}R^2=\frac23MR^2 \).
3. Hole about O: \( \frac12\cdot\frac M3\left(\frac R2\right)^2+\frac M3\left(\frac R2\right)^2=\frac{MR^2}{24}+\frac{MR^2}{12}=\frac{MR^2}{8} \).
4. Remaining: \( \frac23MR^2-\frac18MR^2=\frac{13}{24}MR^2 \).

Teacher's Note:
1. Subtract the moment of inertia of the removed part from that of the whole disc, both about the same axis.
2. The removed piece needs the parallel-axis theorem, since its centre is R/2 from O.

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

Explanation:
1. Process A: torque \( 5R \) acts on the disc, so \( \alpha_A=\frac{5R}{I} \).
2. Process B: the hanging body accelerates downwards, so the tension is less than its weight: \( T=5-ma<5\ \text{N} \).
3. So the torque in B is less than \( 5R \) and \( \alpha_B<\alpha_A \); the tangential acceleration is greater in A.

Teacher's Note:
1. A hanging mass that accelerates has tension smaller than its weight.
2. Tangential acceleration of the rim equals αR.