JEE Physics System Of Particles and Rotational Motion MCQs Set 48

Answer: (d) it keeps the speed of the engine nearly uniform

Explanation:
1. A flywheel has a large moment of inertia.
2. A large I means a given torque change produces only a small change in angular velocity, so fluctuations in torque are smoothed out.

Teacher's Note:
1. The flywheel stores rotational kinetic energy and releases it between power strokes.

Answer: (c) \( \omega \text{ is independent of } r \)

Explanation:
1. In rigid-body rotation every particle sweeps the same angle in the same time.
2. So v = ωr grows with r while ω is the same for all particles.

Teacher's Note:
1. Angular quantities describe the whole body; linear quantities describe individual particles.

Answer: (c) move to a distance l/2 from O

Explanation:
1. The surface is smooth, so no horizontal external force acts and the centre of mass falls vertically.
2. The centre of mass stays above O. When the rod lies flat its centre is at height zero, directly above O, so each end is l/2 from O horizontally; the lower end slides to l/2 from O.

Teacher's Note:
1. No horizontal external force means the horizontal position of the centre of mass is fixed.

Answer: (b) A is true but B is false

Explanation:
1. The definition L = Iω holds in any frame for a rigid body.
2. The relation τ = dL/dt follows from Newton's laws, which hold only in inertial frames. In a non-inertial frame, a pseudo-torque must be added, so B is not true as written.

Teacher's Note:
1. Newton's laws need pseudo-forces, and hence pseudo-torques, in accelerating frames.

Answer: (b) \( \omega=t-3 \)

Explanation:
1. Uniform acceleration means α = dω/dt is constant.
2. For ω = t − 3, α = 1 rad/s², constant. Option (a) gives α = 0 (uniform rotation), (c) gives α = 10t and (d) gives α = 8t + 1, both varying.

Teacher's Note:
1. Check differentiation first: a linear function of time has constant slope.

Answer: (c) sphere 3

Explanation:
1. Let each mass be m. About sphere 1: \( I_1=m(d)^2+m(3d)^2=10md^2 \).
2. About sphere 2: \( I_2=m d^2+m(2d)^2=5md^2 \).
3. About sphere 3: \( I_3=m(3d)^2+m(2d)^2=13md^2 \).
4. The largest is I₃.

Teacher's Note:
1. Distances are measured from the chosen axis to each of the other masses; the end sphere with the far neighbour gives the largest value.

Answer: (c) horizontal and intersecting the axis

Explanation:
1. The particle moves in a horizontal circle at constant speed.
2. Its acceleration is purely centripetal, directed horizontally toward the axis, so the force is horizontal and meets the axis.

Teacher's Note:
1. No tangential acceleration for uniform rotation, so no tangential force.

Answer: (b) horizontal and skew with the axis

Explanation:
1. The particle has a centripetal acceleration ω²r and a tangential acceleration αr.
2. The resultant is horizontal but not directed to the axis. Its line of action does not meet the axis, so it is skew to it.

Teacher's Note:
1. Tangential component means the line of force misses the axis.

Answer: (d) all take the same time

Explanation:
1. With slipping, the kinetic friction is μMg cosθ for every body.
2. The centre of mass acceleration is \( a=g\sin\theta-\mu g\cos\theta \), independent of shape and mass.
3. Hence all take the same time.

Teacher's Note:
1. In slipping motion the shape affects only the spin, not the translation.

Answer: (d) torque of the applied force

Explanation:
1. Torque = F × L is the same whichever end is pivoted.
2. The moment of inertia about the pivot changes when the ends are swapped, so α = τ/I changes.
3. Work done in one turn is τ(2π), the same either way, and it equals \( \tfrac12I\omega^2 \); as I differs, ω differs, and so does L = Iω.

Teacher's Note:
1. Only quantities that do not involve I stay the same.

Answer: (b) The solid sphere reaches the bottom with greater speed

Explanation:
1. Acceleration is \( a=\frac{g\sin\theta}{1+k^2/R^2} \); the solid sphere has the smaller \( k^2/R^2=\tfrac25 \) than the hollow sphere \( \tfrac23 \).
2. So the solid sphere accelerates faster, arrives first and with greater speed.
3. Kinetic energy at the bottom equals mgh for both, and momentum mv is larger for the solid sphere.

Teacher's Note:
1. The body with more mass concentrated near the axis rolls faster.

Answer: (a) remains constant

Explanation:
1. \( L=mvy \) where y is the fixed perpendicular distance of the path from the origin.
2. With v and y both constant, L is constant.

Teacher's Note:
1. Angular momentum equals m v times the perpendicular distance of the line of motion from the point.

Answer: (b) \( M\left[\frac85r^2+2l^2\right] \)

Explanation:
1. Two spheres have their centres on the axis: each gives \( \tfrac25Mr^2 \).
2. The other two are a distance l away: each gives \( \tfrac25Mr^2+Ml^2 \).
3. Total \( =\frac85Mr^2+2Ml^2 \).

Teacher's Note:
1. Use the parallel-axis theorem for each sphere, with the diameter value 2/5 Mr² as I_cm.

Answer: (c) \( \frac34ml^2 \)

Explanation:
1. Two masses lie on the axis and contribute nothing.
2. The third mass is at a perpendicular distance equal to the height \( \frac{\sqrt3}{2}l \).
3. \( I=m\left(\frac{\sqrt3}{2}l\right)^2=\frac34ml^2 \)

Teacher's Note:
1. Only the mass not on the axis contributes.

Answer: (c) \( \frac{v}{\sqrt2} \)

Explanation:
1. Sliding: \( \tfrac12mv^2=mgh \), so \( v=\sqrt{2gh} \).
2. Ring rolling: \( mgh=\tfrac12mv'^2+\tfrac12mv'^2=mv'^2 \), so \( v'=\sqrt{gh}=\frac{v}{\sqrt2} \).

Teacher's Note:
1. For a ring, half the energy goes into rotation.

Answer: (a) \( \frac12 \)

Explanation:
1. For a ring \( I=MR^2 \) and \( v=\omega R \).
2. \( K_{rot}=\tfrac12MR^2\omega^2=\tfrac12Mv^2=K_{trans} \).
3. So rotational energy is half of the total.

Teacher's Note:
1. Equal translational and rotational energies for a ring.

Answer: (b) \( \sqrt{\frac43gh} \)

Explanation:
1. \( Mgh=\tfrac12Mv^2+\tfrac12\left(\tfrac12MR^2\right)\frac{v^2}{R^2}=\tfrac34Mv^2 \).
2. \( v=\sqrt{\frac{4gh}{3}} \).

Teacher's Note:
1. Total kinetic energy of a rolling cylinder is (3/4)Mv².

Answer: (d) 1 m

Explanation:
1. Torque: \( \tau=xF_y-yF_x=2(4)-1(3)=5 \) N m.
2. Magnitude of force: \( F=\sqrt{3^2+4^2}=5 \) N.
3. Moment arm \( =\tau/F=1 \) m.

Teacher's Note:
1. Moment arm = torque magnitude divided by force magnitude.

Answer: (c) \( 5\hat k \) N m

Explanation:
1. The angle between the position vector and the force is 60° − 30° = 30°.
2. \( |\tau|=rF\sin30^\circ=1\times10\times0.5=5 \) N m.
3. Rotating from r to F is anticlockwise, so the direction is +z: \( 5\hat k \).

Teacher's Note:
1. Torque is perpendicular to the plane containing r and F, here along z.

Answer: (b) \( \frac{m_1m_2r^2}{m_1+m_2} \)

Explanation:
1. Distances from the centre of mass: \( \frac{m_2r}{m_1+m_2} \) and \( \frac{m_1r}{m_1+m_2} \).
2. \( I=m_1\frac{m_2^2r^2}{(m_1+m_2)^2}+m_2\frac{m_1^2r^2}{(m_1+m_2)^2}=\frac{m_1m_2r^2}{m_1+m_2} \)

Teacher's Note:
1. This is the reduced mass times r².

Answer: (b) \( l=\sqrt3R \)

Explanation:
1. About the axis: \( \tfrac12MR^2 \). About a transverse axis through the centre: \( M\left(\frac{R^2}{4}+\frac{l^2}{12}\right) \).
2. Equate: \( \frac{R^2}{2}=\frac{R^2}{4}+\frac{l^2}{12}\Rightarrow l^2=3R^2 \).

Teacher's Note:
1. Use the transverse-axis formula for a cylinder.

Answer: (c) directly proportional to the square of the velocity and independent of mass and the angle of the incline

Explanation:
1. Energy is conserved in pure rolling: \( \tfrac12Mv^2+\tfrac12\cdot\tfrac25MR^2\frac{v^2}{R^2}=Mgh \).
2. \( h=\frac{7v^2}{10g} \); mass and angle cancel.

Teacher's Note:
1. Friction does no work in pure rolling, so the energy balance is simple.

Answer: (d) the same, mvd, wherever the origin is taken

Explanation:
1. Let the lines be y = 0 and y = d with the origin at height y₀ and the particles moving in ±x.
2. Contribution of one: \( -mv(0-y_0) \) and of the other \( +mv(d-y_0) \).
3. Sum: \( mvd \), independent of y₀ (the pair is a couple of linear momenta).

Teacher's Note:
1. Equal and opposite momenta on separated lines act like a couple, whose moment is the same about any point.

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

Explanation:
1. No external torque, so angular momentum is conserved.
2. \( MR^2\omega=(MR^2+2mR^2)\omega' \).
3. \( \omega'=\frac{M\omega}{M+2m} \)

Teacher's Note:
1. I rises, so ω must fall.