JEE Physics System Of Particles and Rotational Motion MCQs Set 32

Answer: (d) it helps the engine in keeping the speed uniform

Explanation:
1. A flywheel has a large moment of inertia, so it stores rotational kinetic energy.
2. Because of this large inertia it resists sudden changes in angular speed, smoothing out the jerks produced by the power strokes of the engine.
3. Hence it keeps the speed uniform.

Teacher's Note:
1. Large moment of inertia means large resistance to change of angular velocity.
2. A flywheel does not add power; it only smooths the supply of energy.

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

Explanation:
1. In a rigid body every particle sweeps the same angle in the same time, so all particles have the same angular velocity.
2. The linear speed is v = ωr, so v is proportional to r while ω is the same for every particle.

Teacher's Note:
1. Angular quantities describe the whole rigid body; linear quantities depend on the distance from the axis.
2. Do not confuse ω = v/r with a dependence of ω on r.

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

Explanation:
1. \( \vec L=I\vec\omega \) is only the definition of angular momentum of a rigid body in terms of its moment of inertia, so it holds in any frame.
2. \( \vec\tau=d\vec L/dt \) follows from Newton's laws, which hold only in inertial frames. In a non-inertial frame the pseudo-torque must also be included.
3. So A is true and B is false.

Teacher's Note:
1. Newton's second law and its rotational form are valid only in inertial frames unless pseudo forces are added.
2. Definitions such as L = Iω do not depend on the frame.

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

Explanation:
1. Uniform angular acceleration means \( \alpha=\frac{d\omega}{dt} \) is constant, so ω must be a linear function of t.
2. Only \( \omega=t-3 \) has constant derivative (α = 1 rad/s²).
3. ω = 5 gives α = 0 (uniform rotation, not accelerated), and the other two give α that changes with time.

Teacher's Note:
1. Uniform acceleration means a constant rate of change of angular velocity.
2. Differentiate each option and check whether the result is a constant that is not zero.

Answer: (c) sphere 3

Explanation:
1. About sphere 1: \( I=m(d^2)+m(3d)^2=10md^2 \).
2. About sphere 2: \( I=md^2+m(2d)^2=5md^2 \).
3. About sphere 3: \( I=m(3d)^2+m(2d)^2=13md^2 \).
4. The largest value is about sphere 3.

Teacher's Note:
1. The moment of inertia is the sum of m r² over all particles, so the axis that is farthest from the heavy distribution gives the largest value.
2. The axis through the middle sphere is closest to the others and gives the least.

Answer: (c) horizontal and intersecting the axis

Explanation:
1. Under uniform rotation the particle moves in a horizontal circle at constant speed, so it has only centripetal acceleration.
2. The resultant force is therefore directed horizontally towards the centre of the circle, which lies on the axis, so the line of the force meets the axis.

Teacher's Note:
1. With constant speed on a circle the net force is purely centripetal.
2. The centripetal force points to the centre of the circle, and the centre lies on the axis.

Answer: (b) horizontal and skew with the axis

Explanation:
1. For non-uniform rotation the particle has a centripetal acceleration \( \omega^2 r \) and also a tangential acceleration \( \alpha r \).
2. The resultant force is the vector sum of a radial part (meeting the axis) and a tangential part (perpendicular to the plane containing the axis and the particle).
3. The resultant is horizontal but no longer meets the axis; it is skew to it.

Teacher's Note:
1. A tangential component appears whenever the angular speed changes.
2. The tangential direction is horizontal and perpendicular to the radius, so it does not point at the axis.

Answer: (d) all will take the same time

Explanation:
1. If friction is not sufficient for pure rolling, every object slides while rotating, and kinetic friction \( \mu mg\cos\theta \) acts on each.
2. Linear acceleration \( a=g\sin\theta-\mu g\cos\theta \) is the same for all, whatever the shape.
3. With the same acceleration and same length of incline, all take the same time.

Teacher's Note:
1. In sliding motion the translation depends only on the net force along the incline, not on the moment of inertia.
2. The moment of inertia matters only when rolling without slipping.

Answer: (d) torque of the applied force

Explanation:
1. The torque is \( \tau=F\,l \), where l is the length of the rod; it depends only on F and l.
2. The moment of inertia about the pivot changes when the other end is pivoted (the mass distribution is not symmetric), so angular acceleration, angular velocity and angular momentum all change.
3. Only the torque is independent of the end pivoted.

Teacher's Note:
1. The moment of inertia depends on where the mass lies relative to the axis.
2. Torque depends only on the force and its perpendicular distance from the axis.

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

Explanation:
1. For rolling without slipping \( a=\frac{g\sin\theta}{1+I/mR^2} \).
2. Solid sphere: I/mR² = 2/5, a = (5/7) g sinθ. Hollow sphere: I/mR² = 2/3, a = (3/5) g sinθ.
3. The solid sphere has the larger acceleration, so it arrives first and with the larger speed \( v=\sqrt{\frac{2gh}{1+I/mR^2}} \).
4. Both lose the same potential energy mgh, so their total kinetic energies are equal, and their momenta (mv) are different.

Teacher's Note:
1. The smaller the ratio I/mR², the faster a body rolls down an incline.
2. Kinetic energy at the bottom equals mgh for both, but it is shared differently between translation and rotation.

Answer: (a) remains constant

Explanation:
1. Angular momentum about the origin is \( L=m\,v\,y \), where y is the perpendicular distance of the line of motion from the origin.
2. The line of motion is parallel to the x-axis, so y is fixed and the velocity is constant. Hence L stays constant (and is not zero unless y = 0).

Teacher's Note:
1. L = m v × (perpendicular distance of the line of motion from the origin).
2. A particle moving in a straight line at constant velocity has constant angular momentum about any fixed point.

Answer: (b) \( M\left[\frac{8}{5}r^2+2l^2\right] \)

Explanation:
1. Two spheres have their centres on the axis: each contributes \( \frac25Mr^2 \).
2. The other two are at distance l: each contributes \( \frac25Mr^2+Ml^2 \) (parallel-axis theorem).
3. Total: \( 4\times\frac25Mr^2+2Ml^2=M\left[\frac85r^2+2l^2\right] \).

Teacher's Note:
1. For a solid sphere about a diameter, I = (2/5)MR².
2. Use the parallel-axis theorem for the two spheres that are a distance l away.

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

Explanation:
1. Two masses lie on the axis, so they contribute nothing.
2. The third mass is at the height of the triangle, \( h=\frac{\sqrt3}{2}l \).
3. So \( I=m h^2=\frac34ml^2 \).

Teacher's Note:
1. Masses on the axis have zero distance and contribute zero.
2. The perpendicular distance from the third corner to the opposite side is the altitude of the triangle.

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

Explanation:
1. Sliding (no friction): \( \frac12mv^2=mgh\Rightarrow v=\sqrt{2gh} \).
2. Ring rolling: \( mgh=\frac12mv'^2+\frac12(mR^2)\frac{v'^2}{R^2}=mv'^2\Rightarrow v'=\sqrt{gh} \).
3. Therefore \( v'=\frac{v}{\sqrt2} \).

Teacher's Note:
1. For a ring the kinetic energy is shared equally between translation and rotation.
2. Compare the two energy equations with the same loss of potential energy.

Answer: (a) \( \frac12 \)

Explanation:
1. For a ring \( I=mR^2 \) and \( v=\omega R \).
2. Rotational KE \( =\frac12mR^2\omega^2=\frac12mv^2 \), translational KE \( =\frac12mv^2 \).
3. The rotational part is half of the total.

Teacher's Note:
1. For rolling, KE(rotational)/KE(translational) = I/(mR²).
2. For a ring this ratio is 1, so each share is one half.

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

Explanation:
1. \( Mgh=\frac12Mv^2+\frac12\cdot\frac12MR^2\cdot\frac{v^2}{R^2}=\frac34Mv^2 \).
2. \( v^2=\frac43gh\Rightarrow v=\sqrt{\frac43gh} \).

Teacher's Note:
1. Total kinetic energy of a rolling solid cylinder is (3/4)Mv².
2. Energy conservation links the loss of potential energy to this total.

Answer: (d) \( 1\ \text{m} \)

Explanation:
1. Torque: \( \vec\tau=\vec r\times\vec F=(2\cdot4-1\cdot3)\hat k=5\hat k\ \text{N m} \).
2. Magnitude of force: \( |F|=\sqrt{3^2+4^2}=5\ \text{N} \).
3. Moment arm \( =\frac{\tau}{F}=\frac55=1\ \text{m} \).

Teacher's Note:
1. Moment arm is the perpendicular distance from the point to the line of action, equal to τ/F.
2. Evaluate the cross product by the determinant rule.

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

Explanation:
1. The distances of the masses from the centre of mass are \( r_1=\frac{m_2r}{m_1+m_2} \) and \( r_2=\frac{m_1r}{m_1+m_2} \).
2. \( I=m_1r_1^2+m_2r_2^2=\frac{m_1m_2^2r^2+m_2m_1^2r^2}{(m_1+m_2)^2}=\frac{m_1m_2r^2}{m_1+m_2} \).

Teacher's Note:
1. The moment of inertia about the centre of mass of a two-particle system equals the reduced mass times r².
2. The heavier mass lies closer to the centre of mass.

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

Explanation:
1. About the geometric axis: \( I_1=\frac12MR^2 \).
2. About a transverse axis through the centre: \( I_2=M\left(\frac{R^2}{4}+\frac{l^2}{12}\right) \).
3. Equating: \( \frac{R^2}{2}=\frac{R^2}{4}+\frac{l^2}{12}\Rightarrow l^2=3R^2 \), so \( l=\sqrt3R \).

Teacher's Note:
1. Use I = M(R²/4 + l²/12) for a solid cylinder about a transverse axis through its centre.
2. Equate it to MR²/2 and solve for l.

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

Explanation:
1. The sphere rolls, so its kinetic energy is \( \frac12Mv^2+\frac12\cdot\frac25MR^2\cdot\frac{v^2}{R^2}=\frac{7}{10}Mv^2 \).
2. At the highest point this equals \( Mgh \), so \( h=\frac{7v^2}{10g} \).
3. h is proportional to v² and does not depend on the mass, the radius or the angle of the incline.

Teacher's Note:
1. Mass cancels from both sides of the energy equation.
2. With pure rolling no energy is lost, so the angle of the incline does not matter.

Answer: (d) the same no matter which point is taken as the origin

Explanation:
1. Let the lines be at perpendicular distances \( y_1 \) and \( y_2=y_1+d \) from the origin, on the same side of it. Because the velocities are opposite, \( L=mv\,y_2-mv\,y_1=mvd \).
2. If the origin lies between the lines, the two angular momenta have the same sense and add to \( mv(y_1+y_2)=mvd \) as well.
3. So the total angular momentum is the same about every origin.

Teacher's Note:
1. The two linear momenta are equal and opposite, so the total linear momentum is zero.
2. For a system with zero total linear momentum, angular momentum does not depend on the choice of origin.

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

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

Teacher's Note:
1. Attaching masses gently means no external torque, so L is conserved.
2. The final moment of inertia is that of the ring plus two point masses at radius R.

Answer: (b) \( \frac K2 \)

Explanation:
1. No external torque, so angular momentum L is conserved.
2. Kinetic energy \( K=\frac{L^2}{2I} \). When I doubles, K becomes \( K/2 \).

Teacher's Note:
1. When L is constant, K = L²/2I is inversely proportional to I.
2. The work done by the child on stretching is not stored; K falls as ω falls.

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

Explanation:
1. The full disc would have mass 4M and moment of inertia \( \frac12(4M)R^2=2MR^2 \) about the axis.
2. By symmetry the four quarters contribute equally, so one quarter has \( \frac{2MR^2}{4}=\frac12MR^2 \).

Teacher's Note:
1. Identical quarters share the total moment of inertia equally when the axis passes through the centre of the disc.
2. Always express the full disc's mass in terms of the given mass of the piece.

Answer: (b) up the incline while ascending as well as descending

Explanation:
1. Without friction, gravity would produce only translational deceleration while ascending, and the spin would not change.
2. For the cylinder to keep rolling, its spin must reduce in step with its speed, so friction must supply a torque against the rotation; this requires friction up the incline.
3. While descending, the spin must increase, which again needs friction acting up the incline.

Teacher's Note:
1. The direction of friction in rolling is decided by the angular acceleration needed, not by the direction of motion.
2. The linear acceleration is down the incline in both cases, and the torque must give the matching angular acceleration.