JEE Physics System Of Particles and Rotational Motion MCQs Set 08

Answer: (c) Angular acceleration : radian/s

Explanation:
1. Rotational power has the unit of power, watt = joule/s, so (a) matches. Torque is N m and angular displacement is in radian.
2. Angular acceleration \( \alpha = d\omega/dt \) has the unit rad/s², not rad/s, so (d) does not match.

Teacher's Note:
1. Check the unit by writing the definition. Angular acceleration is a rate of change of angular velocity, so there is one more power of 1/s.

Answer: (d) All of the above

Explanation:
1. \( I = \sum m_ir_i^2 \) depends on the masses, on how they are distributed (shape) and on the distances from the chosen axis.

Teacher's Note:
1. The moment of inertia is not a property of the body alone; it also depends on the axis chosen.

Answer: (c) It increases the moment of inertia of the fly-wheel

Explanation:
1. Mass at the rim is as far from the axis as possible, so \( I = \sum mr^2 \) is large.
2. A large moment of inertia resists changes in the angular speed, which keeps the rotation smooth.

Teacher's Note:
1. A flywheel stores rotational energy and smooths out fluctuations in torque.

Answer: (a) \( L = \sqrt3\,R \)

Explanation:
1. About its own axis \( I = \tfrac{1}{2}MR^2 \). About a perpendicular axis through the centre \( I = \dfrac{ML^2}{12} + \dfrac{MR^2}{4} \).
2. Equating: \( \dfrac{R^2}{4} = \dfrac{L^2}{12} \Rightarrow L^2 = 3R^2 \), so \( L = \sqrt3\,R \).

Teacher's Note:
1. Write the two standard formulae and equate them.

Answer: (c) It increases the moment of inertia of the tyre

Explanation:
1. More mass at the rim increases the moment of inertia.
2. This gives a more uniform rotation of the wheel, as in a flywheel.

Teacher's Note:
1. The same reasoning applies to bicycle wheels with spokes: most of the mass is at the rim.

Answer: (b) \( \dfrac{1}{2}MR^2 + 4mR^2 \)

Explanation:
1. The disc alone has \( \tfrac{1}{2}MR^2 \) and each point mass contributes \( mR^2 \).
2. \( I = \tfrac{1}{2}MR^2 + 4mR^2 \).

Teacher's Note:
1. Add the moments of inertia of the parts about the same axis.

Answer: (c) \( I_x + I_z = I_y \)

Explanation:
1. The axis perpendicular to the plane of the body here is the Y-axis.
2. The theorem says that the moment of inertia about the perpendicular axis equals the sum of those about two perpendicular axes in the plane: \( I_y = I_x + I_z \).

Teacher's Note:
1. Identify the axis that is perpendicular to the plane of the lamina. It is the one on the left-hand side.

Answer: (b) Sphere

Explanation:
1. The theorem applies only to planar (lamina-type) bodies, where all the mass is in one plane.
2. A sphere is a three-dimensional body, so the theorem cannot be used.

Teacher's Note:
1. Remember the condition: planar body only. Ring, disc, and thin blade qualify.

Answer: (a) 10

Explanation:
1. \( I_y = I_x + I_z \Rightarrow 40 = 30 + I_z \).
2. \( I_z = 10 \) kg m².

Teacher's Note:
1. The axis perpendicular to the plane has the largest value among the three.

Answer: (a) \( \sqrt5 : \sqrt3 \)

Explanation:
1. Solid: \( \tfrac{2}{5}mR_s^2 \); hollow: \( \tfrac{2}{3}mR_h^2 \). Equate: \( \dfrac{R_s^2}{R_h^2} = \dfrac{5}{3} \).
2. \( R_s : R_h = \sqrt5 : \sqrt3 \).

Teacher's Note:
1. The solid sphere needs a larger radius to have the same moment of inertia as a hollow one.

Answer: (b) 40 kg m²

Explanation:
1. With side a and mass m, \( I_z = \dfrac{ma^2}{6} = 20 \Rightarrow ma^2 = 120 \).
2. About a diameter in the plane: \( \tfrac{ma^2}{12} = 10 \). The side is at a/2 from it: \( 10 + m\left(\tfrac{a}{2}\right)^2 = 10 + 30 = 40 \) kg m².

Teacher's Note:
1. First the perpendicular axes theorem to get the in-plane axis, then the parallel axes theorem to shift to the side.

Answer: (a) 16I

Explanation:
1. A rod: \( I = \dfrac{ml^2}{12} \). About the centre of the square, each rod is \( I + m\left(\tfrac{l}{2}\right)^2 = \dfrac{ml^2}{12} + \dfrac{ml^2}{4} = \dfrac{ml^2}{3} = 4I \).
2. Four rods: \( 16I \).

Teacher's Note:
1. Use the parallel axes theorem for each rod; the distance of each rod's centre from the square's centre is l/2.

Answer: (c) A ring about its central axis perpendicular to the plane of the ring

Explanation:
1. Values in units of \( MR^2 \): hollow sphere \( \tfrac{2}{3} \), solid sphere \( \tfrac{2}{5} \), disc \( \tfrac{1}{2} \), ring 1.

Teacher's Note:
1. The ring has all its mass at the largest distance, so it has the highest moment of inertia.

Answer: (c) 1.6 kg m²

Explanation:
1. \( I = Mk^2 = 10\times 0.16 = 1.6 \) kg m².

Teacher's Note:
1. The radius of gyration is the distance at which the whole mass could be placed to give the same moment of inertia.

Answer: (a) \( I_3 > I_2 > I_1 \)

Explanation:
1. \( I_1 = \tfrac{2}{5}MR^2 \), \( I_2 = \tfrac{2}{3}MR^2 \), \( I_3 = MR^2 \).
2. So \( I_3 > I_2 > I_1 \).

Teacher's Note:
1. Compare the numerical coefficients 0.4, 0.67 and 1.

Answer: (a) \( \dfrac{2ML^2}{3} \)

Explanation:
1. The rods along x and y are perpendicular to the z-axis and each has \( \dfrac{ML^2}{3} \); the rod along z lies on the axis, so its moment of inertia is zero.
2. Total \( = \dfrac{2ML^2}{3} \).

Teacher's Note:
1. A rod lying along the axis of rotation contributes nothing.

Answer: (b) \( \dfrac{l}{\sqrt2} \)

Explanation:
1. Each particle is at distance \( \dfrac{l}{\sqrt2} \) from the centre.
2. \( I = 4m\dfrac{l^2}{2} = 2ml^2 = (4m)k^2 \Rightarrow k = \dfrac{l}{\sqrt2} \).

Teacher's Note:
1. All four particles are at the same distance from the axis, so that distance is the radius of gyration.

Answer: (a) \( \dfrac{ML^2}{3} \)

Explanation:
1. \( I = \dfrac{ML^2}{12} + M\left(\dfrac{L}{2}\right)^2 = \dfrac{ML^2}{3} \).

Teacher's Note:
1. Use the parallel axes theorem from the centre to the end.

Answer: (d) Mass

Explanation:
1. Torque = I α corresponds to F = m a, so the moment of inertia plays the role of mass.

Teacher's Note:
1. Rotational analogues: torque ↔ force, angular acceleration ↔ acceleration, moment of inertia ↔ mass.

Answer: (c) \( \dfrac{ml^2}{6} \)

Explanation:
1. Each rod has \( \dfrac{ml^2}{12} \) about this axis.
2. Total \( = 2\times\dfrac{ml^2}{12} = \dfrac{ml^2}{6} \).

Teacher's Note:
1. Moments of inertia about the same axis add.

Answer: (a) \( \sqrt5 : \sqrt6 \)

Explanation:
1. Disc about a diameter \( \tfrac{1}{4}MR^2 \), plus \( MR^2 \): \( \tfrac{5}{4}MR^2 \). Ring about a diameter \( \tfrac{1}{2}MR^2 \), plus \( MR^2 \): \( \tfrac{3}{2}MR^2 \).
2. \( \dfrac{k_1^2}{k_2^2} = \dfrac{5/4}{3/2} = \dfrac{5}{6} \), so \( k_1 : k_2 = \sqrt5 : \sqrt6 \).

Teacher's Note:
1. First find the moment of inertia about the diameter in the plane, then shift to the tangent by the parallel axes theorem.

Answer: (b) \( \dfrac{5}{4}ml^2 \)

Explanation:
1. A lies on the line (distance 0). B is at distance l from it. C is at a perpendicular distance \( \dfrac{l}{2} \) (its projection on AB is the midpoint of AB).
2. \( I = 0 + ml^2 + m\dfrac{l^2}{4} = \dfrac{5}{4}ml^2 \).

Teacher's Note:
1. Find the perpendicular distance of every particle from the axis. A particle on the axis does not contribute.

Answer: (b) 0.4 kg m²

Explanation:
1. \( I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(20)(0.2)^2 = 0.4 \) kg m².

Teacher's Note:
1. The length of the cylinder is not needed for the moment of inertia about its geometrical axis.

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

Explanation:
1. \( I = MR^2 + MR^2 = 2MR^2 \) (parallel axes theorem).

Teacher's Note:
1. The tangent is at distance R from the centre.