JEE Physics System Of Particles and Rotational Motion MCQs Set 27

Answer: (d) angular acceleration – radian/second

Explanation:
1. Power is energy per unit time, so joule/second is correct for rotational power.
2. Torque is force times distance, so N m is correct.
3. Angular displacement is an angle, measured in radian.
4. Angular acceleration is rate of change of angular velocity, so its unit is radian/second², not radian/second.

Teacher's Note:
1. Check each unit by writing the defining formula.
2. Angular acceleration = (rad/s)/s.

Answer: (a) the linear velocity of objects at the equator is greater than at other places

Explanation:
1. Every point of the earth completes one rotation in the same time, so all places have the same angular velocity ω.
2. Linear speed v = ωr, where r is the distance from the axis of rotation.
3. r is greatest at the equator, so the linear speed is greatest there and zero at the poles.

Teacher's Note:
1. Angular velocity is common to all points of a rigid rotating body.
2. Linear speed depends on the distance from the axis.

Answer: (b) the particles on the diameter (the axis) do not have any linear acceleration

Explanation:
1. Every particle moves in a circle about the axis with the same angular speed, so (c) is false.
2. Linear speed v = ωr depends on the distance r from the axis, so (d) is false.
3. Particles on the surface move in circles and have centripetal acceleration ω²r, so (a) is false.
4. Particles lying on the axis have r = 0, hence zero linear acceleration.

Teacher's Note:
1. Use v = ωr and a = ω²r.
2. Points on the axis do not move.

Answer: (d) all of the above

Explanation:
1. I = Σmr², where r is the perpendicular distance of each mass element from the axis.
2. So I depends on how the mass is distributed (shape and size) and on the position of the axis.

Teacher's Note:
1. Start from I = Σmr².
2. Shape and distribution of mass fix the r values.

Answer: (b) mass

Explanation:
1. Compare K = ½Iω² with K = ½mv² and τ = Iα with F = ma.
2. I plays the role of mass in rotational motion.

Teacher's Note:
1. Compare the rotational and linear formulas.
2. I is the inertia against rotation.

Answer: (b) Graph (b)

Explanation:
1. By the parallel axes theorem, I = ½MR² + Mx².
2. This is a parabola in x with a non-zero value ½MR² at x = 0 and it is symmetric about the I-axis.
3. Graph (a) would give I = 0 at some x, graph (c) is a straight line and graph (d) falls with x; all are impossible. Graph (b) matches.

Teacher's Note:
1. Write I as a function of x using the parallel axes theorem.
2. Look at the value at x = 0.

Answer: (a) it increases the moment of inertia of the flywheel

Explanation:
1. I = Σmr² is large when the mass lies far from the axis.
2. A large I means the flywheel resists changes in its angular speed and runs smoothly.

Teacher's Note:
1. Mass at large r gives large I.
2. Flywheels are meant to resist speed fluctuations.

Answer: (a) Sphere

Explanation:
1. The perpendicular axes theorem (Iz = Ix + Iy) holds only for planar (laminar) bodies, with the axes x and y in the plane of the body.
2. Disc, ring and rectangular plate are planar, but a sphere is a three-dimensional body.

Teacher's Note:
1. Recall the condition for the theorem: a flat body.
2. Sphere is not flat.

Answer: (d) torque

Explanation:
1. Torque changes the angular momentum: τ = dL/dt.
2. A retarding torque is needed to bring the angular velocity to zero.

Teacher's Note:
1. Torque is the cause of change in rotation.
2. A force alone need not change the rotation.

Answer: (b) it is easier to stop the solid sphere

Explanation:
1. About a diameter, I(ring) = ½MR² and I(sphere) = (2/5)MR².
2. For the same ω, angular momentum L = Iω is smaller for the sphere.
3. The torque needed to stop in a given time is τ = L/t, so the sphere is easier to stop.

Teacher's Note:
1. Compare the moments of inertia about a diameter.
2. Smaller I means smaller L for the same ω.

Answer: (b) more angular acceleration will be produced in (b)

Explanation:
1. The torque is the same in both cases: F × 1 m.
2. In (a) the heavier steel half is far from the pivot, so I(a) is larger; in (b) the heavy half is near the pivot, so I(b) is smaller.
3. Check with m_s > m_w and half-length 0.5 m: I(a) − I(b) = 0.5(m_s − m_w) > 0.
4. α = τ/I, so α is larger in (b).

Teacher's Note:
1. Torque is equal; compare moments of inertia.
2. Mass far from the pivot gives larger I.

Answer: (c) it acquires angular acceleration

Explanation:
1. From τ = Iα, a constant non-zero torque gives a constant angular acceleration α = τ/I.
2. So the angular velocity changes uniformly; it does not stay constant.

Teacher's Note:
1. Use τ = Iα.
2. A torque produces angular, not necessarily linear, acceleration.

Answer: (d) any point on the system or outside it

Explanation:
1. If the net force is zero, the net torque is the same about every point.
2. So if the torque is zero about one point it is zero about all points, inside or outside the body.

Teacher's Note:
1. Net force zero makes torque independent of the reference point.
2. Choose the most convenient point.

Answer: (a) \( T_1 = \dfrac{Mg}{3},\ T_2 = \dfrac{2Mg}{3} \)

Explanation:
1. Take torques about A: T₂ × (3ℓ/4) = Mg × (ℓ/2), so T₂ = 2Mg/3.
2. Vertical balance: T₁ + T₂ = Mg, so T₁ = Mg/3.

Teacher's Note:
1. Choose the pivot at one string to remove one unknown.
2. B is at 3ℓ/4 from A.

Answer: (c) Only 3

Explanation:
1. An extra force at the centre of mass cannot change the torque about the centre of mass, so the torque of the given forces about the centre must already be zero.
2. Rod 1: forces at the left end (up), centre (up) and right end (down) give torque 2F × ℓ/2 ≠ 0. Rod 2: a couple, torque Fℓ ≠ 0.
3. Rod 3: the two end forces (both up) have equal and opposite torques about the centre, so the torque is zero; the net force F upward can be cancelled by a downward force F at the centre. The extra force at the centre does not disturb the torque balance.

Teacher's Note:
1. First make the net torque zero about the centre.
2. Then cancel the net force with the extra force.

Answer: (c) increasing the angle of inclination

Explanation:
1. Let θ be the angle with the ground. Torques about the foot give N × L sinθ = W × (L/2) cosθ, so N = W/(2 tanθ).
2. The friction at the floor must equal N, so f = W/(2 tanθ) and it must not exceed μW.
3. Increasing θ increases tanθ and reduces the friction required. The length L cancels out.

Teacher's Note:
1. Find the friction needed at the floor.
2. The length does not appear in the result.

Answer: (b) shifted to the right of the line of Mg (towards the direction of F)

Explanation:
1. For equilibrium the net torque about the centre of mass must be zero.
2. F passes through the centre of mass, and f acts at the floor level to the left, so f gives a clockwise torque f·h.
3. N must give an equal counter-clockwise torque, which requires its line of action to lie to the right of the centre of mass.

Teacher's Note:
1. Take torques about the centre of mass.
2. Friction at the base produces a torque; N must balance it.

Answer: (c) \( \vec\tau = \dfrac{d\vec L}{dt} \)

Explanation:
1. F = dp/dt is the linear form of the law.
2. Its rotational form is torque = rate of change of angular momentum, τ = dL/dt.

Teacher's Note:
1. Replace F by τ and p by L.
2. The first option is a kinematic equation.

Answer: (d) perpendicular to the orbital plane

Explanation:
1. L = r × p, and the cross product is perpendicular to both r and p.
2. Both r and p lie in the orbital plane, so L is perpendicular to that plane.

Teacher's Note:
1. Use the cross product.
2. It is perpendicular to both vectors.

Answer: (d) the axis of rotation

Explanation:
1. L = r × p is perpendicular to the plane of motion, which is the direction of the axis of rotation.

Teacher's Note:
1. Cross product direction by the right-hand rule.
2. It is perpendicular to the plane of motion.

Answer: (c) does not change

Explanation:
1. L = m v d, where d is the perpendicular distance of the line of motion from the origin.
2. The line of motion is parallel to the x-axis, so d is constant. v is constant, so L is constant.

Teacher's Note:
1. Angular momentum = mv × perpendicular distance.
2. The perpendicular distance stays the same.