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Systems of Particles and Rotational Motion — NEET & JEE MCQs with answers
This chapter extends Newtonian mechanics from single particles to systems of particles and rigid bodies, introducing the centre of mass, torque, angular momentum, and moment of inertia. It develops a parallel between translational motion and rotational motion about a fixed axis, and states the conservation of the total linear momentum of a system of particles and of angular momentum when the external torque is zero.
Topics in this chapter
- 6.1 Introduction
- 6.2 Centre of Mass
- 6.3 Motion of Centre of Mass
- 6.4 Linear Momentum of a System of Particles
- 6.5 Vector Product of Two Vectors
- 6.6 Angular Velocity and Its Relation with Linear Velocity
- 6.7 Torque and Angular Momentum
- 6.8 Equilibrium of a Rigid Body
- 6.9 Moment of Inertia
- 6.10 Kinematics of Rotational Motion About a Fixed Axis
- 6.11 Dynamics of Rotational Motion About a Fixed Axis
- 6.12 Angular Momentum in Case of Rotation About a Fixed Axis
Free practice questions
For $a = 2\hat{i} + 2\hat{j} + 2\hat{k}$ and $b = \hat{i} + \hat{j} + \hat{k}$, the vector product $a \times b$ is:
- A. $6\hat{k}$
- B. $2\hat{i} + 2\hat{j} + 2\hat{k}$
- C. $0$ (null vector)
- D. $\hat{i} + \hat{j} + \hat{k}$
Show answer
Answer: C — Here $a = 2b$, so $a$ and $b$ are parallel ($\theta = 0^\circ$). Hence $a \times b = 0$. (Check: $i = (2)(1) - (2)(1) = 0$, etc.)
NCERT Class 11 Physics · Ch 6 · 6.5 Vector Product of Two Vectors
The 'something else' in addition to translation, present in the rolling motion of a cylinder, is:
- A. Rotational motion
- B. Random motion
- C. Vibrational motion
- D. Oscillatory motion
Show answer
Answer: A — NCERT: rolling motion of a cylinder is 'a combination of rotation about a fixed axis and translation.' The 'something else' referred to earlier is rotational motion.
NCERT Class 11 Physics · Ch 6 · 6.1 Introduction
A solid sphere of mass M and radius R rotates about a diameter. Its radius of gyration is:
- A. $\sqrt{5/2}\,R$
- B. $R/2$
- C. $\sqrt{2/5}\,R$
- D. $\tfrac{2}{5}R$
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Answer: C — I = (2/5)MR² and I = Mk², so k² = (2/5)R², giving k = R√(2/5) = √(2/5) R.
NCERT Class 11 Physics · Ch 6 · 6.9 Moment of Inertia, radius of gyration; Table 6.1
A lever lifts a load $F_1$ four times the effort $F_2$. If the load arm $d_1 = 0.1\,\text{m}$, the effort arm $d_2$ must be:
- A. $4.0\,\text{m}$
- B. $0.025\,\text{m}$
- C. $0.1\,\text{m}$
- D. $0.4\,\text{m}$
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Answer: D — M.A. $= F_1/F_2 = 4 = d_2/d_1$, so $d_2 = 4 \times d_1 = 4 \times 0.1 = 0.4\,\text{m}$.
NCERT Class 11 Physics · Ch 6 · 6.8.1 Mechanical Advantage
A uniform circular disc has radius R. Its centre of mass is located at:
- A. at distance R/4 from the centre
- B. the geometric centre of the disc
- C. on the rim of the disc
- D. a point at distance R/2 from the centre
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Answer: B — For a homogeneous disc the origin (geometric centre) is a point of reflection symmetry, so the centre of mass coincides with the geometric centre of the disc.
NCERT Class 11 Physics · Ch 6 · 6.2 Centre of Mass
Two particles of masses 3 kg and 5 kg form a system whose total linear momentum is 32 kg m/s. If they move together (same velocity), what is the velocity of the centre of mass?
- A. 4 m/s
- B. 8 m/s
- C. 6.4 m/s
- D. 10.7 m/s
Show answer
Answer: A — V = P/M (Eq. 6.15), where M = 3 + 5 = 8 kg. V = 32/8 = 4 m/s.
NCERT Class 11 Physics · Ch 6 · 6.4 Linear Momentum of a System of Particles
The angular momentum of a particle changes from $\mathbf{l}_1 = (2\hat{i} + 3\hat{j})\,\text{kg m}^2\,\text{s}^{-1}$ to $\mathbf{l}_2 = (6\hat{i} + 11\hat{j})\,\text{kg m}^2\,\text{s}^{-1}$ in $2\,\text{s}$. The average torque acting on it is:
- A. $(4\hat{i} + 8\hat{j})\,\text{N m}$
- B. $(1\hat{i} + 2\hat{j})\,\text{N m}$
- C. $(2\hat{i} + 4\hat{j})\,\text{N m}$
- D. $(8\hat{i} + 14\hat{j})\,\text{N m}$
Show answer
Answer: C — $\boldsymbol{\tau} = \dfrac{d\mathbf{l}}{dt} \approx \dfrac{\Delta\mathbf{l}}{\Delta t} = \dfrac{(6-2)\hat{i} + (11-3)\hat{j}}{2} = \dfrac{4\hat{i} + 8\hat{j}}{2} = (2\hat{i} + 4\hat{j})\,\text{N m}$.
NCERT Class 11 Physics · Ch 6 · 6.7 Torque and Angular Momentum
For rotation about a fixed axis, the word 'initial' in $\omega_0$ and $\theta_0$ refers to values of the quantities at
- A. the instant of maximum speed
- B. $t = 0$
- C. $t = 1\ \text{s}$
- D. the instant the body stops
Show answer
Answer: B — NCERT explicitly states the word 'initial' refers to values of the quantities at $t=0$.
NCERT Class 11 Physics · Ch 6 · 6.10 Kinematics of Rotational Motion About a Fixed Axis
A cord is wound round the rim of a flywheel of radius $0.20\ \text{m}$. When $2\ \text{m}$ of the cord is unwound, the angular displacement of the wheel is
- A. $2\ \text{rad}$
- B. $10\ \text{rad}$
- C. $20\ \text{rad}$
- D. $5\ \text{rad}$
Show answer
Answer: B — Angular displacement $\theta = (\text{length of unwound string})/(\text{radius}) = 2\ \text{m} / 0.2\ \text{m} = 10\ \text{rad}$ (Example 6.12).
NCERT Class 11 Physics · Ch 6 · 6.11 Dynamics of Rotational Motion About a Fixed Axis
Two particles of masses 1 kg and 4 kg move toward each other along the x-axis with velocities +10 m/s and −5 m/s respectively. The velocity of the centre of mass is:
- A. −2 m/s (along negative x)
- B. +2 m/s (along positive x)
- C. 0 m/s
- D. −6 m/s
Show answer
Answer: A — Using Eq. (6.8) with signs: $V = (1\times10 + 4\times(-5))/(1+4) = (10-20)/5 = -10/5 = -2$ m/s, i.e. 2 m/s in the negative x-direction.
NCERT Class 11 Physics · Ch 6 · 6.3 Motion of Centre of Mass
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