Class 11 Physics · Original EZT practice
Systems of Particles and Rotational Motion (Class 11 Chapter 6): summary and practice
Rotational motion re-runs the whole of mechanics in angular form: the centre of mass plays the role of the particle, torque plays the role of force, moment of inertia plays the role of mass, and angular momentum is conserved just like linear momentum. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Centre of mass
The centre of mass x_cm = (sum of m x)/(sum of m) is the mass-weighted average position. It moves as if all the mass and all external forces sat there - internal forces between particles never move it.
2. Torque and angular momentum
Torque is the rotational push: tau = r F sin theta, largest when the force is perpendicular to the lever arm. Angular momentum L = I omega plays the role of p = mv, and it is conserved when no external torque acts.
3. Equilibrium of a rigid body
A rigid body balances when net force AND net torque both vanish. The principle of moments (6.8.1) balances a lever: load x load arm equals effort x effort arm. The centre of gravity (6.8.2) is where the whole weight effectively acts.
4. Moment of inertia
I = sum of m r^2 measures how mass is distributed about the axis - the rotational analogue of mass. Table 6.1 lists the results: ring MR^2, disc about its central perpendicular axis (1/2)MR^2, solid sphere (2/5)MR^2. Radius of gyration k writes I = Mk^2.
5. Rotation about a fixed axis
The kinematics mirrors straight-line motion: omega = omega0 + alpha t and friends. The dynamics: torque = I alpha, work becomes torque x angle, rotational KE = (1/2) I omega^2, and L = I omega with its conservation law.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Locate a centre of mass
Choose the method: x_cm = (sum of m x)/(sum of m).
EZT practice question
A 1 kg mass sits at x = 0 and a 3 kg mass at x = 4 m on a light rod. Where is the centre of mass?
Show the worked solution
- x_cm = (1 x 0 + 3 x 4)/(1 + 3).
- = 12/4.
Answer: x_cm = 3 m from the 1 kg mass.
Check: The heavier mass pulls the centre closer to itself: 3 m is nearer to x = 4 than to x = 0.
Watch for this: Do not average the positions - average them weighted by mass.
Now try it yourself
Two 2 kg masses sit at x = 0 and x = 6 m. Centre of mass?
Check your answer
3 m.
Equal masses balance at the midpoint: (0 + 12)/4 = 3 m.
Compute a torque
Choose the method: tau = r F sin theta.
EZT practice question
A 20 N force pushes at 0.5 m from a hinge, perpendicular to the door. Find the torque about the hinge.
Show the worked solution
- tau = r F sin theta.
- Perpendicular means sin theta = 1, so tau = 0.5 x 20.
Answer: tau = 10 N m.
Check: Bigger lever arm or bigger force means more turning effect - 10 N m is sensible for a firm door push.
Watch for this: Forgetting sin theta: a force along the lever arm (theta = 0) produces no torque at all.
Now try it yourself
A 5 N force acts at 0.4 m from the axis, perpendicular to the lever. Torque?
Check your answer
2 N m.
tau = 0.4 x 5 x 1 = 2 N m.
Use the disc moment of inertia
Choose the method: For a disc about its central perpendicular axis, I = (1/2) M R^2 (Table 6.1).
EZT practice question
A uniform disc of mass 2 kg and radius 0.5 m spins about its central axis. Find its moment of inertia.
Show the worked solution
- I = (1/2) M R^2.
- I = 0.5 x 2 x 0.25.
Answer: I = 0.25 kg m^2.
Check: A ring of the same mass and radius would give MR^2 = 0.5 kg m^2 - the disc is easier to spin because its mass sits closer in.
Watch for this: Match the formula to the axis: about a DIAMETER the disc gives (1/4)MR^2 instead.
Now try it yourself
Find I for a ring of mass 0.5 kg and radius 0.2 m about its central axis.
Check your answer
0.02 kg m^2.
Ring: I = MR^2 = 0.5 x 0.04 = 0.02 kg m^2.
Spin up with torque = I alpha
Choose the method: alpha = tau/I, then L = I omega and KE = (1/2) I omega^2.
EZT practice question
The 2 kg disc (I = 0.25 kg m^2) feels a steady 1 N m torque. Find its angular acceleration, and at omega = 4 rad/s find L and the rotational kinetic energy.
Show the worked solution
- alpha = 1/0.25 = 4 rad/s^2.
- L = I omega = 0.25 x 4 = 1 kg m^2/s.
- KE = (1/2) x 0.25 x 16 = 2 J.
Answer: alpha = 4 rad/s^2; L = 1 kg m^2/s; KE = 2 J at omega = 4 rad/s.
Check: The torque needs 1 second to add 4 rad/s of spin - consistent with alpha = 4.
Watch for this: Keep angular and linear quantities apart: alpha pairs with torque and I, not with F and m.
Now try it yourself
Starting from rest with alpha = 4 rad/s^2, what is omega after 3 s?
Check your answer
12 rad/s.
omega = omega0 + alpha t = 0 + 4 x 3 = 12 rad/s.
Conserve angular momentum
Choose the method: No external torque means I omega stays constant.
EZT practice question
A skater spins with arms out and then pulls them in, halving her moment of inertia. What happens to her spin rate?
Show the worked solution
- I omega is constant (no external torque about the axis).
- If I halves, omega must double to keep the product fixed.
Answer: She spins twice as fast.
Check: This is the chapter's own classic: arms in, faster spin; arms out, slower.
Watch for this: Her rotational KE actually rises (KE = L^2/2I) - the extra energy comes from the work she does pulling her arms in.
Now try it yourself
If instead she extends her arms to make I three times larger, the spin becomes?
Check your answer
One third of the original.
omega scales as 1/I when L is fixed.
Balance a lever
Choose the method: Balance torques: m1 g d1 = m2 g d2.
EZT practice question
A 30 kg child sits 2 m from the pivot of a seesaw. Where must a 40 kg friend sit on the other side to balance?
Show the worked solution
- Balance moments: 30 x 2 = 40 x d2.
- 60 = 40 d2.
Answer: d2 = 1.5 m on the other side.
Check: The heavier friend sits closer - 40 x 1.5 = 60 balances 30 x 2.
Watch for this: g cancels from both sides - do not drag 9.8 into a balance problem.
Now try it yourself
A 50 kg person at 1.2 m balances a child at 3 m. Child's mass?
Check your answer
20 kg.
50 x 1.2 = m x 3 gives m = 60/3 = 20 kg.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Centre of mass formula?
Show answer and reason
x_cm = (sum of m x)/(sum of m).
Mass-weighted average position.
2. Torque formula?
Show answer and reason
tau = r F sin theta.
Only the perpendicular part of the force turns the body.
3. Angular momentum of a rotating body?
Show answer and reason
L = I omega.
The rotational analogue of p = mv.
4. Rotational kinetic energy?
Show answer and reason
KE = (1/2) I omega^2.
I and omega replace m and v.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Two 2 kg masses sit at x = 0 and x = 6 m. Find the centre of mass.
- A 5 N force acts at 0.4 m from the axis, perpendicular to the lever. Find the torque.
- Find the moment of inertia of a 0.5 kg ring of radius 0.2 m about its central axis.
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Questions about this practice page
- What is the centre of mass?
- The mass-weighted average position of a body or system. It moves as if all the mass were concentrated there with all external forces applied at that point - internal forces never shift it.
- What is torque?
- The turning effect of a force: tau = r F sin theta, where r is the distance from the axis and theta the angle between the lever arm and the force. Maximum when the force is perpendicular, zero when it points along the arm.
- Why does a skater spin faster when pulling her arms in?
- Angular momentum L = I omega is conserved when no external torque acts. Pulling the arms in reduces I, so omega rises to keep the product constant - the chapter's classic demonstration.
- Does this reprint cover rolling motion?
- Rolling motion appears only as a motivating example in the introduction of the Reprint 2026-27 chapter - the formal sections run 6.1 to 6.12 and end with angular momentum in fixed-axis rotation. There is no separate rolling section in this reprint.
- Are these the NCERT exercise questions for Rotational Motion (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.