System of Particles and Rotational Motion

What to remember

• The centre of mass moves as if all the mass were concentrated there and all external forces acted there: M a(cm) = F(ext). Internal forces cannot change its motion.
• Torque τ = r × F = I α and angular momentum L = I ω. If the net external torque is zero, L is conserved.
• Moment of inertia I = Σ m r². Key values: ring M R², disc ½ M R², solid sphere (2/5) M R², hollow sphere (2/3) M R², rod about its centre M L²/12, rod about its end M L²/3.
• Rolling without slipping: K(total) = ½ M v² (1 + k²/R²). On an incline, a = g sin θ/(1 + k²/R²), so a solid sphere wins and a ring comes last.

Centre of mass and motion of a system of particles

For particles of masses m₁, m₂, … at positions r₁, r₂, …, the centre of mass is R(cm) = (m₁ r₁ + m₂ r₂ + …)/(m₁ + m₂ + …). For two particles on a line at x₁ and x₂: x(cm) = (m₁ x₁ + m₂ x₂)/(m₁ + m₂). The centre of mass lies on the line joining the two particles and closer to the heavier one.

For a rigid body with uniform density, the centre of mass lies at the geometric centre when the body is symmetric.

Motion of the centre of mass: the velocity is V(cm) = (Σ m v)/M and the total linear momentum of the system is P = M V(cm). Newton's second law for a system: F(ext) = M a(cm) = dP/dt. Internal forces cancel in pairs. If F(ext) = 0, then P is constant and the centre of mass stays at rest or moves with constant velocity. A bomb exploding in flight: the fragments fly apart, but the centre of mass continues along the original path.

Worked example 1. Masses 2 kg at x = 0 and 3 kg at x = 5 m. x(cm) = (2 × 0 + 3 × 5)/5 = 3 m from the 2 kg mass. Check by moments about the 3 kg mass: 2 × 3 = 3 × 2 on the two sides.

Angular velocity, angular acceleration and the vector product

Vector product: A × B has magnitude A B sin θ. Its direction is given by the right-hand rule and is perpendicular to both A and B. A × B = − B × A. A × A = 0.

Rotation about a fixed axis: every particle moves in a circle about the axis with the same angular velocity ω = dθ/dt (rad s⁻¹) and the same angular acceleration α = dω/dt (rad s⁻²). The linear speed of a particle at distance r is v = ω r. The tangential acceleration is a(t) = α r, and the centripetal acceleration is ω² r. For uniform angular acceleration:

ω = ω₀ + α t; θ = ω₀ t + ½ α t²; ω² = ω₀² + 2 α θ.

These match v = u + a t, s = u t + ½ a t² and v² = u² + 2 a s with θ in place of s. Also ω = 2π/T = 2π f.

Torque, angular momentum and conservation of angular momentum

Torque τ = r × F. Its magnitude is r F sin θ = F × (perpendicular distance from the axis, the lever arm). SI unit: N m. Dimensions: [M L² T⁻²], the same as work and energy, although torque is a different physical quantity.

Angular momentum of a particle: L = r × p, with magnitude m v r sin θ. For a rigid body rotating about a fixed axis, L = I ω. Unit: kg m² s⁻¹. Dimensions: [M L² T⁻¹].

Rotational form of Newton's second law: τ = dL/dt. Conservation: if the net external torque is zero, the total angular momentum is constant. Examples:

• An ice skater spins faster when she pulls her arms in. I decreases, so ω increases to keep I ω constant. Her kinetic energy increases because she does work in pulling the arms in.
• A diver tucks the body to rotate faster in the air and stretches out to slow down before entering the water.
• A person on a rotating platform holding weights spins faster on drawing the weights in.
• A planet moves faster near the Sun (angular momentum conservation).

Worked example 2. A skater spins at 2 rev s⁻¹ with I = 3 kg m². She pulls in her arms and I becomes 1.5 kg m². New ω: I₁ ω₁ = I₂ ω₂ gives ω₂ = 3 × 2/1.5 = 4 rev s⁻¹. Kinetic energy ratio = (½ I₂ ω₂²)/(½ I₁ ω₁²) = (1.5 × 16)/(3 × 4) = 2, so the kinetic energy doubles.

Equilibrium of a rigid body and centre of gravity

A rigid body is in mechanical equilibrium when:

• 1. The net force on it is zero (translational equilibrium).
• 2. The net torque about any point is zero (rotational equilibrium).

Principle of moments: for a lever in equilibrium, the sum of the clockwise moments equals the sum of the anticlockwise moments about the pivot. For a lever: load × load arm = effort × effort arm. Mechanical advantage = load/effort = effort arm/load arm.

A couple is a pair of equal and opposite forces with different lines of action. It gives a torque F × d (d is the distance between the lines) and no net force.

Centre of gravity is the point where the total weight of the body can be taken to act. In a uniform gravitational field it coincides with the centre of mass. A body is in stable equilibrium if its centre of gravity is lowest, or if a vertical line from it falls inside the base.

Worked example 3. A uniform metre rod of 1 kg rests on a pivot at the 40 cm mark. A load of 2 kg is hung at the 10 cm mark. Take g = 10 m s⁻². Where must a mass of 3 kg hang to balance? The rod's weight acts at 50 cm, 10 cm to the right of the pivot, and gives 10 × 0.1 = 1 N m clockwise. The load gives 20 × 0.3 = 6 N m anticlockwise. Net anticlockwise = 5 N m. The 3 kg mass (30 N) at distance d to the right: 30 d = 5, d = 1/6 m ≈ 16.7 cm, so it hangs at the 56.7 cm mark.

Moment of inertia and the two theorems

Moment of inertia I = Σ m r² (kg m²). It is the rotational analogue of mass. It depends on the mass, on how the mass is distributed and on the position of the axis. Radius of gyration k is defined by I = M k², so k = √(I/M).

Theorem of parallel axes: I = I(cm) + M d². Here I(cm) is the moment of inertia about an axis through the centre of mass and d is the distance between the two parallel axes. It holds for any body. Check: rod about its end = M L²/12 + M (L/2)² = M L²/3.

Theorem of perpendicular axes: for a plane (flat, laminar) body, I(z) = I(x) + I(y), where x and y are two perpendicular axes in the plane and z is perpendicular to the plane through the same point. Check: for a ring, I(z) = M R² gives I(diameter) = ½ M R².

Rotation about a fixed axis, work and power

Linear and rotational quantities correspond as follows.

Rotational kinetic energy K = ½ I ω² = L²/(2 I). Work done by a torque: W = τ θ. Power: P = τ ω. A flywheel has a large I, so it stores energy and keeps the rotation steady.

Worked example 4. A wheel of I = 0.5 kg m² is acted on by a constant torque of 10 N m. Then α = τ/I = 20 rad s⁻². After 3 s from rest, ω = 60 rad s⁻¹. Rotational kinetic energy = ½ × 0.5 × 3600 = 900 J. By the work-energy theorem: θ = ½ α t² = 90 rad and W = τ θ = 10 × 90 = 900 J.

Rolling motion

Rolling without slipping: the point of contact is momentarily at rest, so v(cm) = ω R. Total kinetic energy = translational + rotational = ½ M v² + ½ I ω² = ½ M v² (1 + k²/R²).

For a body of mass M and radius R rolling from rest down an incline of angle θ without slipping:

• Acceleration a = g sin θ/(1 + k²/R²).
• Speed at the bottom after descending a vertical height h: v = √[2 g h/(1 + k²/R²)].
• Friction (static) acts up the incline and makes the body roll. It does no net work and no energy is lost to heat in pure rolling.

Order of acceleration and of speed at the bottom: solid sphere > solid disc (or solid cylinder) > hollow sphere > ring (or hollow cylinder). This order does not depend on mass or radius. A smooth block (no rotation) sliding down a frictionless incline reaches the bottom first, with a = g sin θ.

Formula summary

NEET traps

• The centre of mass of a ring or a hollow sphere lies at the centre, where there is no mass. The centre of mass need not lie inside the body.
• Internal forces (such as in an explosion) do not change the motion of the centre of mass.
• Torque and work have the same dimensions but are different quantities. Torque is a vector in N m; work is a scalar in J.
• Angular momentum is conserved only when the net external torque is zero, not when the net force is zero.
• The moment of inertia depends on the axis. Always check which axis is given.
• The parallel axes theorem needs I about the centre of mass. Using an end axis as the starting point gives a wrong answer.
• The perpendicular axes theorem applies only to flat plane bodies (laminas), not to a solid sphere or a cylinder.
• For a rolling body the friction is static. It does not dissipate energy and it does not oppose the motion in a way that heats the surface.
• On the same incline, a ring is slower than a disc, and a solid sphere is faster than a hollow sphere. The mass and radius do not matter.
• A skater's kinetic energy increases when she pulls in her arms, even though her angular momentum stays the same.

One-liners

• The centre of mass of a uniform rod is at its midpoint.
• Total momentum of a system = M × velocity of the centre of mass.
• v = ω r links linear and angular speed.
• The SI unit of torque is N m.
• τ = dL/dt.
• L = I ω.
• With zero external torque, angular momentum is constant.
• A diver tucks to spin faster because I decreases.
• The principle of moments: clockwise moments = anticlockwise moments.
• A couple produces rotation without a net force.
• I of a ring about its axis is M R².
• I of a solid sphere about a diameter is (2/5) M R².
• I of a rod about its end is M L²/3.
• Rotational kinetic energy = ½ I ω².
• Among ring, disc, hollow sphere and solid sphere rolling down a slope, the solid sphere is fastest.

Practice questions

1. A 1 kg particle is at x = 0 and a 4 kg particle is at x = 10 m. The centre of mass of the two is at x =

   1. 5 m
   2. 8 m
   3. 4 m
   4. 2 m
   Answer

   B. 8 m

   x(cm) = (1 × 0 + 4 × 10)/5 = 8 m. Check by moments: 1 × 8 = 4 × 2.

2. A force of 10 N acts at a point 0.3 m from the axis of rotation, at 30° to the line joining the point to the axis. The torque about the axis is

   1. 0.75 N m
   2. 3 N m
   3. 2.6 N m
   4. 1.5 N m
   Answer

   D. 1.5 N m

   τ = r F sin θ = 0.3 × 10 × 0.5 = 1.5 N m.

3. A flywheel of moment of inertia 2 kg m² rotates at 5 rad s⁻¹. Its angular momentum is

   1. 50 kg m² s⁻¹
   2. 2.5 kg m² s⁻¹
   3. 10 kg m² s⁻¹
   4. 25 kg m² s⁻¹
   Answer

   C. 10 kg m² s⁻¹

   L = I ω = 2 × 5 = 10 kg m² s⁻¹.

4. The moment of inertia of a thin ring of mass 2 kg and radius 0.5 m about an axis through its centre perpendicular to its plane is

   1. 0.5 kg m²
   2. 1.0 kg m²
   3. 0.125 kg m²
   4. 0.25 kg m²
   Answer

   A. 0.5 kg m²

   I = M R² = 2 × 0.25 = 0.5 kg m². Check: k = R, so I = M k².

5. The moment of inertia of a uniform rod of mass 3 kg and length 2 m about an axis through one end and perpendicular to the rod is

   1. 12 kg m²
   2. 1 kg m²
   3. 6 kg m²
   4. 4 kg m²
   Answer

   D. 4 kg m²

   I = M L²/3 = 3 × 4/3 = 4 kg m². Check by parallel axes: 3 × 4/12 + 3 × 1² = 1 + 3 = 4.

6. A uniform disc of mass 2 kg and radius 0.5 m is rotated about an axis through a point on its rim, perpendicular to its plane. The moment of inertia is

   1. 1.5 kg m²
   2. 0.5 kg m²
   3. 0.75 kg m²
   4. 0.25 kg m²
   Answer

   C. 0.75 kg m²

   I = ½ M R² + M R² = 1.5 × 2 × 0.25 = 0.75 kg m².

7. A skater spins at 2 rad s⁻¹ with a moment of inertia of 6 kg m². She pulls in her arms and her moment of inertia becomes 1.5 kg m². Her angular speed becomes

   1. 8 rad s⁻¹
   2. 2 rad s⁻¹
   3. 0.5 rad s⁻¹
   4. 4 rad s⁻¹
   Answer

   A. 8 rad s⁻¹

   I₁ω₁ = I₂ω₂ gives ω₂ = 6 × 2/1.5 = 8 rad s⁻¹.

8. A constant torque of 20 N m acts on a wheel of moment of inertia 5 kg m². The angular acceleration is

   1. 0.25 rad s⁻²
   2. 4 rad s⁻²
   3. 15 rad s⁻²
   4. 100 rad s⁻²
   Answer

   B. 4 rad s⁻²

   α = τ/I = 20/5 = 4 rad s⁻².

9. A wheel of moment of inertia 0.4 kg m² rotates at 10 rad s⁻¹. Its rotational kinetic energy is

   1. 20 J
   2. 2 J
   3. 4 J
   4. 40 J
   Answer

   A. 20 J

   K = ½ I ω² = ½ × 0.4 × 100 = 20 J.

10. A motor exerts a torque of 50 N m on a shaft rotating at 20 rad s⁻¹. The power delivered is

    1. 100 W
    2. 2.5 W
    3. 1000 W
    4. 70 W
    Answer

    C. 1000 W

    P = τ ω = 50 × 20 = 1000 W.

11. A solid sphere of mass 5 kg rolls without slipping at a speed of 2 m s⁻¹ on a level floor. Its total kinetic energy is

    1. 4 J
    2. 14 J
    3. 10 J
    4. 20 J
    Answer

    B. 14 J

    K = ½ M v² (1 + 2/5) = ½ × 5 × 4 × 1.4 = 14 J. Check: translational 10 J + rotational 4 J.

12. A solid disc rolls without slipping down an incline for which sin θ = 0.6. Taking g = 10 m s⁻², its acceleration is

    1. 3 m s⁻²
    2. 2.4 m s⁻²
    3. 6 m s⁻²
    4. 4 m s⁻²
    Answer

    D. 4 m s⁻²

    a = g sin θ/(1 + ½) = 6/1.5 = 4 m s⁻². Check: (2/3) × 6 = 4.

13. A ring starts from rest and rolls without slipping down an incline through a vertical height of 3.6 m. Taking g = 10 m s⁻², its speed at the bottom is

    1. 6 m s⁻¹
    2. 12 m s⁻¹
    3. 6√2 m s⁻¹
    4. 3 m s⁻¹
    Answer

    A. 6 m s⁻¹

    M g h = ½ M v² + ½ M v² = M v², so v² = g h = 36 and v = 6 m s⁻¹.

14. The radius of gyration of a thin ring of radius R about an axis through its centre perpendicular to its plane is

    1. R/√2
    2. R
    3. R/2
    4. 2R
    Answer

    B. R

    I = M R² = M k², so k = R.

15. For a flat plate, the moments of inertia about two perpendicular axes in its plane through a point are 3 kg m² and 5 kg m². The moment of inertia about the axis through the same point perpendicular to the plate is

    1. 4 kg m²
    2. 2 kg m²
    3. 15 kg m²
    4. 8 kg m²
    Answer

    D. 8 kg m²

    Perpendicular axes theorem: I(z) = 3 + 5 = 8 kg m².

16. A uniform lever is balanced when a 20 N load hangs 30 cm from the pivot on one side and a force W acts 10 cm from the pivot on the other side. The value of W is

    1. 600 N
    2. 20 N
    3. 60 N
    4. 6.7 N
    Answer

    C. 60 N

    Principle of moments: 20 × 30 = W × 10 gives W = 60 N.

17. A bomb at rest explodes into two pieces of 1 kg and 3 kg. The 3 kg piece moves at 2 m s⁻¹ east. The speed of the 1 kg piece is

    1. 2 m s⁻¹
    2. 12 m s⁻¹
    3. 6 m s⁻¹
    4. 3 m s⁻¹
    Answer

    C. 6 m s⁻¹

    Momentum conservation: 1 × v = 3 × 2, so v = 6 m s⁻¹ west. The centre of mass stays at rest.

18. Two particles of masses 2 kg and 4 kg are joined by a light rod. An external force of 12 N acts on the system. The acceleration of the centre of mass is

    1. 2 m s⁻²
    2. 3 m s⁻²
    3. 1 m s⁻²
    4. 6 m s⁻²
    Answer

    A. 2 m s⁻²

    a(cm) = F(ext)/M = 12/6 = 2 m s⁻². Internal forces do not matter.

19. A ring and a solid disc of the same mass and radius are released together from rest at the top of the same incline and roll without slipping. Which one reaches the bottom first?

    1. The ring
    2. Both at the same time
    3. The disc
    4. It depends on the common mass
    Answer

    C. The disc

    Disc: a = (2/3) g sin θ. Ring: a = (1/2) g sin θ. Larger a gives a shorter time.

20. What percentage of its total kinetic energy is rotational for a thin-walled hollow sphere rolling without slipping?

    1. 60%
    2. 40%
    3. 29%
    4. 50%
    Answer

    B. 40%

    Rotational fraction = (k²/R²)/(1 + k²/R²) = (2/3)/(5/3) = 2/5 = 40%.

21. A uniform rod of mass 3 kg and length 1.5 m can rotate in a vertical plane about a horizontal axis through one end. It is released from the horizontal position. Taking g = 10 m s⁻², its initial angular acceleration is

    1. 5 rad s⁻²
    2. 20 rad s⁻²
    3. 40 rad s⁻²
    4. 10 rad s⁻²
    Answer

    D. 10 rad s⁻²

    τ = m g L/2 = 3 × 10 × 0.75 = 22.5 N m; I = m L²/3 = 2.25 kg m²; α = 10 rad s⁻². Check: 3g/2L = 30/3 = 10.

22. A wheel starts from rest and gains a uniform angular acceleration so that its angular velocity is 20 rad s⁻¹ after 5 s. The angle through which it turns in these 5 s is

    1. 200 rad
    2. 50 rad
    3. 25 rad
    4. 100 rad
    Answer

    B. 50 rad

    α = 4 rad s⁻²; θ = ½ α t² = 50 rad. Check: θ = average ω × t = 10 × 5 = 50.

23. A planet moves round the Sun. Its nearest distance from the Sun is one-fourth of its farthest distance. The ratio of its speed at the nearest point to that at the farthest point is

    1. 4 : 1
    2. 1 : 4
    3. 2 : 1
    4. 16 : 1
    Answer

    A. 4 : 1

    L = m v r is conserved, so v ∝ 1/r and the ratio is 4 : 1.

24. Given below are two statements: Statement I: The centre of mass of a uniform ring lies at its centre. Statement II: There is some mass of the ring at its centre.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    The centre of mass can lie at a point where there is no mass.

25. Given below are two statements: Statement I: Internal forces can change the velocity of the centre of mass of a system. Statement II: Only external forces can change the motion of the centre of mass.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Internal forces cancel in pairs.

26. Which of the following statements are correct? 1. Torque and work have the same dimensions. 2. Torque is a scalar. 3. The unit of angular momentum is kg m² s⁻¹.

    1. 1 and 3 only
    2. 1 and 2 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    A. 1 and 3 only

    Torque is a vector, so statement 2 is wrong.

27. Given below are two statements: Statement I: Angular momentum is conserved when the net external torque is zero. Statement II: Angular momentum is conserved when the net external force is zero.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Zero net force does not mean zero net torque, as in a couple.

28. Given below are two statements: Statement I: The perpendicular axes theorem can be applied to any three-dimensional body. Statement II: The parallel axes theorem holds for any body.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    The perpendicular axes theorem is for plane laminas only.

29. Which of the following statements are correct? 1. The moment of inertia of a ring about its axis is M R². 2. The moment of inertia of a solid sphere about a diameter is (2/5) M R². 3. The moment of inertia of a rod about its end is M L²/12.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1 and 3 only
    4. 1, 2 and 3
    Answer

    A. 1 and 2 only

    For a rod, M L²/3 is about the end and M L²/12 is about the centre.

30. Given below are two statements: Statement I: A solid sphere rolls down an incline faster than a hollow sphere of the same mass and radius. Statement II: A solid sphere has a smaller k²/R² than a hollow sphere.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Acceleration = g sin θ/(1 + k²/R²), so a smaller k²/R² gives a larger acceleration.

31. Given below are two statements: Statement I: Static friction in pure rolling dissipates energy as heat. Statement II: The point of contact of a rolling body is instantaneously at rest.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Because the contact point is at rest, static friction does no work and produces no heat.

32. Given below are two statements: Statement I: When a skater pulls in her arms, her angular momentum increases. Statement II: When a skater pulls in her arms, her kinetic energy decreases.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Angular momentum stays the same; kinetic energy increases because she does work.

33. Given below are two statements: Statement I: The moment of inertia of a body is the same about all axes. Statement II: The moment of inertia depends on how the mass is distributed about the axis.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    I = Σ m r² depends on the axis chosen.

34. Assertion (A): A figure skater spins faster when she pulls her arms in. Reason (R): The angular momentum of the skater remains constant and her moment of inertia decreases.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    I ω = constant, so a smaller I gives a larger ω.

35. Assertion (A): The perpendicular axes theorem can be applied to a thin circular disc. Reason (R): The parallel axes theorem states that I = I(cm) + M d².

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both statements are true, but the second does not explain the first.

36. Assertion (A): The centre of mass of a ring lies at its centre. Reason (R): The whole mass of the ring is concentrated at its centre.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    The mass is on the rim; the centre of mass is at the centre by symmetry.

37. Assertion (A): A diver tucks her body to spin faster in the air. Reason (R): Tucking in increases the moment of inertia.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    Tucking decreases I, so ω increases.

38. Assertion (A): Torque is a vector quantity. Reason (R): The dimensions of torque are [M L² T⁻²].

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Torque is r × F, a vector. The dimensional formula is true but does not explain the vector nature.

39. Match List I with List II. List I: a. Ring about its axis b. Disc about its axis c. Solid sphere about a diameter d. Hollow sphere about a diameter. List II: i. (2/5) M R² ii. ½ M R² iii. (2/3) M R² iv. M R²

    1. a-ii, b-iv, c-i, d-iii
    2. a-i, b-ii, c-iv, d-iii
    3. a-iv, b-ii, c-iii, d-i
    4. a-iv, b-ii, c-i, d-iii
    Answer

    D. a-iv, b-ii, c-i, d-iii

    Standard values.

40. Match List I with List II. List I: a. Torque b. Angular momentum c. Angular velocity d. Moment of inertia. List II: i. [M L²] ii. [T⁻¹] iii. [M L² T⁻²] iv. [M L² T⁻¹]

    1. a-iii, b-iv, c-i, d-ii
    2. a-iii, b-iv, c-ii, d-i
    3. a-i, b-iv, c-ii, d-iii
    4. a-iv, b-iii, c-ii, d-i
    Answer

    B. a-iii, b-iv, c-ii, d-i

    Torque [M L² T⁻²]; L [M L² T⁻¹]; ω [T⁻¹]; I [M L²].

41. Match List I with List II. The List II values give the acceleration of each rolling body down an incline as a fraction of g sin θ. List I: a. Solid sphere b. Solid disc c. Hollow sphere d. Ring. List II: i. 1/2 ii. 3/5 iii. 5/7 iv. 2/3

    1. a-iv, b-iii, c-ii, d-i
    2. a-iii, b-iv, c-i, d-ii
    3. a-iii, b-iv, c-ii, d-i
    4. a-ii, b-iv, c-iii, d-i
    Answer

    C. a-iii, b-iv, c-ii, d-i

    a/(g sin θ) = 1/(1 + k²/R²): 5/7, 2/3, 3/5 and 1/2.

42. Match List I with List II (rotational analogues). List I: a. Mass b. Force c. Linear momentum d. ½ m v². List II: i. Torque ii. I ω iii. Moment of inertia iv. ½ I ω²

    1. a-iii, b-i, c-iv, d-ii
    2. a-i, b-iii, c-ii, d-iv
    3. a-iii, b-ii, c-i, d-iv
    4. a-iii, b-i, c-ii, d-iv
    Answer

    D. a-iii, b-i, c-ii, d-iv

    Mass ↔ I; force ↔ torque; momentum ↔ angular momentum; translational K ↔ rotational K.

43. Match List I with List II. List I: a. Moment of inertia of a 2 kg ring of radius 0.5 m about its axis b. Torque of a 10 N force acting perpendicular to an arm of 0.2 m c. Angular momentum for I = 3 kg m² and ω = 4 rad s⁻¹ d. Kinetic energy for I = 2 kg m² and ω = 3 rad s⁻¹. List II: i. 12 kg m² s⁻¹ ii. 9 J iii. 0.5 kg m² iv. 2 N m

    1. a-iii, b-iv, c-i, d-ii
    2. a-iii, b-iv, c-ii, d-i
    3. a-iv, b-iii, c-i, d-ii
    4. a-i, b-iv, c-iii, d-ii
    Answer

    A. a-iii, b-iv, c-i, d-ii

    0.5 kg m²; 2 N m; 12 kg m² s⁻¹; ½ × 2 × 9 = 9 J.

44. Match List I with List II. List I: a. Uniform rod b. Ring c. Solid sphere d. A 1 kg and a 3 kg particle 4 m apart. List II: i. 3 m from the 1 kg particle ii. at its midpoint iii. at the centre, where there is no mass iv. at the centre of the sphere

    1. a-ii, b-iii, c-iv, d-i
    2. a-iii, b-ii, c-iv, d-i
    3. a-ii, b-iii, c-i, d-iv
    4. a-iv, b-iii, c-ii, d-i
    Answer

    A. a-ii, b-iii, c-iv, d-i

    (1 × 0 + 3 × 4)/4 = 3 m from the 1 kg particle.