Exercises
1. Give the location of the centre of mass of: (i) sphere (ii) cylinder (iii) ring (iv) cube — each of uniform mass density. Does CM necessarily lie inside the body?
Answer
For uniform mass density, the CM coincides with the geometric centre:
(i) Sphere: CM at the geometric centre of the sphere
(ii) Cylinder: CM at the midpoint of the axis (centre of the cylinder)
(iii) Ring: CM at the centre of the ring (geometric centre — not on the ring itself)
(iv) Cube: CM at the geometric centre (intersection of body diagonals)
No, the CM does not necessarily lie inside the body. For a ring, the CM is at the centre, which is empty space. Similarly for a hollow sphere or a doughnut-shaped body (torus).
2. In HCl molecule, separation between nuclei = 1.27 Å. Cl is 35.5 times as massive as H. Find CM of molecule.
Answer
Let H nucleus be at origin (x = 0); Cl nucleus at x = 1.27 Å
m_H = m; m_Cl = 35.5m
x_CM = (m_H × x_H + m_Cl × x_Cl) / (m_H + m_Cl)
= (m × 0 + 35.5m × 1.27) / (m + 35.5m)
= (35.5 × 1.27) / 36.5
= 45.085 / 36.5 = 1.235 Å from the H atom (very close to the Cl atom)
3. A child sits at one end of a trolley moving at speed V on smooth floor. Child runs about on the trolley. What is the speed of CM of (trolley + child) system?
Answer
The floor is smooth (frictionless), so there is no external horizontal force on the (trolley + child) system.
By Newton's 2nd law: F_ext = M × a_CM
⇒ 0 = M × a_CM
⇒ a_CM = 0
Therefore the CM continues with constant velocity = V (unchanged), regardless of the child's motion on the trolley.
4. Show that the area of the triangle formed between vectors a and b is half the magnitude of a × b.
Answer
The cross product |a × b| = |a||b|sinθ, where θ is the angle between vectors a and b.
This equals the area of the parallelogram formed by a and b (base × height = |a| × |b|sinθ).
Area of triangle = (1/2) × area of parallelogram = (1/2)|a × b|
5. Show that a·(b × c) equals the volume of the parallelepiped formed on vectors a, b, c.
Answer
b × c gives a vector perpendicular to the base (the parallelogram formed by b and c), with magnitude equal to the area of that base parallelogram: |b × c| = |b||c|sinθ_bc
a·(b × c) = |a| × |b × c| × cosφ, where φ is the angle between a and (b × c).
|a| cosφ = height of parallelepiped (component of a along the direction perpendicular to base).
Therefore a·(b × c) = base area × height = volume of parallelepiped
6. Find components of angular momentum l of a particle with position r = (x, y, z) and momentum p = (p_x, p_y, p_z). Show that for motion in x-y plane, l has only z-component.
Answer
Angular momentum l = r × p
l = |î ĵ k̂|
|x y z|
|pₓ p_y p_z|
lₓ = yp_z − zp_y
l_y = zp_x − xp_z
l_z = xp_y − yp_x
For motion in x-y plane: z = 0 and p_z = 0
→ lₓ = y(0) − 0(p_y) = 0
→ l_y = 0(p_x) − x(0) = 0
→ l_z = xp_y − yp_x ≠ 0 in general
Therefore l has only a z-component for motion in the x-y plane.
7. Two particles each of mass m and speed v travel in opposite directions along parallel lines separated by distance d. Show that angular momentum is same whatever point it is calculated about.
Answer
Let particle 1 travel along y = 0 in +x direction with speed v, and particle 2 along y = d in −x direction.
Choose any point O at (x₀, y₀).
l₁ = r₁ × p₁: |r₁ × p₁| = m × v × y₀ (perpendicular distance from O to line y=0) in +z direction
l₂ = r₂ × p₂: for particle 2 moving in −x direction along y = d:
|r₂ × p₂| = m × v × (d − y₀) in +z direction (since −x direction and +y₀ distance contribute positively)
Total l = mvȳ₀ + mv(d − y₀) = mvd
This is independent of y₀ (and also of x₀), confirming it's the same for any choice of reference point.
8. A non-uniform bar of weight W is suspended by two strings at angles 36.9° and 53.1° to vertical. Bar is 2 m long. Find distance d of CM from left end.
Answer
Tension in string 1 (at left, 36.9°): T₁ cos36.9° balances vertical component at left
Tension in string 2 (at right, 53.1°): T₂ cos53.1° balances vertical component at right
For equilibrium: T₁ sin36.9° = T₂ sin53.1° (horizontal balance)
T₁ × 0.6 = T₂ × 0.8 → T₁/T₂ = 4/3
Vertical: T₁ cos36.9° + T₂ cos53.1° = W
T₁ × 0.8 + T₂ × 0.6 = W → T₁ = (4/3)T₂
(4/3)T₂ × 0.8 + T₂ × 0.6 = W
⇒ T₂(1.067 + 0.6) = W
⇒ T₂ = W/1.667
T₁ = (4/3) × W/1.667 = 4W/5 (approximately); T₂ = 3W/5
Taking moments about left end (where T₁ acts):
T₂ × sin53.1° × 2 = W × d (assuming strings connect at ends)
Actually: taking torques about the left suspension point:
T₂ × (horizontal component along bar) × L − W × d = 0
d = T₂ × L / W × ... Let me use the simpler approach:
Taking moments about the left support: W × d = T₂ cosθ₂ × 2 (where the right string acts at right end)
W × d = (3W/5) × 2
⇒ d = 6/5 = 1.2 m from the left end
9. A car of mass 1800 kg has wheelbase 1.8 m. CM is 1.05 m behind front axle. Force on each front and back wheel?
Answer
Weight W = Mg = 1800 × 10 = 18000 N
Distance of CM from front axle = 1.05 m; from rear axle = 1.8 − 1.05 = 0.75 m
Taking moments about rear axle:
2F_front × 1.8 = W × 0.75
F_front (both) = 18000 × 0.75 / 1.8 = 7500 N; F_front (each) = 7500/2 = 3750 N
F_rear (both) = 18000 − 7500 = 10500 N; F_rear (each) = 10500/2 = 5250 N
10. Equal torques applied to a hollow cylinder and a solid sphere of same mass and radius. Which acquires greater angular speed after a given time?
Answer
Moment of inertia: hollow cylinder I_c = MR² (about axis); solid sphere I_s = (2/5)MR²
Since I_s < I_c, and same torque τ applied:
Angular acceleration: α = τ/I
α_s = τ/I_s = τ/[(2/5)MR²] = 5τ/(2MR²)
α_c = τ/I_c = τ/(MR²)
α_s/α_c = 5/2 = 2.5
After same time t: ω = αt
⇒ ω_s/ω_c = α_s/α_c = 5/2
The solid sphere acquires greater angular speed (2.5 times that of the hollow cylinder).
11. A solid cylinder of mass 20 kg, radius 0.25 m rotates at 100 rad s⁻¹. Find KE and angular momentum.
Answer
Moment of inertia of solid cylinder = ½MR² = ½ × 20 × (0.25)² = ½ × 20 × 0.0625 = 0.625 kg m²
Rotational KE = ½Iω² = ½ × 0.625 × (100)² = ½ × 0.625 × 10000 = 3125 J
Angular momentum L = Iω = 0.625 × 100 = 62.5 kg m² s⁻¹
12. (a) Child on turntable at 40 rev/min reduces moment of inertia to 2/5 of initial. Find new angular speed.
Answer
(b) Show new KE > initial KE. Account for increase.
Ans.
(a) By conservation of angular momentum (no friction): I₁ω₁ = I₂ω₂
I₂ = (2/5)I₁; ω₁ = 40 rev/min
ω₂ = I₁ω₁/I₂ = I₁ × 40 / (2/5)I₁ = 40 × (5/2) = 100 rev/min
(b) Initial KE = ½I₁ω₁² = ½I₁(40)²
New KE = ½I₂ω₂² = ½(2/5)I₁(100)² = ½I₁ × (2/5) × 10000 = ½I₁ × 4000
Initial KE = ½I₁ × 1600
New KE / Initial KE = 4000/1600 = 2.5
⇒ New KE = 2.5 × Initial KE
⇒ New KE > Initial KE
The increase in KE comes from the work done by the child's muscles in pulling the arms inward against the centrifugal tendency.
13. A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. Force = 30 N. Find angular acceleration and linear acceleration of rope.
Answer
Moment of inertia of hollow cylinder = MR² = 3 × (0.40)² = 3 × 0.16 = 0.48 kg m²
Torque = F × R = 30 × 0.40 = 12 N·m
Angular acceleration α = τ/I = 12/0.48 = 25 rad s⁻²
Linear acceleration of rope = αR = 25 × 0.40 = 10 m s⁻²
14. Engine maintains rotor at 200 rad s⁻¹ by transmitting torque of 180 N·m. Power required?
Answer
Power = τ × ω = 180 × 200 = 36,000 W = 36 kW
15. From a uniform disc of radius R, a circular hole of radius R/2 is cut, centered at R/2 from center. Find CM of remaining body.
Answer
Let surface density = σ; mass of full disc = σπR²
Mass of removed disc = σπ(R/2)² = σπR²/4
Mass of remaining body = σπR² − σπR²/4 = (3/4)σπR²
CM of full disc at x = 0 (center); CM of removed disc at x = R/2
Using: M_full × x_full = M_remaining × x_remaining + M_removed × x_removed
σπR² × 0 = (3σπR²/4) × x_remaining + (σπR²/4) × (R/2)
0 = (3/4)x_remaining + R/8
x_remaining = −R/6
CM of the remaining body is at R/6 from the centre, on the opposite side from the hole.
16. A metre stick balanced at 50 cm. Two coins (5 g each) placed at 12 cm mark. Stick balances at 45 cm. Find mass of stick.
Answer
Two coins at 12 cm mark: total mass = 10 g; stick balances at 45 cm (new pivot)
Taking moments about 45 cm:
10 × (45 − 12) = M × (50 − 45)
10 × 33 = M × 5
M = 330/5 = 66 g
17. Oxygen molecule: mass = 5.30 × 10⁻²⁶ kg, I = 1.94 × 10⁻⁴⁶ kg m². Mean speed = 500 m/s. KE_rotation = (2/3) KE_translation. Find average angular velocity.
Answer
KE_translation = ½mv² = ½ × 5.30 × 10⁻²⁶ × (500)² = ½ × 5.30 × 10⁻²⁶ × 2.5 × 10⁵
= 6.625 × 10⁻²¹ J
KE_rotation = (2/3) × KE_translation = (2/3) × 6.625 × 10⁻²¹ = 4.417 × 10⁻²¹ J
½Iω² = 4.417 × 10⁻²¹
ω² = 2 × 4.417 × 10⁻²¹ / (1.94 × 10⁻⁴⁶) = 8.834 × 10⁻²¹ / 1.94 × 10⁻⁴⁶
= 4.553 × 10²⁵
ω = √(4.553 × 10²⁵) = 6.75 × 10¹² rad s⁻¹
Average angular velocity ω ≈ 6.7 × 10¹² rad s⁻¹
Important Extra Questions
Q. Define moment of inertia. State the parallel axis theorem and perpendicular axis theorem.
Ans. Moment of inertia I = Σmᵢrᵢ² — the sum of the product of mass of each particle and the square of its distance from the axis of rotation. It plays the role of mass in rotational motion.
Parallel Axis Theorem: I = I_cm + Md², where I_cm is MI about an axis through CM parallel to the given axis, M is the total mass, and d is the distance between the axes.
Perpendicular Axis Theorem (for laminas): I_z = I_x + I_y, where x and y axes are in the plane of the lamina and z is perpendicular to it.
Q. Derive an expression for rotational kinetic energy.
Ans.
Consider a rigid body rotating about an axis with angular velocity ω. Each particle i at distance rᵢ from the axis has speed vᵢ = ωrᵢ.
KE = Σ½mᵢvᵢ² = Σ½mᵢ(ωrᵢ)² = ½ω² Σmᵢrᵢ² = ½Iω²
where I = Σmᵢrᵢ² is the moment of inertia.
Q. State the law of conservation of angular momentum with an example.
Ans. If the net external torque on a system is zero, the total angular momentum of the system remains constant.
Mathematically: If τ_ext = 0, then dL/dt = 0 → L = Iω = constant
Examples:
1. A figure skater spins faster when she pulls her arms in (I decreases, ω increases to keep Iω = constant).
2. A diver curls into a ball (I decreases) to spin faster, and stretches out (I increases) to slow down for entry.
Q. A flywheel of moment of inertia 4 kg m² is rotating at 240 rpm. A constant torque of 50 N·m is applied to stop it. Find time to stop and number of rotations.
Ans.
ω₀ = 240 rpm = 240 × 2π/60 = 8π rad s⁻¹
Angular retardation α = −τ/I = −50/4 = −12.5 rad s⁻²
Time to stop: ω = ω₀ + αt → 0 = 8π − 12.5t → t = 8π/12.5 ≈ 2.01 s
Rotations: θ = ω₀t + ½αt² = 8π × 2.01 + ½(−12.5)(2.01)² = 50.52 − 25.25 = 25.27 rad
Number of rotations = 25.27/(2π) ≈ 4.02 rotations
Q. What is rolling motion? Write the expression for total kinetic energy of a rolling body.
Answer
Rolling motion is a combination of translational motion of the CM and rotational motion about the CM. For a body rolling without slipping, v_CM = ωR, where R is the radius.
Total KE = KE_translational + KE_rotational = ½Mv²_CM + ½Iω²
Substituting ω = v/R: KE = ½Mv² + ½I(v/R)² = ½Mv²(1 + I/MR²)
For a solid cylinder: I = ½MR²
⇒ KE = ½Mv² × 3/2 = ¾Mv²
For a solid sphere: I = 2MR²/5
⇒ KE = ½Mv² × 7/5 = 7Mv²/10