Class 11 - Unit # 04 : Rotational & Circular Motion - Solved Numericals

 

1. Torque Applied by a Wrench

A car mechanic applies a force of 800 N to a wrench for the purpose of loosening a bolt. He applies the force perpendicular to the arm of the wrench. The distance from the bolt to the mechanic's hand is 0.40 m. Find out the magnitude of the torque applied.

Data:

• Force Applied (F) = 800 N
• Moment Arm / Distance (r) = 0.40 m (Note: Corrected from typographical 'in' to 'm' to align with standard metric units)
• Angle (θ) = 90° (Perpendicular force implies sin 90° = 1)
• Torque (τ) = ?

Solution:

According to the definition of Torque: τ = F × r × sin(θ)

Substitute values: τ = 800 × 0.40 × sin(90°)

Calculate value: τ = 320 × 1 = 320 N·m

Result: The magnitude of the torque applied is 320 N·m.

2. Rotational Mechanics of a Rolling Tire

A car accelerates uniformly from rest and reaches a speed of 22 m/s in 9 s. If the diameter of a tire is 58 cm, find (a) the number of revolutions the tire makes during this motion, assuming no slipping, and (b) the final rotational speed of the tire in revolutions per second.

Data:

• Initial Velocity (v_i) = 0 m/s (from rest)
• Final Velocity (v_f) = 22 m/s
• Time elapsed (t) = 9 s
• Diameter of Tire (d) = 58 cm = 0.58 m ⇒ Radius (r) = d / 2 = 0.29 m
• (a) Number of Revolutions (N) = ?
• (b) Final Rotational Speed (f_f) = ?

Solution:

(a) For Total Distance and Revolutions:

Average Linear Velocity: v_avg = (v_i + v_f) / 2 = (0 + 22) / 2 = 11 m/s

Total Linear Distance: S = v_avg × t = 11 × 9 = 99 m

Circumference of the tire: C = 2 × π × r = 2 × 3.1416 × 0.29 ≈ 1.822 m

Number of Revolutions: N = S / C = 99 / 1.822 ≈ 54.33 rev

(b) For Final Rotational Speed:

Final Linear Velocity: v_f = r × ω_f ⇒ Final Angular Velocity: ω_f = v_f / r

ω_f = 22 / 0.29 ≈ 75.86 rad/s

Convert to rev/s: f_f = ω_f / 2π = 75.86 / (2 × 3.1416) ≈ 12.07 rev/s

Result: The tire makes approximately 54 revolutions and its final rotational speed is 12 rev/s.

3. Kinematics of a Workshop Grindstone

An ordinary workshop grindstone has a radius of 7.5 cm and rotates at 6500 rev/min. (a) Calculate the magnitude of centripetal acceleration at its edge in m/s² and convert it into multiples of g. (b) What is the linear speed of a point on its edge?

Data:

• Radius (r) = 7.5 cm = 0.075 m
• Rotational Speed (f) = 6500 rev/min = 6500 / 60 ≈ 108.33 rev/s
• Angular Velocity (ω) = 2 × π × f = 2 × 3.1416 × 108.33 ≈ 680.68 rad/s
• (a) Centripetal Acceleration (a_c) = ?
• (b) Linear Speed (v) = ?

Solution:

(b) For Linear Speed first (simplifies part a):

v = r × ω = 0.075 × 680.68 ≈ 51.05 m/s

(a) For Centripetal Acceleration:

a_c = v² / r = (51.05)² / 0.075 ≈ 34748 m/s² (Alternatively: a_c = rω² = 0.075 × 680.68² ≈ 34749 m/s²)

In multiples of g (taking g = 9.8 m/s²): a_c (in g) = 34749 / 9.8 ≈ 3546g

Result: The centripetal acceleration is 3.47 × 10⁴ m/s² (3546g) and the linear speed is 51.05 m/s.

4. Orbital Radius of a Satellite

A satellite is orbiting the Earth with an orbital velocity of 3200 m/s. What is its orbital radius?

Data:

• Orbital Velocity (v) = 3200 m/s
• Mass of Earth (M) = 6.0 × 10²⁴ kg
• Gravitational Constant (G) = 6.67 × 10⁻¹¹ N·m²/kg²
• Orbital Radius (r) = ?

Solution:

The formula for orbital velocity is: v = √(G × M / r)

Squaring both sides: v² = G × M / r

Rearranging for r: r = G × M / v²

Substitute values: r = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (3200)²

r = (4.002 × 10¹⁴) / (1.024 × 10⁷)

Calculate value: r ≈ 3.91 × 10⁷ m (or 39,100 km)

Result: The orbital radius of the satellite is 3.91 × 10⁷ m.

5. Low Earth Orbit Satellite Parameters

A satellite orbits the Earth at a height of 100 km above the surface of the Earth. Determine the speed, acceleration, and orbital period of the satellite.

Data:

• Height above Earth surface (h) = 100 km = 100,000 m = 1 × 10⁵ m
• Radius of Earth (R) = 6.4 × 10⁶ m
• Orbital Radius (r) = R + h = (64 × 10⁵) + (1 × 10⁵) = 6.5 × 10⁶ m
• Mass of Earth (M) = 6.0 × 10²⁴ kg
• Gravitational Constant (G) = 6.67 × 10⁻¹¹ N·m²/kg²

Solution:

1. Speed of the Satellite (v):

v = √(G × M / r)

v = √[(6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (6.5 × 10⁶)] = √[4.002 × 10¹⁴ / (6.5 × 10⁶)]

v = √[6.157 × 10⁷] ≈ 7846.6 m/s

2. Centripetal Acceleration (a_c):

a_c = v² / r = (7846.6)² / (6.5 × 10⁶) ≈ 9.47 m/s²

3. Orbital Period (T):

T = (2 × π × r) / v

T = (2 × 3.1416 × 6.5 × 10⁶) / 7846.6 ≈ 40841 / 7846.6 ≈ 5205 s

In hours: T = 5205 / 3600 ≈ 1.45 hours

Result: The orbital speed is 7846.6 m/s, the acceleration is 9.47 m/s², and the time period is 1.45 hours.

6. Rotational Inertia of a Composite System

A thin disk with a 0.3 m diameter and a total moment of inertia of 0.45 kg·m² is rotating about its center of mass. There are three rocks with masses of 0.2 kg on the outer part of the disk. Find the total moment of inertia of the system.

Data:

• Moment of inertia of disk (I_disk) = 0.45 kg·m²
• Diameter of disk (d) = 0.3 m ⇒ Radius (r) = 0.15 m
• Mass of each rock (m) = 0.2 kg
• Number of rocks (n) = 3
• Total Moment of Inertia (I_total) = ?

Solution:

Treating the rocks as point particles fixed at distance r on the edge:

Moment of inertia of a single rock: I_rock = m × r²

I_rock = 0.2 × (0.15)² = 0.2 × 0.0225 = 0.0045 kg·m²

Total moment of inertia for 3 rocks: I_rocks = 3 × 0.0045 = 0.0135 kg·m²

Combined system total: I_total = I_disk + I_rocks

I_total = 0.45 + 0.0135 = 0.4635 kg·m²

Result: The total moment of inertia of the system is 0.4635 kg·m².

7. Ideal Highway Curve Banking Angle

What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit, assuming everyone travels at the limit?

Data:

• Radius of curve (r) = 1.20 km = 1200 m
• Speed Limit (v) = 105 km/h = 105 × (1000 / 3600) ≈ 29.17 m/s
• Acceleration due to gravity (g) = 9.8 m/s²
• Banking Angle (θ) = ?

Solution:

The formula for the ideal banking angle is: tan(θ) = v² / (r × g)

Substitute the values: tan(θ) = (29.17)² / (1200 × 9.8)

tan(θ) = 850.89 / 11760 ≈ 0.07235

Taking inverse tangent: θ = tan⁻¹(0.07235) ≈ 4.14°

Result: The ideal banking angle for the highway turn is 4.14°.

8. Maximum Safe Speed on a Flat Circular Turn

A 1500-kg car moving on a flat, horizontal road negotiates a curve. If the radius of the curve is 35.0 m and the coefficient of static friction between the tires and dry pavement is 0.523, find the maximum speed the car can have and still make the turn successfully.

Data:

• Mass of the car (m) = 1500 kg
• Radius of curve (r) = 35.0 m
• Coefficient of static friction (μ_s) = 0.523
• Acceleration due to gravity (g) = 9.8 m/s²
• Maximum safe speed (v_max) = ?

Solution:

For a flat horizontal curve, the maximum static friction force provides the necessary centripetal force:

F_c = f_s ⇒ (m × v²) / r = μ_s × m × g

The mass (m) cancels out from both sides: v² = μ_s × r × g

Isolating velocity: v = √(μ_s × r × g)

v = √(0.523 × 35.0 × 9.8)

v = √(179.389) ≈ 13.39 m/s

Result: The maximum speed the car can have to safely complete the turn is 13.39 m/s.

9. Moment of Inertia of Co-planar Coplanar Particles

A system of coplanar point particles is arranged in a configuration where each particle has a mass of 0.3 kg. What is the total moment of inertia of the system about the designated configuration axis if the perpendicular distances are r₁ = 1m, r₂ = 2m, r₃ = 3m, r₄ = 4m?

Data:

• Mass of each particle (m) = 0.3 kg
• Distances from the axis: r₁ = 1 m, r₂ = 2 m, r₃ = 3 m, r₄ = 4 m
• Total Moment of Inertia (I_total) = ?

Solution:

The general expression for discrete rotational inertia is: I = Σ m_i r_i²

Expand for the full set: I_total = m(r₁² + r₂² + r₃² + r₄²)

I_total = 0.3 × (1² + 2² + 3² + 4²)

I_total = 0.3 × (1 + 4 + 9 + 16)

I_total = 0.3 × 30 = 9.0 kg·m²

Result: The net moment of inertia of the point mass particle system is 9.0 kg·m².

10. Rotational Momentum and Stopping Torque of a Cylinder

(a) What is the angular momentum of a 2.9 kg uniform cylindrical grinding wheel of radius 20 cm when rotating at 1550 rpm? (b) How much torque is required to stop it in 6 s?

Data:

• Mass of cylinder (m) = 2.9 kg
• Radius (r) = 20 cm = 0.2 m
• Rotational speed = 1550 rpm = 1550 / 60 ≈ 25.83 rev/s
• Initial angular velocity (ω_i) = 2 × π × f = 2 × 3.1416 × 25.83 ≈ 162.3 rad/s
• Final angular velocity (ω_f) = 0 rad/s (stopped)
• Time interval (Δt) = 6 s

Solution:

(a) For Angular Momentum (L):

The moment of inertia of a solid cylinder is: I = ½ m r²

I = 0.5 × 2.9 × (0.2)² = 0.058 kg·m²

Angular momentum: L = I × ω_i = 0.058 × 162.3 ≈ 9.41 kg·m²/s

(b) For Stopping Torque (τ):

Using the rotational impulse-momentum theorem: τ = ΔL / Δt = (L_f - L_i) / Δt

Torque magnitude: τ = 9.41 / 6 ≈ 1.57 N·m

Result: The angular momentum is 9.41 kg·m²/s and the magnitude of the torque required to stop it is 1.57 N·m.

11. Orbital vs. Spin Angular Momentum of the Earth

Determine the angular momentum of the Earth (a) about its own rotation axis (assume the Earth is a uniform sphere), and (b) in its orbit around the Sun (treating Earth as a point particle orbiting the Sun). Given: Mass = 6 × 10²⁴ kg, Radius = 6.4 × 10⁶ m, Orbital distance = 1.5 × 10⁸ km.

Data:

• Mass of Earth (M) = 6.0 × 10²⁴ kg
• Radius of Earth (R) = 6.4 × 10⁶ m
• Orbital Distance (r) = 1.5 × 10⁸ km = 1.5 × 10¹¹ m
• Spin Period (T_spin) = 1 day = 24 × 3600 s = 86400 s
• Orbital Period (T_orbit) = 1 year ≈ 365.25 days ≈ 3.156 × 10⁷ s

Solution:

(a) Angular Momentum about its rotation axis (Spin Angular Momentum - L_spin):

Moment of inertia of a uniform sphere: I = (2/5) M R²

I = 0.4 × (6.0 × 10²⁴) × (6.4 × 10⁶)² = 9.83 × 10³⁷ kg·m²

Spin angular velocity: ω_spin = 2π / T_spin = 2 × 3.1416 / 86400 ≈ 7.27 × 10⁻⁵ rad/s

L_spin = I × ω_spin = (9.83 × 10³⁷) × (7.27 × 10⁻⁵) ≈ 7.15 × 10³³ kg·m²/s

(b) Angular Momentum in its orbit around the Sun (Orbital Angular Momentum - L_orbit):

Treating Earth as a point particle: I_orbit = M r²

I_orbit = (6.0 × 10²⁴) × (1.5 × 10¹¹)² = 1.35 × 10⁴⁷ kg·m²

Orbital angular velocity: ω_orbit = 2π / T_orbit = 2 × 3.1416 / (3.156 × 10⁷) ≈ 1.99 × 10⁻⁷ rad/s

L_orbit = I_orbit × ω_orbit = (1.35 × 10⁴⁷) × (1.99 × 10⁻⁷) ≈ 2.69 × 10⁴⁰ kg·m²/s

Result: The angular momentum of the Earth about its axis is 7.15 × 10³³ kg·m²/s, and in its orbit around the Sun is 2.69 × 10⁴⁰ kg·m²/s.