§10.7 — Worked Problems

Seven worked problems combining all of Chapter 10's tools — from kinematics to gyroscopes.

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Problem 1 (§10.1) — Accelerating Wheel

A wheel starts from rest and rotates with constant angular acceleration α = 3 rad/s². After 5 seconds:

• Final angular velocity?
• Total angle rotated (in revolutions)?

Solution:

Angular velocity: \[ \omega_f = \omega_0 + \alpha t = 0 + 3 \times 5 = 15 \text{ rad/s} \]

Angle: \[ \theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2} \times 3 \times 25 = 37.5 \text{ rad} \]

Number of revolutions: \[ N = \frac{37.5}{2\pi} = \frac{37.5}{6.28} \approx 5.97 \text{ revolutions} \]

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Problem 2 (§10.2) — Comparing Moments of Inertia

Three cylinders of equal mass (M = 2 kg) and radius (R = 0.4 m):

• Solid cylinder: \( I_1 = \frac{1}{2}MR^2 \)
• Hollow cylinder: \( I_2 \approx 0.9MR^2 \)
• Hoop: \( I_3 = MR^2 \)

Which is hardest to spin?

Solution:

\[ I_1 = \frac{1}{2} \times 2 \times (0.4)^2 = 0.16 \text{ kg·m}^2 \]

\[ I_2 = 0.9 \times 2 \times (0.4)^2 = 0.288 \text{ kg·m}^2 \]

\[ I_3 = 2 \times (0.4)^2 = 0.32 \text{ kg·m}^2 \]

The hoop (\( I_3 = 0.32 \)) is hardest! All mass is at distance \( R \).

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Problem 3 (§10.3) — Rotational Dynamics

A disk with moment of inertia I = 0.5 kg·m² experiences three torques:

• \( \tau_1 = +8 \) N·m (counterclockwise)
• \( \tau_2 = -3 \) N·m (clockwise)
• \( \tau_3 = +2 \) N·m (counterclockwise)

Angular acceleration?

Solution:

Net torque: \[ \tau_{\text{net}} = 8 - 3 + 2 = 7 \text{ N·m} \]

Angular acceleration: \[ \alpha = \frac{\tau_{\text{net}}}{I} = \frac{7}{0.5} = 14 \text{ rad/s}^2 \]

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Problem 4 (§10.4) — Rotational Energy

Solid cylinder: M = 3 kg, R = 0.2 m, angular velocity ω = 10 rad/s.

• Rotational kinetic energy?
• If this energy were linear, what velocity would be needed?

Solution:

Moment of inertia: \[ I = \frac{1}{2} \times 3 \times (0.2)^2 = 0.06 \text{ kg·m}^2 \]

Rotational energy: \[ K_{\text{rot}} = \frac{1}{2} \times 0.06 \times 10^2 = 3 \text{ J} \]

Equivalent linear velocity: \[ \frac{1}{2}mv^2 = 3 \implies v^2 = \frac{6}{3} = 2 \implies v \approx 1.41 \text{ m/s} \]

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Problem 5 (§10.5) — Rolling Without Slip

Billiard ball: R = 0.028 m (radius), center-of-mass velocity v = 2 m/s.

• Angular velocity for no-slip rolling?
• Revolutions per second?

Solution:

Angular velocity (no-slip): \[ \omega = \frac{v}{R} = \frac{2}{0.028} \approx 71.4 \text{ rad/s} \]

Revolutions per second: \[ f = \frac{\omega}{2\pi} = \frac{71.4}{6.28} \approx 11.4 \text{ Hz} \]

The ball spins 11.4 times per second!

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Problem 6 (§10.5) — Sphere Rolling Down Incline

A sphere of mass M = 1 kg and radius R = 0.1 m rolls without slipping down an incline at angle θ = 30°.

Linear acceleration of center of mass?

Solution:

For sphere: \( I = \frac{2}{5}MR^2 \)

Acceleration: \[ a = \frac{g\sin\theta}{1 + I/(MR^2)} = \frac{g\sin\theta}{1 + \frac{2}{5}} = \frac{g\sin\theta}{1.4} \]

Substituting: \[ a = \frac{9.8 \times \sin 30°}{1.4} = \frac{9.8 \times 0.5}{1.4} = \frac{4.9}{1.4} \approx 3.5 \text{ m/s}^2 \]

(For comparison, sliding without rotation would give 9.8 × 0.5 = 4.9 m/s²!)

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Problem 7 (§10.6) — Gyroscope Precession

A gyroscope:

• Moment of inertia: I = 0.01 kg·m²
• Angular velocity: ω = 50 rad/s
• Mass: m = 0.2 kg
• Distance from pivot to center: r = 0.25 m

Precession rate? Precession period?

Solution:

Precession rate: \[ \Omega_p = \frac{mgr}{I\omega} = \frac{0.2 \times 9.8 \times 0.25}{0.01 \times 50} = \frac{0.49}{0.5} = 0.98 \text{ rad/s} \]

Period: \[ T_p = \frac{2\pi}{\Omega_p} = \frac{6.28}{0.98} \approx 6.4 \text{ seconds} \]

The gyroscope completes one precession cycle every 6.4 seconds.

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Key Summary

Topic	Equation	Problem
Kinematics	\( \omega = \omega_0 + \alpha t \)	#1
Moment of inertia	\( I = \int r^2 dm \)	#2
Dynamics	\( \tau = I\alpha \)	#3
Energy	\( K = \frac{1}{2}I\omega^2 \)	#4
Rolling	\( v = R\omega \)	#5, #6
Precession	\( \Omega_p = \frac{mgr}{I\omega} \)	#7

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Three Practical Tips

1. Signs matter: Counterclockwise is +, clockwise is −.

2. Units: radians, never degrees: Never plug degrees into formulas!

3. Distance from axis matters: Both moment and torque depend on distance.

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Preview of §10.8

Dozens of unsolved problems for practice.

📚 See also: Halliday Vol 1, Ch 10 — Full rotation chapter.

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