§11.5 — Worked Problems

Seven worked problems testing all tools from Chapter 11 — from rolling to gyroscopes.

Problem 1 (§11.1) — Solid Cylinder on Incline

A solid cylinder (mass M = 2 kg, radius R = 0.1 m) is released without slipping on an incline (\( \theta = 30° \)).

a) Acceleration of center of mass? b) Velocity after dropping h = 0.5 m?

Solution:

a) Acceleration:

\[ a = \frac{g\sin\theta}{1 + \frac{I}{MR^2}} \]

For solid cylinder: \( I = \frac{1}{2}MR^2 \)

\[ a = \frac{g\sin\theta}{1 + \frac{1}{2}} = \frac{2g\sin\theta}{3} = \frac{2 \times 10 \times 0.5}{3} = \frac{10}{3} \approx 3.33 \text{ m/s}^2 \]

b) Velocity from energy:

\[ mgh = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I\omega^2 \]

Substituting \( I = \frac{1}{2}MR^2 \) and \( v_{cm} = \omega R \):

\[ mgh = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2} \times \frac{1}{2}MR^2 \times \left(\frac{v_{cm}}{R}\right)^2 \]

\[ mgh = \frac{1}{2}Mv_{cm}^2 + \frac{1}{4}Mv_{cm}^2 = \frac{3}{4}Mv_{cm}^2 \]

\[ v_{cm} = \sqrt{\frac{4gh}{3}} = \sqrt{\frac{4 \times 10 \times 0.5}{3}} = \sqrt{\frac{20}{3}} \approx 2.58 \text{ m/s} \]

Problem 2 (§11.1) — Three-Object Race

Three objects (solid cylinder, sphere, hoop) — all mass m = 1 kg, radius R = 0.1 m — released down an incline.

Question: Which reaches the bottom first?

Solution:

Acceleration of each:

\[ a = \frac{g\sin\theta}{1 + k} \]

where \( k = \frac{I}{mR^2} \):

Result: Sphere first, then cylinder, then hoop.

Problem 3 (§11.2) — Friction Force in Rolling

A disk (mass m = 3 kg, radius R = 0.2 m) rolls on a horizontal surface. A horizontal force F = 10 N is applied perpendicular to the axle.

Question: Acceleration of center of mass and friction force?

Solution:

No-slip condition: \( v = \omega R \) → \( a = \alpha R \)

Translational equation: \[ F - f = ma \]

Rotational equation: \[ \tau = fR = I\alpha = I\frac{a}{R} \]

\[ f = \frac{Ia}{R^2} \]

For disk: \( I = \frac{1}{2}mR^2 \)

\[ f = \frac{\frac{1}{2}mR^2 \times a}{R^2} = \frac{1}{2}ma \]

Substituting into translational equation:

\[ F - \frac{1}{2}ma = ma \]

\[ F = \frac{3}{2}ma \]

\[ a = \frac{2F}{3m} = \frac{2 \times 10}{3 \times 3} = \frac{20}{9} \approx 2.22 \text{ m/s}^2 \]

\[ f = \frac{1}{2} \times 3 \times \frac{20}{9} = \frac{10}{3} \approx 3.33 \text{ N} \]

Problem 4 (§11.3) — Conservation of Angular Momentum

A figure skater spins with arms extended:

Pulls arms to body:

Question: Final angular velocity?

Solution:

Angular momentum is conserved:

\[ L_1 = L_2 \]

\[ I_1\omega_1 = I_2\omega_2 \]

\[ \omega_2 = \frac{I_1}{I_2}\omega_1 = \frac{1.5}{0.4} \times 4 = 3.75 \times 4 = 15 \text{ rad/s} \]

Angular velocity increases 3.75-fold!

Problem 5 (§11.3) — Two Colliding Disks

Disk A: \( I_A = 0.5 \) kg·m², \( \omega_A = 8 \) rad/s (spinning) Disk B: \( I_B = 0.3 \) kg·m², \( \omega_B = 0 \) (at rest)

Disks collide and stick together.

Question: Final angular velocity and energy lost?

Solution:

Angular momentum:

\[ L_{\text{before}} = I_A\omega_A + I_B\omega_B = 0.5 \times 8 + 0 = 4 \text{ kg·m}^2\text{/s} \]

\[ L_{\text{after}} = (I_A + I_B)\omega_f = 0.8\omega_f \]

\[ \omega_f = \frac{4}{0.8} = 5 \text{ rad/s} \]

Energy:

\[ K_{\text{before}} = \frac{1}{2} \times 0.5 \times 8^2 = 16 \text{ J} \]

\[ K_{\text{after}} = \frac{1}{2} \times 0.8 \times 5^2 = 10 \text{ J} \]

\[ \Delta K = 16 - 10 = 6 \text{ J} \text{ (converted to heat)} \]

Problem 6 (§11.4) — Precessing Gyroscope

A gyroscope:

Question: Precession rate?

Solution:

Angular momentum: \[ L = I\omega = 0.01 \times 200 = 2 \text{ kg·m}^2\text{/s} \]

Precession rate: \[ \Omega = \frac{mgr}{L} = \frac{0.5 \times 10 \times 0.3}{2} = \frac{1.5}{2} = 0.75 \text{ rad/s} \]

Precession period: \[ T = \frac{2\pi}{\Omega} = \frac{2\pi}{0.75} \approx 8.38 \text{ seconds} \]

Problem 7 (§11.3) — Sphere on Frictionless Point

A solid sphere (mass m = 2 kg, radius R = 0.15 m) rotates about a point. Initial angular velocity ω₀ = 10 rad/s. Friction gradually stops it.

Question: Time to stop and number of revolutions?

Solution:

Friction torque: \[ \tau = -f \cdot R = I\alpha \]

For sphere: \( I = \frac{2}{5}mR^2 = \frac{2}{5} \times 2 \times (0.15)^2 = 0.018 \) kg·m²

Assume: \( f = 0.1 \) N

\[ \alpha = \frac{-0.1 \times 0.15}{0.018} = \frac{-0.015}{0.018} \approx -0.83 \text{ rad/s}^2 \]

Time to stop (\( \omega_f = 0 \)): \[ \omega_f = \omega_0 + \alpha t \implies 0 = 10 - 0.83t \implies t \approx 12 \text{ s} \]

Number of revolutions: \[ \theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 10 \times 12 - \frac{1}{2} \times 0.83 \times 144 \approx 60 \text{ rad} \]

\[ N = \frac{60}{2\pi} \approx 9.5 \text{ revolutions} \]

Comparison: Chapters 10 and 11

Quantity Linear (Ch 7) Rotational (Ch 10) Rolling (Ch 11)
Definition \( F = ma \) \( \tau = I\alpha \) \( F = ma \) + no-slip
Momentum \( p = mv \) \( L = I\omega \) Combined \( L \) and \( p \)
Energy \( K = \frac{1}{2}mv^2 \) \( K = \frac{1}{2}I\omega^2 \) \( K = \frac{1}{2}m(v^2 + \omega^2R^2) \)
Conservation \( p \) (if \( F_{\text{net}}=0 \)) \( L \) (if \( \tau_{\text{net}}=0 \)) Combined \( E \) and \( L \)

📚 See also: Halliday Vol 1, Ch 11, §11.5 — Worked problems on rolling, torque, and angular momentum. 🔗 Cross-reference: Chapters 7–10 for comparison.

⇧ Back to chapter

Have a question? 🤔

If something isn't clear or you have a question, ask it here. The answer will be published on this page.