Question

A solid sphere of mass 50 kg and radius 20 cm is rotating about its diameter with an angular velocity of 420 rpm. The angular momentum of the sphere is:

• A. \( 8.8 \) Js
• B. \( 70.4 \) Js
• C. \( 17.6 \) Js
• D. \( 35.2 \) Js

Hint:

For rotating bodies, use \( L = I \omega \), where \( I \) depends on the shape of the object. Convert rpm to rad/s using \( \omega = {rpm} \times \frac{2\pi}{60} \).

Answer 1

The Correct Option is D

Step 1: Formula for Angular Momentum
The angular momentum \( L \) of a rotating body is given by: \[ L = I \omega. \] where: - \( I \) is the moment of inertia about the axis of rotation,
- \( \omega \) is the angular velocity in rad/s.
Step 2: Moment of Inertia of a Solid Sphere
For a solid sphere rotating about its diameter, \[ I = \frac{2}{5} M R^2. \] Substituting the given values:
- \( M = 50 \) kg,
- \( R = 20 \) cm = \( 0.2 \) m.
\[ I = \frac{2}{5} \times 50 \times (0.2)^2. \] \[ I = \frac{2}{5} \times 50 \times 0.04. \] \[ I = \frac{2}{5} \times 2 = 0.8 { kg m}^2. \] Step 3: Angular Velocity Calculation
The given angular velocity is 420 rpm. Converting to rad/s: \[ \omega = 420 \times \frac{2\pi}{60}. \] \[ \omega = 420 \times \frac{\pi}{30}. \] \[ \omega = 14\pi { rad/s}. \] Approximating \( \pi \approx 3.14 \): \[ \omega = 14 \times 3.14 = 43.96 { rad/s}. \] Step 4: Calculating Angular Momentum
\[ L = I \omega = 0.8 \times 43.96. \] \[ L = 35.2 { Js}. \] Step 5: Conclusion
Thus, the angular momentum of the sphere is: \[ 35.2 { Js}. \]