Question

A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad/s. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

• A. 62.5 T-s
• B. 70.4 T-s
• C. 79.6 T-s
• D. 60.5 T-s

Hint:

The rotational kinetic energy is calculated using the moment of inertia and angular velocity. The angular momentum is the product of the moment of inertia and angular velocity.

Answer 1

The Correct Option is A

The rotational kinetic energy \( K_{\text{rot}} \) of a solid cylinder is given by: \[ K_{\text{rot}} = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia of the solid cylinder, and \( \omega \) is the angular velocity.
The moment of inertia for a solid cylinder rotating about its axis is: \[ I = \frac{1}{2} m r^2 \] Substituting the given values: \[ I = \frac{1}{2} \times 20 \times (0.25)^2 = 0.625 \, \text{kg} \cdot \text{m}^2 \] \[ K_{\text{rot}} = \frac{1}{2} \times 0.625 \times (100)^2 = 3125 \, \text{J} \] The angular momentum \( L \) is given by: \[ L = I \omega = 0.625 \times 100 = 62.5 \, \text{kg} \cdot \text{m}^2/\text{s} \]
Thus, the correct answer is (a).