# A solid sphere of uniform density is attached to a massless rod at the sphere s surface, and is...

## Question:

A solid sphere of uniform density is attached to a massless rod at the sphere s surface, and is spinning around the rod at {eq}1.0 \ rad / s {/eq}. The sphere is {eq}1.0 {/eq} meter in radius, and {eq}1.0 \ kg {/eq} in mass. What is the rotational kinetic energy of the sphere?

a) {eq}\displaystyle \frac {2} {5} \ J {/eq}

b) {eq}\displaystyle \frac {1} {5} \ J {/eq}

c) {eq}\displaystyle \frac {3} {5} \ J {/eq}

d) {eq}\displaystyle \frac {7} {10} \ J {/eq}

e) {eq}\displaystyle \frac {3} {10} \ J {/eq}

## Rotational Kinematic Energy

Moment of inertia is the measure of the resistance of a rigid body to change in its rotational state of motion. The rotational kinetic energy of the object is the energy an object possesses due to its rotational motion. The rotational kinetic energy of the object rotating with angular velocity {eq}\omega {/eq} is given by: {eq}I \ = \ \dfrac{1}{2} \ I \times \omega^2 \\ {/eq}

Where:

{eq}I {/eq} - is the moment of inertia of the object.

## Answer and Explanation:

**Given**:

- Angular velocity of the sphere: {eq}\omega \ = \ 1.0 \ rad s^{-1} \\ {/eq}

- Radius of the sphere: {eq}r \ = \ 1.0 m {/eq}

- Mass of the sphere: {eq}M \ = \ 1 \ kg \\ {/eq}

The moment of inertia of the sphere about its axis of rotation is: {eq}I \ = \ \dfrac{2}{5} \ M \times r^2 \ = \ \dfrac{2}{5} \times 1 \times 1^2 \ = \ \dfrac{2}{5} \ kg m^2 \\ {/eq}

The kinetic energy of the rotating sphere is: {eq}KE \ = \ \dfrac{1}{2} \ I \times \omega^2 \ = \ \dfrac{1 \times 2}{2 \times 5} \ = \ \dfrac{1}{5} \ J \\ {/eq}

The correct option is b.

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