# A 2.5kg solid homogenous ball is rolling without slipping at 3 m per second to the right on a...

## Question:

A 2.5kg solid homogenous ball is rolling without slipping at 3 m per second to the right on a level surface. The diameter of the ball is 80 cm. What is the total kinetic energy of the ball?.

## Total Kinetic Energy:

The total kinetic energy of a rolling body is given by the equation;

{eq}K = \dfrac{1}{2} M V_{cm}^2+ \dfrac{1}{2} I \omega^2 {/eq}

where, {eq}V_{cm} {/eq} is the velocity of the center of mass of the body, {eq}\omega {/eq} is the angular velocity about the axis of rotation and I is the moment of inertia about the same axis of rotation.

## Answer and Explanation:

**Given**

- The mass of solid homogeneous ball is {eq}m = 2.5\ kg {/eq}

- Radius of the solid homogeneous ball is {eq}r = 0.8\ m {/eq}

- The velocity of centre of mass of the ball is {eq}v = 3\ m/s {/eq}

** Let**

- The angular velocity of the ball is {eq}\omega {/eq}

The velocity of centre of mass of the ball is related to the angular velocity as;

{eq}\begin{align} v &= \omega r\\ \implies \omega &=\dfrac{v}{r}\\ \end{align} {/eq}

The translational kinetic energy of the ball is;

{eq}\begin{align} K.E._{t} &=\dfrac{1}{2}mv^2\\ \end{align} {/eq}

The rotational kinetic energy of the ball is;

{eq}\begin{align} K.E._{r}&=\dfrac{1}{2}I\omega^2\\ &=\dfrac{1}{2} (dfrac{2}{5}m r^2)\omega^2\\ &=\dfrac{1}{2} (dfrac{2}{5}m r^2)(\dfrac{v}{r})^2\\ &=\dfrac{1}{5}m v^2\\ \end{align} {/eq}

**Thus, the total kinetic energy (K.E.) due to the translational and rotational motion of the ball is; **

**{eq}\begin{align} K.E. &=K.E._{t}+K.E._{r}\\ &= \dfrac{1}{2}mv^2+\dfrac{1}{5}m v^2\\ &= \dfrac{7}{10}mv^2\\ &= \dfrac{7}{10}\times 2.5 \times (3)^2\\ &= 15.75\ J\\ \end{align} {/eq}**

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