Question

A solid sphere rolling without slipping on a horizontal surface. If the translational speed of the sphere is 2.00 m/s, what is its total kinetic energy?

Answer 1

Answer:

The total kinetic energy is 2.8m J. (NOTE: m is mass of the sphere)

Explanation:

The total kinetic energy of a sphere is given by the sum of the rotational kinetic energy and the translational kinetic energy. That is,

$K_{Total} = K_{R} + K_{T}$

The rotational kinetic energy $K_{R}$ is given by

$K_{R} = \frac{1}{2}I\omega^{2}$

Where $I$ is the moment of inertia

and $\omega$ is the angular velocity

The translational kinetic energy $K_{T}$ is given by

$K_{T} = \frac{1}{2}mv^{2}$

Where $m$ is the mass

and $v$ is the translational speed (velocity)

∴ $K_{Total} = \frac{1}{2}I\omega^{2} + \frac{1}{2}mv^{2}$

But, the moment of inertia $I$ of a sphere is given by

$I = \frac{2}{5}mr^{2}$

Where $m$ is mass

and $r$ is radius

∴ $K_{Total} = \frac{1}{2}\times \frac{2}{5}mr^{2} \omega^{2} + \frac{1}{2}mv^{2}$

$K_{Total} = \frac{1}{5}mr^{2} \omega^{2} + \frac{1}{2}mv^{2}$

Also, $\omega = \frac{v}{r}$

∴ $\omega^{2} = \frac{v^{2} }{r^{2} }$

Then,

$K_{Total} = \frac{1}{5}mr^{2} \times \frac{v^{2} }{r^{2} } + \frac{1}{2}mv^{2}$

$K_{Total} = \frac{1}{5}mv^{2} + \frac{1}{2}mv^{2}$

∴ $K_{Total} = \frac{7}{10}mv^{2}$

From the question, $v = 2.00 m/s$

Then,

$K_{Total} = \frac{7}{10}m(2.00)^{2}$

$K_{Total} = \frac{7}{10}m\times 4.00$

$K_{Total} = 2.8m J$

Hence, the total kinetic energy is 2.8m J. (NOTE: m is mass of the sphere)