Physics, asked by Triptiyadav, 3 months ago

a hollow sphere of mass M and radius r is in pure rolling on the horizontal surface the ratio of difference and translational kinetic energy and rotational kinetic energy to the total kinetic energy of the hollow sphere will be​

Answers

Answered by sonuvuce
0

The ratio of difference of translational kinetic energy and rotational kinetic energy to the total kinetic energy is 1/5

Explanation:

Given:

A hollow sphere of mass m and radius r is in pure rolling on horizontal surface

To find out:

The ratio of difference of translational kinetic energy and rotational kinetic energy to the total kinetic energy

Solution:

We know that the moment of inertia of hollow sphere

I=\frac{2}{3}mr^2

If the sphere is in pure rolling and its velocity is then its angular velocity

\omega=\frac{v}{r}

The total kinetic energy will be the sum of rotational and translational kinetic energy

\implies K_T=\frac{1}{2}I\omega^2+\frac{1}{2}mv^2

\implies K_T=\frac{1}{2}(\frac{2}{3}mr^2(\frac{v}{r})^2+\frac{1}{2}mv^2

\implies K_T=\frac{1}{2}(\frac{2}{3}+1)mv^2

\implies K_T=\frac{5}{6}mv^2

The difference of translational and rotational kinetic energy is

K=\frac{1}{2}mv^2-\frac{1}{2}I\omega^2

\implies K=\frac{1}{2}mv^2-\frac{1}{2}(\frac{2}{3}mr^2(\frac{v}{r})^2

\implies K=\frac{1}{2}(1-\frac{2}{3})mv^2

\implies K=\frac{1}{6}mv^2

Therefore, the ratio

\frac{K}{K_T}=\frac{1/6 mv^2}{5/6mv^2}

\implies K/K_T=1/5

Hope this answer is helpful.

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