Question

hollow sphere of mass m and radius r is in pure rolling on the horizontal surface. the ratio of difference of translational kinetic energy and rotational kinetic energy to the total kinetic energy of hollow sphere
a.5/6
b.1/5
c. 3/5
d.2/3​

Answer 1

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

Therefore, option (b) is correct.

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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