Question

If a sphere is rolling. the ratio of the translational energy to total kinetic energy is given by

Answer 1

Let the sphere have mass m and linear velocity v and angular velocity w and radius R

Translation kinetic energy=1/2mv^2

Rotational kinetic energy=1/2Iw^2 (where I is moment of inertia of sphere)

=1/2*2/5*mR^2*w^2

=1/2*2/5*mv^2

Ratio= (1/2mv^2)/ (2/10mv^2)=10/4=5/2

Feel free to suggest edits

Answer 2

The ratio of the translational energy to total kinetic energy is 5:7.

Explanation:

The transnational kinetic energy pf the sphere is given by :

$K_t=\dfrac{1}{2}mv^2$

The total kinetic energy of the sphere is the sum of rotational kinetic energy and the translational energy.

$E=\dfrac{1}{2}I\omega^2+\dfrac{1}{2}mv^2$

I is the moment of inertia of the sphere,

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

Since, $v=r\omega$

Total energy becomes :

$E=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \dfrac{v^2}{r^2}+\dfrac{1}{2}mv^2\\\\E=\dfrac{7}{10}mv^2$

The ratio of the translational energy to total kinetic energy is given by :

$\dfrac{K_t}{E}=\dfrac{(1/2)mv^2}{(7/10)mv^2}\\\\\dfrac{K_t}{E}=\dfrac{5}{7}$

So, the ratio of the translational energy to total kinetic energy is 5:7.

Learn more,

Rotational motion

https://brainly.in/question/7391638