Question

When a sphere rolls without slipping, the ratio of its kinetic energy of translation to its total kinetic energy is?

Answer 1

Answer:

KE (translational) = 1/2 M V^2

KE (rotational) = 1/2 I w^2 = 1/2 * 2/5 M R^2 * w^2 = 1/5 M R^2 w^2

KE (rotational) = 1/5 M V^2  since V = R * w

So the ratio is (1/2) / (1/5) = 5 / 2

Answer 2

Given :

Sphere rolls without slipping .

To Find :

The ratio of its kinetic energy of translation to its total kinetic energy .

Solution :

We know ,translation kinetic energy is given by :

$K.E_t=\dfrac{mv^2}{2}$

Rotational energy is given by :

$K.E_r=\dfrac{I\omega^2}{2}$

Here , $I=\dfrac{2MR^2}{5}$ and $\omega=\dfrac{v}{R}$

We need to find , $r=\dfrac{K.E_t}{K.E_t+K.E_r}$

Putting all give value we get :

$r=\dfrac{}{K.E_t+K.E_r}\\\\r=\dfrac{\dfrac{mv^2}{2}}{\dfrac{mv^2}{2}+\dfrac{I\omega^2}{2} }\\\\\\r=\dfrac{\dfrac{mv^2}{2}}{\dfrac{mv^2}{2}+\dfrac{\dfrac{2MR^2}{5}(\dfrac{v}{R})^2}{2}}\\\\\\r=\dfrac{5}{7}$

Therefore , the ratio of its kinetic energy of translation to its total kinetic energy is $\dfrac{5}{7}$ .

Learn More :

Rotation Energy

https://brainly.in/question/15967938