Question

A uniform ball of mass m rolls without sliding on a fixed horizontal surface. The velocity of the lowest point of the ball with respect to the centre of the ball is v. Find out the total kinetic energy of the ball

Answer 1

Answer:

$K.E\ =\ \dfrac{7mv^2}{10}$

Explanation:

Given,

• Mass of the ball = m
• Final velocity of the ball at lowest point of the surface = v

Moment of inertia of the rotating ball = $I\ =\ \dfrac{2}{5}mR^2}$

Angular speed of the rotating ball = $w\ =\ \dfrac{v}{R}$

Now the Kinetic energy due to the rotation,

$K.E_1\ =\ \dfrac{1}{2}Iw^2\ =\ \dfrac{1}{2}\times \dfrac{2}{5}mR^2\times \dfrac{v^2}{R^2}\ =\ \dfrac{1}{5}mv^2$

Again, the kinetic energy due to the translational motion,

$K.E_2\ =\ \dfrac{1}{2}mv^2$

Now total kinetic energy at the lowest point of the surface

$K.E_t\ =\ K.E_1\ +\ K.E_2\\\Rightarrow K.E_t\ =\ \dfrac{1}{5}mv^2\ +\ \dfrac{1}{2}mv^2\\\Rightarrow K.E_t\ =\ \dfrac{7mv^2}{10}$