Question

A solid sphere of mass 4.0 kg and radius 0.12 m starts from rest at the top of a ramp inclined 15°, and rolls to the bottom. The upper end of the ramp is 1.2 m higher than the lower end. What is the linear speed of the sphere when it reaches the bottom of the ramp

Answer 1

Answer:

v = 4.1 m/s

Explanation:

As per mechanical energy conservation law we can say that initial total gravitational potential energy of the sphere is equal to final kinetic energy of rolling

so we will say

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

now for pure rolling condition we know that

$v = R\omega$

so we will have

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2})$

$mgh = \frac{7}{10}mv^2$

now we will have

$v^2 = \sqrt{\frac{10}{7}gh}$

$v^2 = \sqrt{\frac{10}{7}(9.8)(1.2)}$

$v = 4.1 m/s$