Question

A 34 kg bowling ball with a radius of 22 cm starts from rest at the top of an incline 4.4 m in height. Find the translational speed of the bowling ball after it has rolled to the bottom of the incline. (Assume that the ball is a uniform solid sphere.)The acceleration of gravity is 9.81 m/s 2 . Answer in units of m/s.

Answer 1

Answer:

7.85 m/s.

Explanation:

Given,

Mass of the bowling ball, m = 34 Kg

radius, r = 0.22 cm

height of the inclination, h = 4.4 m

transnational velocity = ?

Moment of inertia of bowling ball,

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

Using conservation of energy

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

We know that $v = r\omega$

$mgh = \dfrac{1}{2}( \dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2$

$m gh = 0.7 m v^2$

$v =\sqrt{\dfrac{gh}{0.7}}$

$v =\sqrt{\dfrac{9.81\times 4.4}{0.7}}$

$v = 7.85\ m/s$

Speed of the bowling ball is equal to 7.85 m/s.