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alexandr1967 [171]
2 years ago
8

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 sp

eed 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.
Physics
1 answer:
liberstina [14]2 years ago
3 0

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.

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