Answer:

Explanation:
Given that,
Mass of the bowling ball, m = 5 kg
Radius of the ball, r = 11 cm = 0.11 m
Angular velocity with which the ball rolls, 
To find,
The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.
Solution,
The translational kinetic energy of the ball is :



The rotational kinetic energy of the ball is :



Ratio of translational to the rotational kinetic energy as :

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2