Kinetic Energy of a Bowling Ball

[PeachBanana]

Homework Statement

A bowling ball of mass 7.2 kg and radius 9.1 cm rolls without slipping down a lane at 3.0 m/s.

What is the total kinetic energy of the bowling ball?

Homework Equations

KE rot. = 1/2 Iω^2

The Attempt at a Solution

KE initial = 0 J (I'm assuming it started from rest).
I = 2/5 (7.2 kg)(0.091 m)^2
I = 0.0238492 kg * m^2
I used the formula for a uniform sphere to calculate this.

3.0 m/s / 0.091 m = 0.3296 rad./s

KE final = 1/2 (0.0238492 kg*m^2)(0.3296 rad./s)^2
KE final = 1.3 * 10 ^ -3 J

I'm thinking I messed up at the "I" variable because my answer is tiny.

 

[Doc Al]

3.0 m/s / 0.091 m = 0.3296 rad./s

Recheck this calculation.

Also: Don't forget the translational KE.

 

[PeachBanana]

DocAl - I redid the calculation and read the exponent of 3.296 as -1 and not +1. I added in the translational kinetic energy and got the right answer of 45 J.

 

[Doc Al]

Good!

 

[asteriatic]

Your calculation for the moment of inertia (I) looks correct. However, the value for ω (angular velocity) should be in units of radians per second (rad/s), not just radians. So your calculation for the final kinetic energy should be:

KE final = 1/2 (0.0238492 kg*m^2)(0.3296 rad/s)^2 = 0.001954 J

This is still a relatively small value, but it makes sense because the bowling ball is not moving very fast. If you compare this to the kinetic energy of a car traveling at 60 mph (26.8 m/s), which would have a mass of around 1000 kg, the kinetic energy would be approximately 360,000 J. So even though the bowling ball has a smaller mass, its slower speed results in a much smaller kinetic energy.