Question

Quick question! If a ball has a transnational Kinetic Energy of 65 Joules at the bottom of a ramp, it will have a rotational Kinetic Energy of?

Thanks, I appreciate any help!

Answer 1

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Answer 2

Assuming the ball is of uniform density and not slipping proceed as follows.

Using the normal symbols the kinetic energy of the ball is \[\frac 1 2 m v^2\]If the ball is rolling and not slipping the angular speed of the ball and the linear speed of the centre of mass of the ball are linked via the following equation \[v = r \omega\] the rotational kinetic energy of the ball about its centre of mass is \[\frac 1 2 I_C \omega^2\] where the moment of inertia of the ball about ist centre of mass is \[I_C = \frac 2 5 m r^2\]
If you put all of this together you will find that the radius of the ball cancels out and you will get for the rotational kinetic energy of the ball a numerical constant times the translational kinetic energy of the ball.

Answer 3

The problem also stated that the ball has a PE of 0 and a KE of 100. I apologize, but I'm a bit confused on how to put all of it together. Thank You!

Answer 4

You put it all together by carefully reading what's written above and combining the equations provided.