We can roughly model a gumnastic tumbler as a uniform solid cylinder of mass 75 kg and diameter 1.0 m. If the tumbler rolls forward at .50 rev/s. A) how much total kinetic energy does he have and B) what percent of his total kinetic enery is rotational?

Answer 1

I = 1/2MR^2

Answer 2

if the angular rotation is 1/2 rev/sec, then the forward speed is 1/2*2PI*.5 m/s

Look up the moment of inertia I for a cylinder.

TranKE=1/2 m v^2
RotationalKE=1/2 I w^2 where w is PI r/sec

Answer 3

I for the cylinder is I = 1/2MR^2

I still don't understand how to get the total KE through TranKE=1/2 m v^2

because 1/2 (75 kg)(.50 rev/s) = 9.375 and the answer is coming up in J in the back of the book

Answer 4

Total KE=translationalKE + rotationalKE
= 1/2 m v^2 + 1/2 I w^2