Question

We can roughly model a gymnastic tumbler as a uniform solid cylinder of mass 65.0 kg and diameter 1.00 m.

Part A If this tumbler rolls forward at 0.300 rev/s , how much total kinetic energy does he have?

Part B What percent of his total kinetic energy is rotational?

Answer 1

Part one:
Total KE = Rot KE + Trans KE
Trans KE = (1/2) m* v^2
Rot KE = (1/2) I* w^2

So the Total KE = (1/2) I* w^2 + (1/2) m* v^2
I = (1/2) m * r^2 for Solid Cylinder

and we know that v = w * r

Total KE = (1/2) ((1/2) m * r^2)* w^2 + (1/2) m* (w * r)^2
Total KE = (1/4) m * r^2 * w^2 + (1/2) m* w^2 * r^2
Total KE = (3/4) m * r^2 * w^2 [EQ 1]

We are almost ready,
We need the radius,
r = 1/2 D =0.5 m
We need the rotational speed to be in rad/sec
w = 0.300 rev / sec * ( 2 pi rad / 1 rev) = 1.89 rad / sec
m = 65.0 kg

Total KE = (3/4) * (65 kg) * (0.5 m)^2 * (1.89 rad/sec)^2
Total KE = 43.54 J

Now for the next part, you can do it the easy way or the hard way.

The hard way:
Find Rotational KE from
Rot KE = (1/2) I* w^2
I = (1/2) m * r^2 for Solid Cylinder
Rot KE = (1/2) ((1/2) m * r^2)* w^2
Rot KE = (1/4) m * r^2* w^2 [EQ 2]

Now you could solve this mathematically for Rot KE and then divide by the number we got above for Total KE to get the percentage.

But......(the easy way). From above [EQ 1]we know that

Total KE = (3/4) m * r^2 * w^2

and from [EQ 2]

Rot KE = (1/4) m * r^2* w^2

Ratio = Rot KE / Total KE

Ratio = ((1/4) m * r^2* w^2) / ((3/4) m * r^2 * w^2)

Ratio = (1/4) / (3/4)

Ratio = 1/3

Percentage = 1/3 * 100%

Percentage = 33.333 %