Question

Problem 9.43: Gymnastics. - Enhanced - with Solution

We can roughly model a gymnastic tumbler as a uniform solid cylinder of mass 75.0 kg and diameter 1.10 m .

Part A: If this tumbler rolls forward at 0.700 rev/s , how much total kinetic energy does he have?

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

Answer 1

(A)

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 =0.55 m

We need the rotational speed to be in rad/sec

w = 0.700 rev / sec * ( 2 pi rad / 1 rev) = 4.40 rad / sec

m = 75.0 kg

Total KE = (3/4) * (75 kg) * (0.55 m)^2 * (4.40 rad/sec)^2

Total KE = 329.4 J

(B)

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 %