You stand on a frictional platform that is rotating at 1.3 rev/s. Your arms are
ID: 2066879 • Letter: Y
Question
You stand on a frictional platform that is rotating at 1.3 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 8.6 kg · m2. When you pull the weights in toward your body, the moment of inertia decreases to 4.4 kg · m2.(a) What is the resulting angular speed of the platform?
rev/s
(b) What is the change in kinetic energy of the system?
J
(c) Where did this increase in energy come from? (Select all that apply.)
gravity
your internal energy
kinetic energy of the platform
air resistance
mass of the weights
Explanation / Answer
a) First, we convert the initial speed to radians per second,
1.3 rev/s = 2*1.3 rad/s = 8.2 rad/s
Next, we apply conservation of angular momentum
I11 = I22
2 = I11/I2 = (8.6 kg*m2)(8.2 rad/s)/(4.4 kg*m2) = 16.0 rad/s
b) Then, we compute the change in kinetic energy, which is:
I222/2 - I121/2 = (4.4 kg*m2)(16.0 rad/s)2/2 - (8.6 kg*m2)(8.2 rad/s)2/2 = 274 J
This change in energy comes from the effort that you put in in order to pull the weights closer to yourself, so the answer is "your internal energy."