Question

A student sits on a freely rotating stool holding two weights, each of mass 3.00 kg (Fig. P10.48). When his arms are extended horizontally, the weights are 1.00 m from the axis of rotation and he rotates with an angular speed of 0.750 rad/s. The moment of inertia of the student plus stool is 3.00 kg · m2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.300 m from the rotation axis. (a) Find the new angular speed of the student. (b) Find the kinetic energy of the rotating system before and after he pulls the weights inward.

Answer 1

what I assume is:

the kinetic energy initially has to be proportional to the kinetic energy final.

Thus I set

K1 = K2

I calculated the moment of inertia for the final location to be 0.27kgm^2

so I get

1/2(0.27)(x)^2 = 1/2(3)(0.75)^(2)
x = 2.5 rad/s
thus I get a speed of 2.5rad/s

does this seem correct?

Answer 2

would I use angular momentum instead?

Answer 3

Yeah I think I use angular momentum as kinetic energy changes in regards to speed

Answer 4

so kinetic energy would differ I assume between both positions of the weights

Answer 5

So I get Momentum1 = momentum2

so

3(0.75) = (0.27)(w)
w = 8.2rad/sec

Answer 6

So I get Momentum1 = momentum2
so 3(0.75) = (0.27)(w)
w = 8.33rad/sec

Answer 7

Hi!
Your second-thought is the right one.

What does cause a change in angular momentum? External torque.
Is there external torque here? No (stool pivot is friction-free and air resistance is negligible)
Hence: Angular momentum is conserved.

What does cause a change in kinetic energy? Work of external AND internal actions.
Are external forces working here? No
Are internal forces working here? Yes: the student brings weights inwards by power of muscles.
Hence: kinetic energy is not conserved.

For instance, if moment of inertia is doubled, angular speed will be halved by conservation of angular momentum. KE will be doubled.

Answer 8

thanks for a detailed explanation :)