Finding angular velocity after block is moved from middle to outside of disc

  • #1

Homework Statement



A 200 g, 42.0-cm-diameter turntable rotates on frictionless bearings at 56.0 rpm. A 20.0 g block sits at the center of the turntable. A compressed spring shoots the block radically outward along a frictionless groove in the surface of the turntable.

What is the turntable's rotation angular velocity when the block reaches the outer edge?

Homework Equations



Ei=Ef
.5Iw^2 (initial) = .5Iw^2 (final)

For finding the final rotational inertia:
I(final) = Icm + Md^2 = .5MR^2 + md^2

The Attempt at a Solution



I believe that energy is conserved in this process so the equation above is valid. I calculated the final rotational inertia to be (1/2)*(0.2kg)*(0.21m)^2 + (0.02kg)*(0.21m)^2 = 0.005292

The initial energy is (.5)*(.5*0.2kg*0.21m^2)*(56rpm)^2 = 6.915

The final energy is (.5)*(0.005292)*wfinal^2

Solving for wfinal I get 53.6 rpm. I've done the calculations repeatedly and I can't come up with an alternate way of doing it but this answer is not correct. Any ideas?

Thank you so much in advance!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi snoworskate! :smile:
… I believe that energy is conserved in this process so the equation above is valid

I can't come up with an alternate way of doing it but this answer is not correct. Any ideas?

never never never use conservation of energy if you can use conservation of momentum (or angular momentum).

Momentum (or angular momentum) is always conserved (if there is zero net force or torque, as in this case).

Energy usually isn't conserved, and virtually never is in exam questions unless the question gives a pretty clear hint that it is. :wink:
 
  • #3
Wow, that was MUCH easier. Thanks so much, I'll remember that!
 

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