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

AI Thread Summary
The discussion centers on calculating the angular velocity of a turntable after a block is moved from the center to the edge. Initially, the user attempted to use energy conservation principles, calculating the final rotational inertia and energy, but arrived at an incorrect answer. A key insight shared is that conservation of angular momentum should be used instead of energy conservation in this scenario, as angular momentum is conserved when no net external forces or torques are acting. This approach simplifies the problem significantly and leads to the correct solution. The user expresses gratitude for this clarification, indicating a shift in understanding.
snoworskate
Messages
8
Reaction score
0

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!
 
Physics news on Phys.org
hi snoworskate! :smile:
snoworskate said:
… 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:
 
Wow, that was MUCH easier. Thanks so much, I'll remember that!
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Replies
5
Views
5K
Replies
3
Views
10K
Replies
7
Views
2K
Replies
2
Views
4K
Replies
5
Views
6K
Replies
3
Views
2K
Replies
9
Views
3K
Back
Top