# Conservation Laws in Rotational Motion

1. May 14, 2014

### Draggu

1. The problem statement, all variables and given/known data

15. A solid sphere of mass 6.0 kg is mounted on a vertical axis and can rotate freely without friction. A massless cord is wrapped around the middle of the sphere and passes over a 1.0 kg pulley and is attached to block of mass 4.0 kg, as shown. What is the speed of the block after it has fallen 80 cm? Treat the pulley as solid cylinder.

2. Relevant equations

E initial = E after
ωp = angular velocity of pulley
ωs = angular velocity of sphere

Forces
-------

Pulley: Iα = Rp(T2-T1)
(rp^2/2)(a/rp) = Rp(T2-T1)
=(a)/2 = T2-T1

Block: mg-T2 = ma
=39.2 - T2 = 4a

Sphere: Iα = Rs(T1)
T1 = (2/5)(6)(rs)^2
T1= 2a/5

a = 8m/s^2
T2=7.2N
T1=3.2N

v0 = 0

v^2 = 2(8)(.8)
v=3.57m/s

So initially I thought energy would be used in this situation, but now I'm trying to think if it's even necessary.

Last edited: May 14, 2014
2. May 14, 2014

### Simon Bridge

There is more than one way to approach the problem fersure.
Just pick the one you feel comfy with.

3. May 14, 2014

### Draggu

When I try with the energy conservation method, I get a different number. Perhaps you could offer some insight into this?

mgh = ΔKE block + ΔKE pulley + ΔKE sphere
mgh = (0.5)(4)(v22-v12) + (0.5)(Ipulley)(ω2212) + (0.5)(Isphere)(ω2212)
=(0.5)(4)(v22) + (0.5)(0.5)(1kg)(Rp2)(ω22) + (0.5)(2/5)(6kg)(Rs2)(ω22)
(4kg)(9.8)(0.8m) = 2(v22) + (1/4)(Rp2)(vp2/Rp2) + (6/5)(Rs2)(vs2/Rs2)
31.36 = 2(vb2) + (1/4)(vp2) + (6/5)(vs2)

Not too sure where to go from here.

4. May 15, 2014

### Simon Bridge

relate vp and vs to the speed of the block.

5. May 15, 2014

### Draggu

How would I go about doing that? Assuming the velocities are the same, I get 3.01m/s.. which is different than the value from forces. Hmm, so you say to relate them. Which means either vb = vp = vs, or vb = vp + vs?

Last edited: May 15, 2014
6. May 15, 2014

### haruspex

vb = vp = vs, as you did to obtain the (correct) result in post #5.
In the OP, these equations are wrong:
Check the dimensions of each side.

7. May 15, 2014

### Draggu

I re-did the question and figured out what went wrong. I'm extremely silly... in the net force for the sphere I wrote "2a/5" instead of "12a/5" heh. That was enough to mess everything up! Confirmed, got the same answers using both methods. Thanks guys.

8. May 15, 2014

### Simon Bridge

Well done. :)

9. Jan 27, 2015

### Grabo

Sorry, but this makes absolutely zero sense to me. Without radius, I am hooped. I have tried both the energy
and force approach and gotten stuck midway through.
I have no clue how you solved for this...how did you go from Ia=r(sphere)(T1) to T1=(2/5)(6)r(sphere)^2. Shouldn't it go to T1=mr^2a/r(sphere). I am so lost. Could somebody just explain to me how you got the tensions + how to not get stuck since radius is not given. Please!

10. Jan 27, 2015

### haruspex

Why would it do that? What is I in terms of m and r for a solid sphere? Please post your working.