How Does Energy Conservation Apply to a Rotating Pulley System?

[snoggerT]

Homework Statement

(a) If R = 12 cm, M = 500 g, and m = 50 g (below), find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles.

- The pulley is setup on the corner of a desk with the object hanging from it. M is the Pulley, m is the object, R is the radius of the pulley.

Homework Equations

mg-T=ma
-RT=1/2MR^2(alpha)

The Attempt at a Solution

I don't know where to start on this problem. I don't know how to combine the 2 equations or relate them. The only thing I know is that if I can find alpha, I can find the tangential acceleration. That would be the same as the acceleration of the object going down, so I could use equations of motion to find out the speed. At least I think that's how it would work. please help.

 

[Doc Al]

Hint: How are alpha and the tangential acceleration related?

 

[snoggerT]

Hint: How are alpha and the tangential acceleration related?

- They are related by the radius of the disk with the equation alpha=a(tangential)/r, but I'm having problems finding alpha since I am missing alpha and T in my equations. Could I substitute a/r in for alpha to cancel out one of the R^2 in the 1/2MR^2 and get TR=1/2MRa? That would all me to get a=2T/M and plug into mg-T=ma. That would then allow me to solve for T and then for alpha. Is that legal to do?

 

[Doc Al]

Absolutely. It's not only legal, it's the only way to solve the problem!

Since the pulley and block are connected, their motions (and accelerations) are not independent. [itex]a = \alpha R[/itex] describes how they are connected.

 

[snoggerT]

That worked for that problem, but now I have another problem that is more complicated. I think I can work it out if I just get a hint to start it off. Here's the problem:
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In the figure below, two blocks, of mass m1 = 320 g and m2 = 610 g, are connected by a massless cord that is wrapped around a uniform disk of mass M = 500 g and radius R = 12.0 cm. The disk can rotate without friction about a fixed horizontal axis through its center; the cord cannot slip on the disk. The system is released from rest.

Find the magnitude of the acceleration of the blocks.
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I don't think the way I solved the last problem would work because I have 2 T's now. How would you even begin this one?

 

[Doc Al]

Yes, there are two T's, but you have three things to analyze and thus three equations. Set up an equation for each and link them together. Be very careful with signs. If one mass goes up, the other must go down.

 

[snoggerT]

Yes, there are two T's, but you have three things to analyze and thus three equations. Set up an equation for each and link them together. Be very careful with signs. If one mass goes up, the other must go down.

- my equations would be:
m2g-T2=m2a (the right side going down)
T1-m1g=m1a (the left side going up)
T2R-T1R=1/2MR^2(alpha)

correct? I know how to combine the first 2 equations, but my problem is coming in how to combine the 3rd equation.

 

[Doc Al]

Use the same trick as before! Then you'll have three equations and three unknowns: a and the two T's.

 

[snoggerT]

Use the same trick as before! Then you'll have three equations and three unknowns: a and the two T's.

- I can't seem to figure this one out like the last one. Were the 3 equations I listed above correct? it's the two T's in the 3rd equation that are throwing me off.

 

[Doc Al]

- I can't seem to figure this one out like the last one. Were the 3 equations I listed above correct? it's the two T's in the 3rd equation that are throwing me off.

They are perfectly correct.

- my equations would be:
m2g-T2=m2a (the right side going down)
T1-m1g=m1a (the left side going up)
T2R-T1R=1/2MR^2(alpha)

In that third equation, replace alpha with a/R:

[tex]m_2g - T_2 = m_2a[/tex]
[tex]T_1 - m_1g = m_1a[/tex]
[tex]T_2R - T_1R = 1/2 MR^2(a/R) = (1/2)MRa[/tex]

There are many ways to solve these simultaneous equations. Don't be afraid to play around with them. Try multiplying the first two by R and adding them together. Then you can use the third to eliminate the two Ts and solve for a.

You can always just systematically eliminate the variables.

 

[snoggerT]

They are perfectly correct.


In that third equation, replace alpha with a/R:

[tex]m_2g - T_2 = m_2a[/tex]
[tex]T_1 - m_1g = m_1a[/tex]
[tex]T_2R - T_1R = 1/2 MR^2(a/R) = (1/2)MRa[/tex]

There are many ways to solve these simultaneous equations. Don't be afraid to play around with them. Try multiplying the first two by R and adding them together. Then you can use the third to eliminate the two Ts and solve for a.

You can always just systematically eliminate the variables.

- Thanks, I ended up dividing the 3rd equation by R to single out my T's. Then was able to find my a and solve the rest of the problem. I didn't know you could just divide/multiply equations by the variables like that to get what you wanted. I need to spend a lot more time on this board though, I've been struggling pretty bad with my physics class. The problems I've gotten help with on here make sense now.