# Connected masses on a cylinder

1. Jul 11, 2010

### bobred

1. The problem statement, all variables and given/known data
Find the equation of motion, see diagram. Masses are P1 2m and P2 3m.

2. Relevant equations
Particle 1
$$\texttt{T}_1=\texttt{T}_1\texttt{e}_\theta$$
$$\texttt{N}_1=\texttt{N}_1\texttt{e}_r$$
$$\texttt{W}_1=-2mg\sin\theta{e}_r-2mg\cos\theta{e}_\theta$$

Particle
$$\texttt{T}_2=\sin\theta{e}_r+\cos\theta{e}_\theta$$
$$\texttt{W}_2=-3mg\sin\theta{e}_r-3mg\cos\theta{e}_\theta$$

3. The attempt at a solution
Using Newton's 2nd law $$m\texttt{a}=\texttt{F}$$

P1
$$-2mR\dot{\theta}^2\texttt{e}_r+2mR\ddot{\theta}\texttt{e}_\theta=-2mg\sin\theta{e}_r-2mg\cos\theta{e}_\theta+\texttt{N}_1\texttt{e}_r+\texttt{T}_1\texttt{e}_\theta$$

P2
$$-3mg=\sin\theta{e}_r+\cos\theta{e}_\theta-3mg\sin\theta{e}_r-3mg\cos\theta{e}_\theta$$

Does what I have done look ok? The string is a model string so $$\texttt{T}_1=-\texttt{T}_2$$. How can I find $$\texttt{N}$$ and $$\texttt{T}$$?

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Last edited: Jul 12, 2010
2. Jul 11, 2010

### hikaru1221

What's wrong with the dimension of the second equation of P2?
Does the problem provide any initial condition? Under some certain circumstances, the problem may be less complicated.
Let's consider the problem in general case. P2 swings (and maybe goes up or down) and thus, the section connected to P2 in general is not in vertical position. Therefore, in general, the angle coordinate of P2 is not theta.
I suggest a way to look at the problem, though I'm not sure if it's an easy way. The system's position is described by 3 coordinates: theta for P1, alpha and r for P2.

EDIT: I'm sorry, 2 coordinates are enough: theta and either alpha or r.

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Last edited: Jul 11, 2010
3. Jul 12, 2010

### bobred

Hi

The system is initially at rest with P1 at B. P2 hangs vertically below A.

For P2 should I be using rectangular components instead of polar.

James

Last edited: Jul 12, 2010
4. Jul 12, 2010

### hikaru1221

Okay, so that's much easier. Then the equation for P2 is: $$3mg - T = 3m\ddot{y}$$

We also have this: $$y=R\theta$$

Therefore, $$3mg - T = 3mR\ddot{\theta }$$ (1)

We also have 2 equations for P1:

$$2mR \ddot{\theta }=T-2mgcos \theta$$ (2)

$$2mR \dot{\theta }^2=2mgsin \theta - N$$ (3)

From (1) and (2), get rid of T and do the integration, you will have $$\dot{\theta }(\theta )$$. Combine with (3), you should have $$N(\theta )$$.

Actually if you apply the energy conservation law, things are much easier

5. Jul 12, 2010

### bobred

I take it that P2 has a non-polar coordinate system i.e. x/y?

Thanks

6. Jul 12, 2010

### hikaru1221

Yes, that's the easiest way to analyze P2's motion. In this problem particularly, you even only need 1 coordinate (either x or y) for P2.

7. Jul 12, 2010

### bobred

Thanks, great help.

James

8. Jul 13, 2010

### bobred

Hi

I have completed the question, thanks for your help.
But I'm asked to show that

$$\texttt{N}=\frac{6mg}{5}(3\sin\theta-2\theta)$$

I have integrated and substituted in to eqn 3, I get

$$\texttt{N}=2mg\sin\theta-2mR\ddot{\theta}^2$$
$$\texttt{N}=2mg\sin\theta-2mR(\frac{g}{5R}(3\theta-2\sin\theta))$$
$$\texttt{N}=\frac{2mg}{5}(3\sin\theta-3\theta)$$

Can you see where I am going wrong?

Thanks

Last edited: Jul 13, 2010
9. Jul 13, 2010

### hikaru1221

I got the 1st answer. Can you show me your work?

EDIT: I think you got wrong $$\dot{\theta }^2$$
I got: $$\dot{\theta }^2 = \frac{2}{5mR}(3mg\theta - 2mgsin\theta )$$

Last edited: Jul 13, 2010
10. Jul 13, 2010

### bobred

Sorry, I was a bit too hasty, I missed the 2g out, duh!

Thanks

11. Jul 13, 2010

### bobred

Hi

One thing I had forgotten about was the constant of integration, which I have found from the initial conditions of $$\theta=0$$ to be $$\dot{\theta}^2$$. We are also told that the system starts from rest so am I right in thinking that rate of rotation $$\dot{\theta}^2=0$$?

Thanks

12. Jul 13, 2010

### hikaru1221

Yes

13. Jul 13, 2010

### bobred

Thanks for all your help.

James

14. Jul 18, 2010

### squinted

Hi
Been working thru this problem and I get to here and can't figure out how you arrived at the result.. could you explain it further??
Thanks

15. Jul 18, 2010

### bobred

Using

$$\ddot{\theta}=\dfrac{g}{5R}\left(3-2\cos\theta\right)$$

Whoops, see below.

Last edited: Jul 18, 2010
16. Jul 18, 2010

### bobred

Using

$$\ddot{\theta}=\dfrac{g}{5R}\left(3-2\cos\theta\right)$$ multiply both sides by $$\dot{\theta}$$

Using the fact that $$\dfrac{d}{dt}\left(\dot{\theta}^{2}\right)=2\dot{\theta}\ddot{\theta}$$ insert this into the above equation and integrating both sides gives

$$\dot{\theta}^{2}=\dfrac{2g}{5R}\left(3\theta-2\sin\theta\right)+2c$$

From the given initial conditions c = 0

$$N=2mgsin \theta - 2mR \dot{\theta }^2$$ inserting $$\dot{\theta}^{2}=\dfrac{2g}{5R}\left(3\theta-2\sin\theta\right)$$ into tis equation gives

$$\texttt{N}=\frac{6mg}{5}(3\sin\theta-2\theta)$$

James

17. Jul 18, 2010

### squinted

Thanks James :)

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