Cylinders (or pulley) cord-motion

[fluidistic]

Homework Statement

Hi PF,
See figure for a clear view of the situation.
(There's a cylinder attached to a ceiling so that it cannot move. There's a cord on it and the cord goes down until another cylinder is attached to it. So that the only cylinder that can move is the one at the bottom. I call it "second cylinder". There's friction between the cord and cylinders such that the cord cannot slide on the cylinders. The radius of both cylinder is R and their mass is worth M.)


1)What is the acceleration of the center of mass of the second cylinder?
2)How much is worth the tension in the rope?
3)What's the velocity of the second cylinder when it has moved a distance of 10R?


2. The attempt at a solution
I'm not even able to find part a).
For the second cylinder, $$\frac{d \vec P}{dt}=\vec F_e=M \vec a_{cm}=\vec P + \vec T_2 \Leftrightarrow \vec a_{cm} =\frac{\vec P + T_2}{M}=\vec g +\frac{\vec T_2}{M}$$.
Now I must find $$\vec T_2$$ in order to answer part a) so by doing it I'd solve part b).
We have that $$T_2=T_1$$.
For the first cylinder, $$\frac{d \vec P}{dt}=0\Leftrightarrow \vec F_e=0\Leftrightarrow \vec P + \vec T_1+ \vec N \Leftrightarrow \vec N =-\vec P - \vec T_1$$.
Now I chose a center of momentum in the center of mass of the first cylinder. I have that $$\vec L= \vec L_{\text{spin}}=I\vec \omega$$. (I do that in order to find out the angular acceleration of the first cylinder and then the acceleration of the point where is exerted $$T_1$$).
Hence $$\frac{d \vec L}{dt}=I\vec \alpha = Ri \wedge -T_1j=-Rt_1k \Leftrightarrow \vec \alpha =-\frac{RT_1}{I}k=-\frac{RT_1}{\frac{MR^2}{2}}k=-\frac{2T_1}{MR}k$$.
In a rigid body we have that the velocity of a point of the body is worth $$\vec v_p =\vec v_{cm}+\vec \omega \wedge \vec r \Rightarrow \vec a_p = \vec a_{cm}+ \vec \alpha \wedge \vec r+ \vec \omega \wedge \vec v = \vec \alpha \wedge \vec r = \vec \alpha \vec r =-\frac{2T_1}{MR}j \wedge Ri=-\frac{2T_1}{M}j$$.
Thus $$\vec T_1 =-2 T_1j$$ which is impossible.

 

[Doc Al]

I'm not even able to find part a).
For the second cylinder, $$\frac{d \vec P}{dt}=\vec F_e=M \vec a_{cm}=\vec P + \vec T_2 \Leftrightarrow \vec a_{cm} =\frac{\vec P + T_2}{M}=\vec g +\frac{\vec T_2}{M}$$.

First off, what's P (I assume momentum) and how can dP/dt have a term equal to P?

How does the acceleration of the cm relate to the angular accelerations of the cylinders?

 

[fluidistic]

First off, what's P (I assume momentum) and how can dP/dt have a term equal to P?

How does the acceleration of the cm relate to the angular accelerations of the cylinders?

Oops... Sorry the first P is the linear momentum and it's not equal to something with it but the other P is the weight. (In Spanish weight is peso so I forgot to translate this.)
I don't understand the question

How does the acceleration of the cm relate to the angular accelerations of the cylinders?

isn't it that the acceleration of the center of mass of a cylinder is equal to the acceleration of this cylinder?

 

[Doc Al]

Oops... Sorry the first P is the linear momentum and it's not equal to something with it but the other P is the weight. (In Spanish weight is peso so I forgot to translate this.)

Weight acts down while tension acts up, so they have different signs.

I don't understand the question isn't it that the acceleration of the center of mass of a cylinder is equal to the acceleration of this cylinder?

There are three accelerations in this problem: α₁, α₂, and a_cm. How are they related?

 

[fluidistic]

There are three accelerations in this problem: α1, α2, and acm. How are they related?

If $$\alpha _1$$ stands for the angular acceleration of the first cylinder, then I think that $$\alpha _1=\alpha _2$$. They are related as this : $$a_{cm}=\alpha R$$. My problem is that I have $$\alpha$$ in function of $$T_1$$. And $$T_1$$ is an unknown.

 

[Doc Al]

If $$\alpha _1$$ stands for the angular acceleration of the first cylinder, then I think that $$\alpha _1=\alpha _2$$.

OK.

They are related as this : $$a_{cm}=\alpha R$$.

That's not correct.

My problem is that I have $$\alpha$$ in function of $$T_1$$. And $$T_1$$ is an unknown.

You should have all the equations needed to solve for the accelerations and T.

 

[fluidistic]

Hmm sorry I don't know why it isn't true that $$a_{cm}=\alpha R$$.

 

[Doc Al]

Hmm sorry I don't know why it isn't true that $$a_{cm}=\alpha R$$.

Answer these questions: What's the acceleration of the rope? What's the acceleration of the bottom pulley with respect to the rope?

 

[fluidistic]

Answer these questions: What's the acceleration of the rope? What's the acceleration of the bottom pulley with respect to the rope?

I don't know what is the acceleration of the rope. My result was $$\frac{2T_1}{M}$$.

 

[Doc Al]

I don't know what is the acceleration of the rope. My result was $$\frac{2T_1}{M}$$.

That's fine. Now answer my second question.

 

[fluidistic]

I'm not sure at all. I'd guess it's 0m/s^2. So $$a_{cm}=\frac{2T}{M}$$...

 

[Doc Al]

I'm not sure at all. I'd guess it's 0m/s^2. So $$a_{cm}=\frac{2T}{M}$$...

If the speed of the cm with respect to the rope is 0, that means that the bottom cylinder is not rotating or unwinding. Do you really think that?

(What torque acts on the bottom cylinder? Does it have an angular acceleration about its center of mass?)

 

[fluidistic]

If the speed of the cm with respect to the rope is 0, that means that the bottom cylinder is not rotating or unwinding. Do you really think that?

Ahahah. ok no.

(What torque acts on the bottom cylinder? Does it have an angular acceleration about its center of mass?)

I chose the origin as being the center of mass of the bottom cylinder and the torque is $$Ri \wedge T_2 j$$. Thus $$\alpha_2 = \frac{2T_2}{MR}$$ which is the same than $$\alpha _1$$ in magnitude but opposite in direction.
Hmm, I've no idea about how to continue from now.

 

[Doc Al]

Try this: Does the rope accelerate down or up? With respect to the rope, does the bottom cylinder accelerate down or up?

 

[fluidistic]

Both accelerates down.

 

[fluidistic]

Thank you for your help Doc Al. I got the answers (now I can see answer in an answer sheet).
I restarted all the problem. I imagined it as being horizontal (I don't know why it helped me a lot), I supposed that $$\vec \alpha _1=- \vec \alpha _2$$ and also that the point on the cord of the second cylinder or whatsoever it is, moves at twice the one of the first cylinder. That was the key to find out $$T$$. From it it was very easy to find $$a_{cm}$$ hence I find the problem badly asked. (They should have inverted part a) with part b)).