# Rotational Dynamics

1. Jan 12, 2013

### golanor

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

A hollow cylinder of internal radius r and external radius R and mass M is connected by a string to a weight m.
What are the angular and linear acceleration of the cylinder? Assume that the cylinder is rolling without slipping.
What is the acceleration of the weight and what is the tension on the string?

2. Relevant equations
*everything*

3. The attempt at a solution
First, by using dimensional analysis i can see that the solution should be something along the lines of:
$$a=\frac{m*r}{M*R}g$$
I tried to do newton's second law, for the cylinder i get that:
$$\sum F=T=\text{Ma}$$
torque:
$$\sum \tau =T*R=\text{I\alpha }$$
moment of inertia around the center of mass:
$$I=M\left(R^2-r^2\right)$$
and the relation between the angular and linear acceleration:
$$\alpha =\frac{a}{r}$$

I have no idea what i did wrong or what do i need to do next.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Jan 12, 2013

### haruspex

Because the cylinder is rolling, you need to be a bit careful about the moments. If you take moments about the centre of mass of the cylinder then you have to include the torque from friction at the point of contact. Alternatively, you can take moments about the point of contact and use the parallel axis theorem for the moment of inertia.
You've left friction out of the linear acceleration equation for the cylinder.
You also need a linear acceleration equation for the mass m.

Last edited: Jan 12, 2013
3. Jan 13, 2013

### golanor

$$\sum F=m*g-T=m*a$$

Shouldn't the friction be 0, since the cylinder is rolling without slipping?

4. Jan 13, 2013

### haruspex

Yes.
If the friction is zero it's sure to slip. 'Rolling' tells you that static friction is enough to prevent slipping.

5. Jan 13, 2013

### golanor

I understand..just got confused by something we proved in class.
So, the linear acceleration should be: $$\sum F=T-f=M*a_1$$
and the torque around the center of mass: $$\sum \tau =R(T+f)=\text{I\alpha }$$
Around the point of contact: $$\sum \tau =2R*T=\text{I\alpha }$$
And in this case, the moment of inertia will be $$I=M\left(R^2-r^2\right)+M\left(R^2\right)$$

6. Jan 13, 2013

### haruspex

A couple of problems with the first term on the RHS there. What would the MI be about the cylinder's centre if r were only a little less than R?

7. Jan 13, 2013

### golanor

Well, i forgot the 1/2 there. it's basically the MI of the large cylinder minus the one we took out. They are co-centric so there is no distance to add. if i look at extreme cases - r<<R it's as if the cylinder is complete, r=R - there is no mass. = the equation makes sense.
$$I=\frac{1}{2}M(2R^{2}-r^{2})$$ - this is relative to the contact point.
i know it's wrong tho, since when i put it into the equation i get $$r^{2}=0$$
the thing is, i don't know what i'm doing wrong.

8. Jan 13, 2013

### MrWarlock616

How do you know the mass of the cylinder you took out is the equal to the mass of the original cylinder?

9. Jan 13, 2013

### golanor

They have the same density.
$$I=\rho \left(\int _0^h\int _0^{2\pi }\int _0^Rr^3drd\phi dz+\int _0^h\int _0^{2\pi }\int _0^rr^3drd\phi dz\right)=\frac{M}{\pi *h\left(R^2-r^2\right)}h*\frac{\pi }{2}\left(R^4-r^4\right)=\frac{M}{2}\left(R^2+r^2\right)$$

10. Jan 13, 2013

### haruspex

That's better. (And of course +MR2 for parallel axis.)

11. Jan 14, 2013

### golanor

The point I missed was the relation between the cylinder's acceleration and the weight's acceleration.