# Mass falling and pulling a cylinder without sliding

1. May 1, 2016

### Karol

1. The problem statement, all variables and given/known data
A cylinder of mass M and radius r is at distance l0 from the edge. mass m is hanging from a rope. no friction at the weightless pulley. the cylinder starts rotating without sliding.
What's the acceleration of the COM of the cylinder.
What's the velocity of the COM of the cylinder when it reaches the edge.

2. Relevant equations
Torque and acceleration of a rigid body: $M=I\alpha$
Moment of inertia of a solid cylinder: $I_c=\frac{mr^2}{2}$
Shteiner's theorem: $I_b=I_a+md^2$

3. The attempt at a solution
Relation between angular and linear acceleration of COM:
$$2r\alpha=\dot{x_A}\;\rightarrow\; \alpha=\frac{\dot{x_A}}{2r}$$
Tension in rope is T:
$$\left\{ \begin{array}{l} mg-T=m\dot{x_A} \\ 2rT=(I_c+Mr^2)\frac{\dot{x_A}}{2r} \end{array}\right.$$
$$\rightarrow \dot {x_A}=\frac{4mgr^2}{I_c+(M+4m)r^2}$$
The acceleration of the COM: $\dot{x_c}=\frac{1}{2}\dot{x_A}$
The velocity at the edge:
$$v^2=2\dot{x_c}l_0\;\rightarrow\; v^2=\dot{x_A}l_0$$

2. May 1, 2016

### ehild

Think it over.

3. May 2, 2016

### haruspex

Presumably everywhere you wrote $\dot x_A$ you mean $\ddot x_A$, xA being the displacement of m.
ehild seems to take exception to your equation relating that to angular acceleration, but it looks right to me.
I agree with your answers, but you should simplify them using the formula you quote for Ic.

4. May 2, 2016

### kinemath

The angular acceleration should be r instead of 2r

5. May 2, 2016

### ehild

I see, the equations are written for the acceleration of point A, with respect to the point of contact.
The problem can be solved by conservation of energy, too, taking into account that the vertical displacement (and velocity) of m is twice the horizontal displacement/velocity of the CM.

Last edited: May 2, 2016
6. May 2, 2016

### haruspex

r is a radius, not an acceleration. What do you mean?

7. May 2, 2016

### Karol

$$2r\alpha=\ddot x_A \;\rightarrow\; \alpha=\frac{\ddot{x}_A}{2r}$$
$$\left\{ \begin{array}{l} mg-T=m\ddot x_A \\ 2rT=(I_c+Mr^2)\frac{\ddot x_A}{2r} \end{array}\right.$$
$$\rightarrow \ddot x_A=\frac{4mgr^2}{I_c+(M+4m)r^2}$$
$$\ddot x_c=\frac{1}{2}\ddot x_A=\frac{4mgr^2}{2\left[ \frac{Mr^2}{2}+(M+4m)r^2 \right]}=\frac{4mg}{3M+2m}$$
$$v^2=2\ddot x_c l_0\;\rightarrow\; v^2=\ddot x_A l_0 \;\rightarrow \; v=2\sqrt{\frac{2mgl_0}{3M+2m}}$$
$$mgl_0=\frac{1}{2}I_B\omega^2=\frac{1}{2}\left( I_c+Mr^2 \right)\omega^2=\frac{\omega^2}{2}\left[ \frac{Mr^2}{2}+Mr^2 \right]$$
$$\rightarrow \omega^2=\frac{4mgl_0}{3Mr^2}$$
$$v=\omega r=...=2\sqrt{\frac{mgl_0}{3M}}$$
They differ

8. May 2, 2016

### ehild

You forgot the KE of the hanging mass. Lo is the distance traveled by the CM. There are mistakes in the other derivation, too.

9. May 2, 2016

### Karol

$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}I_B\omega^2$$
$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}(I_C+Mr^2)\omega^2$$
$$v=2\omega r\;\rightarrow\; -mgl_0+\frac{1}{2}m\cdot 4\omega^2r^2=\frac{1}{2}\left[ \frac{Mr^2}{2}+Mr^2 \right]\omega^2$$
$$\rightarrow\omega=\frac{2}{r}\sqrt{\frac{mgl_0}{M}},\;\; v=\omega r\;\rightarrow\; v=2\sqrt{\frac{mgl_0}{M}}$$

10. May 2, 2016

### ehild

When distributing factor 2 in the denominator, you divided 4m by 2 instead of multiplying.
It would be otherwise correct.

11. May 3, 2016

### Karol

$$\ddot x_c=\frac{1}{2}\ddot x_A=\frac{4mgr^2}{2\left[ \frac{Mr^2}{2}+(M+4m)r^2 \right]}=\frac{4mg}{3M+8m}$$
$$v^2=2\ddot x_c l_0 \;\rightarrow \; v=2\sqrt{\frac{2mgl_0}{3M+8m}}$$
But still differs from the energy method in post #9 which gave:
$$v=2\sqrt{\frac{mgl_0}{M}}$$

12. May 3, 2016

### ehild

I do not follow your notations. The KE of the cylinder is the sum of those of translational and rotational motion. If the CM travels with speed V, ω=V/r. KE(cylinder)= 3/4MV2.
The falling block moves with speed of 2V, and the change of its height is 2l0.

13. May 3, 2016

### dean barry

Ive assumed that linear distance moved by the hanging mass and the cylinder will be the same.
The driving force (F) for the system is = m * g (Newtons)
Assuming its a solid cylinder rolling without slipping, then in this problem, its effective mass = 1.5 * M
So the linear acceleration (a) is from : F / ( ( 1.5 * M ) + m )

14. May 3, 2016

### haruspex

Not a good start.

15. May 3, 2016

### Karol

In:
$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}I_B\omega^2$$
v is the speed of the falling mass, it's kinetic energy contribution. on the right side, the $\frac{1}{2}I_B\omega^2$ is the combined translational+rotational energy of the cylinder.
Later, in:
$$\rightarrow\omega=\frac{2}{r}\sqrt{\frac{mgl_0}{M}},\;\; v=\omega r\;\rightarrow\; v=2\sqrt{\frac{mgl_0}{M}}$$
The v is for the center of the cylinder. indeed not a wise notation, but i assumed it will be understood.

16. May 3, 2016

### ehild

But your result is wrong. You included the KE of the falling block in the first equation, but ignored later.

17. May 3, 2016

### Karol

I don't have and don't know how to include the KE of the falling block later. i made:
$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}I_B\omega^2$$
The connection between v, the velocity of the falling block at the edge of the table, and ω, the angular velocity at the edge is $v=2\omega r$, and i am left only with ω.
Now i translate ω that i found into vc by $v_c=\omega r$. where and how to take into consideration KE?

18. May 3, 2016

### ehild

When you collect the terms with ω2, there should be a term with m. But it is missing from your formula $\omega=\frac{2}{r}\sqrt{\frac{mgl_0}{M}}$
That is correct.

19. May 3, 2016

### Karol

$$v=2\omega r\;\rightarrow\; -mgl_0+\frac{1}{2}m\cdot 4\omega^2r^2=\frac{1}{2}\left[ \frac{Mr^2}{2}+Mr^2 \right]\omega^2$$
$$\left( \frac{3}{4}Mr^2-2mr^2 \right)\omega^2=-mgl_o\;\rightarrow\; \omega=\frac{2}{r}\sqrt{\frac{-mgl_0}{3M-8m}}$$
Not good, negative under the square root.

20. May 4, 2016

### haruspex

That says the gain in energy of the descending mass equals the gain in energy of the rotating cylinder. I'm sure you did not mean that.