# Velocity of a block attached to a pulley and cylinder

Tags:
1. Nov 25, 2014

### whonut

1. The problem statement, all variables and given/known data
A cylinder and pulley turn without friction about stationary horizontal axles that pass through their centres. A light rope is wrapped around the cylinder, passes over the pulley, and has a 3.00 kg box suspended from its free end. There is no slipping between the rope and the pulley surface. The uniform cylinder has mass 5.00 kg and radius 40.0 cm. The pulley is a uniform disk with mass 2.00 kg and radius 20.0 cm. The box is released from rest and descends as the rope unwraps from the cylinder. EDIT: Question is: What is the velocity of the block when it has descended 1.5m?

2. Relevant equations

$$\vec{\tau} = \vec{r} \times \vec{F} = I\vec{\alpha}$$
$$I = \frac{1}{2}MR^2$$
$$\alpha=\frac{a}{R}$$
3. The attempt at a solution
I have expressions for the angular accelerations of the 2 discs in terms of the tensions. $Var_C$ stands for some variable with respect to the cylinder, $Var_B$ is the same for the block and $Var_P$ is the same for the pulley.

$$\alpha_C = \frac{2T_2)}{M_{C}R_C}$$
$$\alpha_P = \frac{2(T_{1}-T_2)}{M_{P}R_P}$$

Where I get stuck is how to somehow combine those angular accelerations in order to then convert into a linear acceleration for the block. I've solved this problem by energy conservation but I really need to understand this method.

Hope I've not many faux pas, thanks in advance.

Last edited: Nov 25, 2014
2. Nov 25, 2014

### Staff: Mentor

Your expressions for the angular accelerations of the pulley and cylinder look fine. But what is the question? What are you trying to calculate?

3. Nov 25, 2014

### whonut

Oh hell. Copied the set up and not the question! Apologies. I've edited the original post.

4. Nov 25, 2014

### Staff: Mentor

Ah. So you're looking for the linear acceleration of the box then.

What's the relationship between the angular acceleration of a pulley (or wheel, or cylinder,...) and the tangential acceleration? HInt: it involves the radius.

5. Nov 25, 2014

### whonut

Yet another error in my OP. I put that in but screwed up the LaTeX. It's $a=\alpha{r}$, right? My first thought was that the linear acceleration of the box was just the sum of the 2 cylinders' linear accelerations but I can't get the tension terms to go away. Is it just my algebra?

6. Nov 25, 2014

### Staff: Mentor

The tangential accelerations of the pulley and cylinder are "tied" to that of the rope. Use the angular vs linear tangential relationship to convert the angular accelerations their tangential counterparts.

You need another equation for acceleration, namely that of the box. After all, it's the box's weight that's providing the motivating force for the system. Then you'll have three equations in three unknowns (a, T1, T2).

7. Nov 25, 2014

### whonut

I am so bad at forums! It's $T_1-mg=ma$, I believe. Can I solve it from there?

Last edited: Nov 25, 2014
8. Nov 25, 2014

### Staff: Mentor

Note that you might also consider a conservation of energy approach to the problem. When the box has descended by the given amount, the change in gravitational PE will go into the box's KE and the rotational KE of the pulley and cylinder.

9. Nov 25, 2014

### Staff: Mentor

You'll have to show how you got there. But off hand I'd say no... You don't know what T1 is.

10. Nov 25, 2014

### whonut

I've done the energy thing. Much more straightforward but less apparently less general. I got it from the fact that T1 and mg are the only forces on the block & result in the acceleration. Am I wrong there?

11. Nov 26, 2014

### Staff: Mentor

It's actually a very powerful method which allows you to skip over all the intervening kinematics in going from initial to final state.

The only assumptions you need to make are that the change in gravitational PE for the box ends up as kinetic energy for all three moving items, and that the rope stays taught and links all their speeds.

If you show your work we can check it over.