# Calculate the speed of the block

1. Dec 8, 2004

### gunnar

Hi. I have a problem.
Two metal disks with different radius and different weight are welded together. The smaller disk has a string wrapped around it with a block on the other end. I have to calculate the speed of the block after 2 meters. I'm trying to use this equation: v=sqrt(2gh/1+M/2m) to solve this. Where M is the total mass of the disks and block while m is the mass of the block. This is not working. Do you guys know a way around this, I guess I need some kind of a relation between the two disks.

2. Dec 8, 2004

### Galileo

I don't understand the entire problem. How are the disks set in motion?
Are they allowed to roll on a table with the block hanging over a pulley at the end of the table? Or something else?

3. Dec 8, 2004

### Evil_Kyo

If i've understood correctly the two disks act as a pulley with the mass at the end of the string. If it is the case you can work it as a typical solid rigid problem calculating the angular acceleration and relating it with the linear acceleration of the mass. Then it becomes a "semi-free" fall problem.

Or you can take the simpler approach of conservation of energy:

Supposing the pulley is fixed so its potential energy remains constant, the initial energy of the system is

$$mgh_1$$

and the final would be

$$\frac{1}{2}(I_1+I_2)\omega^2+\frac{1}{2}mv^2+mgh_2$$

$$I_1$$ and $$I_2$$ are the moments of inertia of the disks.

you can relate $$v$$ and $$\omega$$ by $$v=\omega r$$

where $$r$$ is the radius of the disk where the string is wrapped.

I think you can work the rest. I think there's no major mistake, feel free to correct me.

Last edited: Dec 8, 2004
4. Dec 8, 2004

### gunnar

You understod the probmlem I think but I'm not quite following you Evil_kyo. I only need one kind of h since it's falling from 0 to 2 m. I have been trying to solve this with conservation of energy. I can't see from you equation what the tangential speed v should be, is the equation equal to zero or mgh?

With thanks,

Gunnar

5. Dec 8, 2004

### prasanna

The only kind of h that you need is nothing but Evil_Kyo's $$h_1 - h_2$$(taking origin at the pulley)
Since your $$h_1 = 0$$ and $$h_2 = -2$$,

Your only kind of h = $$\0-(-2) = 2$$

Tangential speed cannot be mgh!!!

Last edited: Dec 9, 2004
6. Dec 8, 2004

### Evil_Kyo

As prasanna has noted i haven't selected a origin for the potential energy.
$$h_1$$ and $$h_2$$ mean the initial and final position of the falling mass attached to the pulley. You don't know it but you know its difference $$h_1-h_2$$ that equals the length traveled by the falling mass 2 metres.

Conservation of energy states that the initial energy of the system equals the final energy, then

$$mgh_1=\frac{1}{2}(I_1+I_2)\omega^2+\frac{1}{2}mv^2+mgh_2$$

rearranging

$$mg(h_1-h_2)=\frac{1}{2}(I_1+I_2)\omega^2+\frac{1}{2}mv^2$$

now we have the equation relating the angular speed $$\omega$$ of the pulley and the linear speed $$v$$ of the falling mass.

$$v=\omega r$$

Imagine the pulley rotates an angle $$\omega$$ in a unit of time, then the lenght of string unrolled is equal to the lenght of arc subtended by this angle $$s=\omega r$$. If the pulley rolls $$\omega$$ in a unit of time the mass falls $$\omega r$$ in a unit of time, then $$\omega r$$ is the speed $$v$$ of the mass.

Using this equation we can simplify

$$mg(h_1-h_2)=\frac{1}{2}(I_1+I_2)(\frac{v}{r})^2+\frac{1}{2}mv^2$$

$$mg(h_1-h_2)=\frac{1}{2}(\frac{I_1+I_2}{r^2}+m)v^2$$

$$v^2=\frac{2mgr^2(h_1-h_2)}{I_1+I_2+mr^2}$$

It's finished if there is no major mistake, i hope this helps you.

7. Dec 8, 2004

### gunnar

It works, of course. Thanks a lot.
I love this place.