# Energy Problem

1. Jun 9, 2005

### kitz

Hi,

I'm having a bit of trouble finding the right equation for this:

The pulley in the figure has radius 0.160 m and a moment of inertia 0.480. The rope does not slip on the pulley rim.
Use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor.

To do this, would the following formula be correct?

$$m_{1}gh = 1/2m_{1}v^2 + 1/2I\omega^2 + m_{2}gh$$

The initial energy would just be potential, right? The 2kg block on the floor would have 0 potential energy, and 0 kinetic energy...

I'm getting an answer of 2.9352, is this correct?

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2. Jun 9, 2005

### Staff: Mentor

Yes, the initial energy is entirely PE of $m_1$. But don't forget that when the mass falls, both masses (and the pulley) will have KE. And you'll need to relate the rotational speed of the pulley to the translational speed of the masses.

3. Jun 9, 2005

### kitz

Thank you sir!

Isn't adding 1/2I*omega^2 relating the rotational and the translational speeds? If not, how do I relate tralsational and rotational to the PE of the 2-kg mass?

Thank you!

Here it is better written:

$$m_{1}gh = \displaystyle{\frac{1}{2}}m_{1}v^2 + \displaystyle{\frac{1}{2}}I\omega^2 + m_{2}gh$$

and in my calculations, I substituted in $$\displaystyle{\frac{v}{R}} for \omega$$

Last edited: Jun 9, 2005
4. Jun 9, 2005

### Staff: Mentor

By "relating translational and rotational speeds" I just meant that you need to use $v = \omega R$, which you did.

What about the kinetic energy of $m_2$?

5. Jun 9, 2005

### kitz

would that just be:

$$m_{1}gh = \displaystyle{\frac{1}{2}}m_{1}v^2 + \displaystyle{\frac{1}{2}}I\omega^2 + m_{2}gh+\displaystyle{\frac{1}{2}}m_{2}v^2$$

6. Jun 9, 2005

### Staff: Mentor

That's the one.

7. Jun 9, 2005

### kitz

Thank you very much!!!!!

After plugging the numbers in, I got 2.814

Thank you, sir!