Calculating Speed of 4kg Block using Energy Methods

[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?

 

[Doc Al]

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...

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.

 

[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 $$\displaystyle{\frac{v}{R}} for \omega$$

 

[Doc Al]

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?

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

Here it is better written:

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

What about the kinetic energy of $m_2$?

 

[kitz]

would that just be:

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

 

[Doc Al]

That's the one.

 

[kitz]

Thank you very much!

After plugging the numbers in, I got 2.814


Thank you, sir!