# Uniform Pully with a single block, do not understand where mass relates?

1. Oct 26, 2008

### StingerManB

OK, my largest issue with dealing with uniform cylindrical pulleys and a single block (object) is understanding where mass of the block relates in any of the equations.

Specifically, I was doing some problems earlier with no issues at all. Then I came across one asking about the mass of the block hanging from the pulley. The book and my notes do not speak of this.
Known:
mass of the pulley
Tension,

This seems like a pretty straight-forward topic, yet I cannot seem to locate much on it.
Can anyone please explain the topic?
Thank you in advance, I have already found much help from these forums, and without them would probably not pass calculus based physics!

2. Oct 26, 2008

### StingerManB

I tried solving for acceleration then using:

ma = mg - T, rearranged as T = m(g-a) but this did not give me a correct solution either. hmmmmm.

3. Oct 27, 2008

### Staff: Mentor

Are you talking about a cylinder that has a massless rope wrapped around it from which a block is hanging? And you're supposed to find the acceleration of the block?
Generally the tension is not known, but you can figure it out.

That looks like a reasonable equation for the block. Note that you have two unknowns, tension and acceleration. You need a second equation--one for the cylinder--to solve for the acceleration.

4. Oct 27, 2008

### StingerManB

Ok, thank you. I will look for some info on this. I am not too familiar with the equations of a pulley.
To clarify the problem, I am looking for the mass of a block hanging from a single pulley.
From the problem I am given the mass of the pulley, the radius of the pulley, and the tension in the cable.
Do you know, specifically, which equation I am missing for the cylinder?
Thank you again for the help. I need to know this subject in and out...

5. Oct 27, 2008

### Staff: Mentor

OK. Instead of the tension being unknown, the mass is unknown. You still have two unknowns and need two equations.
Yes, the rotational equivalent of Newton's 2nd law: Net Torque = Iα

6. Oct 27, 2008

you have PM.