# Spinning Mass and Finding Omega

1. Sep 3, 2011

### BryMan92

1. The problem statement, all variables and given/known data
http://imageshack.us/f/41/physicsc.png/
My professor gave the following hints:
Write down what you know about how r (position of mass from hole)
varies with time. Then write Newton's 2nd law in polar coordinates.
For part (a), the differential equation that you are looking for comes
from the F_theta = ma_theta equation. Part (b) involves integrating
the result from part (a), and Part (c) uses the F_r = ma_r equation.

2. Relevant equations
F(θ)=m[rθ"+r'θ']

3. The attempt at a solution
For part a, I labeled the force of tension pointing towards the hole. And that's the only force I had.

For part B, I tried doing this:
r=vt (same v of the rope)
r'=v
r'=0

θ=wt
θ'=w
θ"=0

Then, I plugged the above results into the equation:
F(theta)=[rθ"+2r'θ']
F=0+2(-v)w

I then got:
θ''=θ'
m(-v)w=-bw(given)
bw(given)/m=w

Would this be right? I do not thinkso...

Thanks all!

2. Sep 5, 2011

### Spinnor

You know that angular momentum will be constant.

angular momentum = m*omega_0*(r_0)^2 = m*omega(t)*(r(t))^2

but r(t) = [r_0 - v*t]

so omega(t) = omega_0*(r_0)^2/[r_0 - v*t]^2

Also the centripetal force = m*v^2/r = m*omega^2*r

We know omega and r so we should know the centripetal force.

?

Last edited: Sep 5, 2011
3. Sep 5, 2011

### BryMan92

Oh, that's right. I misrepresented my r(t) function.

We had to use Diffy Q's to solve the equation, so I did manage to do:
rΘ"+2rΘ'=0

Then ended up with:
m*omega_0*(r_0)^2 = m*omega(t)*(r(t))^2
(Which I thought was wicked cool equation because I completely forgot about conversation of angular momentum).

Thank you for your help! It's good to know that I could have made that assumption from the start, and thank you for fixing my r(t) function!