# Torque/Rotation Problem

1. Mar 26, 2014

### sreya

1. The problem statement, all variables and given/known data

A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, as shown in the figure (see the figure (Figure 1) ). The pulley is a uniform disk with mass 11.8kg and radius 55.0cm and turns on frictionless bearings. You measure that the stone travels a distance 12.7m during a time interval of 2.80s starting from rest.

Find the mass of the stone.

Find the tension in the wire.

Figure 1

2. Relevant equations

Work Kinetic Energy Theorem
Kinematics
Torque(maybe?)

3. The attempt at a solution

$a_{stone}=\frac{2\Delta X}{t^2}=3.24$
$T=m_{stone}g-m_{stone?}a$
$\Delta K = W_{wheel} + W_g=0$
$\Delta K = \frac{I\omega_f^2}{2}+W_g$
$\omega = \alpha t$ ??

And that's where I got stuck, I don't know how to derive the final angular velocity

Take 2:

Okay so after going back to drawing board my thinking is this: the stone acts as a torque on the wheel so

$\tau=rFsin\theta=Tr=\frac{1}{2}*M_{wheel}r^2\alpha$

$T=\frac{M_{wheel}r\alpha}{2}$

$T=mg-ma=>\frac{M_{wheel}r\alpha}{2}=m(g-a)$

$a=a_t=\alpha r => \alpha = a/r$

$\frac{M_{wheel}a}{2(g-a)}=m$

solve for m

That ended up being the right answer

Last edited: Mar 26, 2014
2. Mar 26, 2014

### dauto

You left out the part of the question that asks a question. What are you trying to calculate?

3. Mar 26, 2014

### sreya

4. Mar 26, 2014

### dauto

The first two lines of your solution look right to me. The third line doesn't. Why would ΔK be zero? The final angular velocity is related to the final speed by the no slip condition v=ωR.

5. Mar 26, 2014

### sreya

Conservation of energy? I wasn't sure if I could apply that due to the torque but it's the only way I could think of to figure out mass.

Edit:

Okay so after going back to drawing board my thinking is this: the stone acts as a torque on the wheel so

$\tau=Tr=\frac{1}{2}*M_{wheel}r^2\alpha$

$T=\frac{M_{wheel}r\alpha}{2}$

$T=mg-ma=>\frac{M_{wheel}r\alpha}{2}=m(g-a)$

$a=a_t=\alpha r => \alpha = a/r$

$\frac{M_{wheel}a}{2(g-a)}=m$

solve for m

That ended up being the right answer

Last edited: Mar 26, 2014