# Angular acceleration on a pulley

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1. May 9, 2016

### df102015

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

A mass of 6.1 kg tied to a string is wrapped around a disk as shown. If the disk has a mass of 8.2 kg and a radius of 2.3 m, how fast will the disk be rotating when the weight has fallen 7.4 m and was released from rest?

2. Relevant equations
Θ = S/R = x/R
Θ = 0.5 α t^2
τ = I α
τ = T R
T = mg - ma
I = 0.5 M R^2

3. The attempt at a solution
These are the equations that my teacher gave me. So far, when i combined them i get this...

α = τ/I = T R / 0.5 M R^2 = (mg - ma) R / 0.5 M R^2 = (mg - ma) / 0.5 M R

Which, when numbers are substituted in becomes...

(6.1 x 9.81 - 6.1a) / 0.5 x 8.2 x 2.3 = α

but i still do not have acceleration (a) and the equation
a = Δv / t
gives me two other things that i do not have... change in velocity and time. And the equations using theta that he gave me seem almost useless because the value of theta is not given, although it looks as if it is 90 degrees, but i could be wrong. When i calculate theta using x/R i got 3.2174 which i assume is in pi radians, but could just be wrong altogether. Can somebody please just point me in the right direction, help me with some different equations if they are needed, it is the last problem that i need to finish on the homework. Also i believe my teacher may have mentioned there is a way to solve by using energy equations, if that is easier then please help using that. Thanks so much!

2. May 9, 2016

### TSny

Energy is a nice way to solve it. But you've already done most of the work in using the method of forces and torques. All you need is another relation between α and a. I bet you've covered that relation. It's similar to the equation s = Rθ.

3. May 9, 2016

### df102015

is the equation i am looking for α = a/R ?

4. May 9, 2016

### TSny

Yes, that's it.

5. May 9, 2016

### df102015

Thanks so much!!!! You have no idea how much i appreciate it!

6. May 9, 2016

### Sahil Kukreja

Another way to solve is by conservation of energy:-

loss in G.P.E. of box = gain in K.E. of box + gain in K.E. of Disc.

$mgh=\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2\$
$\omega\$ is clockwise
and for no slipping $v=r \omega\$