# Dual pulley

1. Dec 16, 2003

### Boltak

Hi everyone...

I've been struggling with a physics problem for 2 days now... I'm really not sure what to do.

Problem -
A pair of 10.0 kg masses are suspended from (massless) strings wrapped around a dual pulley. The radius of the smaller shaft on the pulley is half the radius of the larger shaft on the pulley. If the total mass of the pulley is also 10.0kg, and the pulley is considered to be a uniform disk, what is the acceleration (magnitude and direction) of mass ($$m_{1}$$) connected to the larger shaft?

Basically... one 10.0 kg mass is attached on the right side of the small shaft, while other 10.0kg mass is attached to the left side of the bigger shaft.

picture at http://members.cox.net/lorddreg/p9.jpg

I was assuming $$m_{1}$$ would most likely accelerate down.

This is what I have so far... I'm not sure if it is correct or not ...

For $$m_{1}$$ ...
$$\sum{F = m * a_{y} = -T_{1} + m_{1} * g}$$

For $$m_{2}$$ ...
$$\sum{F = m * a_{y} = T_{2} - m_{2} * g}$$

I would appreciate any help!

Thank you

Last edited by a moderator: Apr 20, 2017
2. Dec 16, 2003

### StephenPrivitera

You must consider torques about the axis through the center.
The net external torque is
$$m_1R - m_2r=I_{net}\alpha$$
so
$$\alpha=\frac{m_1R - m_2r}{I_{net}}$$
where
$$I_{net}=\frac{1}{2}MR^2+\frac{1}{2}mr^2+m_1R^2+m_2r^2$$
you know r=1/2R, so you should be able to find the relation between m and M (the mass of the individual pulleys)
also, a_1=&alpha;R and a_2=&alpha;r
so there you have it