# Atwood's machine with two connected discs

Tags:
1. Sep 3, 2016

### lukast

1. The problem statement, all variables and given/known data
The system looks like this:

I have two discs which are connected.
Disc 1 has $R_1$(radius) and $M_1$(mass)
Disc 2 has $R_2$ and $M_2$
$R_2 > R_1$
$M_2 > M_1$

on both discs weights are attached on opposite sides.
On smaller $m_1$ is pulling and on bigger $m_2$

$m_2 > m_1$

i need to calculate angular acceleration and accelerations of both weights

2. Relevant equations

$F = m * a$
$\alpha = \frac{a}{R}$
$M = I * \alpha$
$Q = m * g$

3. The attempt at a solution

The force that would cause acceleration of system is equal to :

$F = Q_2 - Q_1$
$F = m_2*g - m_1*g$

The force that will cause tangential acceleration of discs would be equal to :

$I * \alpha = M$
$\frac{1}{2} m * R^2 * \frac{a}{R} = F * R$
$\frac{1}{2}*m * a = F$

Now we know that Disc 2 and $m_2$ will have the same accelerations and Disc 1 and $m_1$ will have the same accelerations. We also now that Disc 1 and Disc 2 also have same angular acceleration. So from that i thought i can write this:

$a_1*(\frac{1}{2} * M_1 + m_1) + a_2(\frac{1}{2}*M_2 + m_2) = F$ (force that cause acceleration)

$a = \alpha * R$

$\alpha * R_1 *(\frac{1}{2} * M_1 + m_1) + \alpha * R_2 *(\frac{1}{2}*M_2 + m_2) = F$

and i have everything to get an alpha but its wrong

if somebody solve this on different way, could please explain why is my approach wrong

Last edited: Sep 3, 2016
2. Sep 3, 2016

### Staff: Mentor

Can you explain how you derived this?

I suggest you set up three force equations: one for each hanging mass and one for the disk. You can combine those equations, adding what you know about the relationships between the angular and linear accelerations.