# Find the magnitude of the acceleration of the two masses.

1. Mar 28, 2010

### mparsons06

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

An Atwood machine consists of two masses (16.4 kg and a 2.9 kg) strung over a 0.43 m diameter pulley whose moment of inertia is 0.031 kg·m2. If there is a constant friction torque of 0.18 N·m at the bearing, compute the magnitude of the acceleration of the two masses.

Can someone just tell me where to start please?

2. Mar 28, 2010

### Redsummers

Not sure if there's more information than the needed in that problem, but first of all you should think for instance of the relation of the moment of inertia and torque.

$$\tau = I\alpha$$

3. Mar 28, 2010

### Redsummers

Maybe I wasn't explicit enough, but to begin with this problem you will first need to state the tension of the string (T=mg-ma). With that you will be able to find the 'applied torque' in terms of the acceleration:
$$\tau_a = (mg - ma)r$$

And once you have this, you will be able to easily use the formula I wrote in the last post.

Note that:
$$\tau_a - \tau_f = \tau$$

and:
$$\alpha = a/r$$.

Last edited: Mar 28, 2010
4. Mar 28, 2010

### mparsons06

5. Mar 28, 2010

### Redsummers

Oh you're right! then you have to think also as there are two tensions, let me develop the corresponding equations: (T1, being from the heavier mass, m1)

$$(T_1 - T_2)r - \tau_f = I\alpha$$

and according to Newton's concerning the tensions:

$$m_1g - T_1 = m_1a$$

$$m_2g - T_2 = m_2a$$

Thus,

$$(T_1 - T_2)r = I\alpha + \tau_f$$

$$(T_1 - T_2) = \frac{Ia}{r} + \tau_f$$

$$T_1 - T_2 = \frac{Ia}{r} + \tau_f$$

And now you would have to plug the tension equations stated above, and do the maths to get the acceleration.

6. Mar 28, 2010

### mparsons06

Where does the part about "constant friction torque of 0.18 N·m at the bearing" come in to play?

7. Mar 29, 2010

### Redsummers

oh the constant friction torque is simply represented as
$$\tau_f$$
It is, as the name represents the friction done by the pulley to the string.

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