# Pulleys with inertia and friction

1. Nov 10, 2013

### johnschmidt

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

http://essbaum.com/images/Pulley_diagram.png [Broken]

We have a bonus question where we are asked to find the mass of $m_{1}$ necessary to accelerate $m_{2}$ upward at $1g$, and we have to include friction in the pulley bearings as well as the pulley inertias. The pulleys are flat disks, and the rope doesn't have mass or stretch.

$m_{2}$ = 1kg
$a_{2}$ = 1g (positive is upward)
$r_{p1}$ (radius of pulley 1) = 10cm
$m_{p1}$ (mass of pulley 1) = 1kg
$\tau_{frictionofp1}$ (friction torque of pulley 1 bearing) = 0.4Nm
$r_{p2}$ (radius of pulley 2) = 20cm
$m_{p2}$ (mass of pulley 2) = 2kg
$\tau_{frictionofp2}$ (friction torque of pulley 2 bearing) = 0.6Nm

2. Relevant equations

Torque $\tau_{3}$ is associated with tension $T_{3}$ and torque $\tau_{2}$ is associated with tension $T_{2}$
$I$ = moment of inertia
$\alpha$ = angular acceleration

3. The attempt at a solution

First I solve the torques of pulley 2:

$\tau_{2} - \tau_{3} = I_{p2} \alpha_{p2} + \tau_{friction of p2}$
$T_{2} r_{p2} - T_{3} r_{p2} = \frac{m_{p2} r_{p2}^2}{2} \frac{a_{2}}{r_{p2}} + \tau_{frictionofp2}$
$T_{2} - T_{3} = \frac{m_{p2} a{2}}{2} + \tau_{frictionofp2}$

Now solve for $T_{3}$

$T_{3} = m_{2} g + m_{2} a_{2}$

Substituting $T_{3}$ into the previous equation gives

$T_{2} - m_{2} g - m_{2} a_{2} = \frac{m_{p2} a{2}}{2} + \tau_{frictionofp2}$

Solve for $T_{2}$

$T_{2} = m_{2} g + m_{2} a_{2} + \frac{m_{p2} a{2}}{2} + \tau_{frictionofp2}$

Great! Now I have $T_{2}$ in terms of known variables.

But now I get stuck. How do I proceed with pulley 1? I can follow the same procedure as above and find the difference between $T_{1}$ and $T_{2}$, but that seems to neglect that pulley 1 is in motion downward.

Similarly if I solve for the forces on pulley 1 (for example $T_{1} + T_{2} = \left(m_{1} + m_{p1}\right) g - \left(m_{1} + m_{p1}\right) a_{1}$ then I am ignoring the forces that go into rotating the inertia of pulley 1...

?

Last edited by a moderator: May 6, 2017
2. Nov 10, 2013

### haruspex

Shouldn't there be a denominator 1/rp2 in the last term?
Both equations are valid. Taking moments about the centre of pulley 1 gives you the torque equation, T1 and T2 acting oppositely, while linear momentum gives an equation with T1 and T2 acting together. Do you think that gives you too many equations?

3. Nov 11, 2013

### johnschmidt

Yes! Good catch.

I guess one shouldn't complain about having too many equations ;-)

Ok, then let's solve for T1 in terms of T2.

$\tau_{1} - \tau_{2} = I_{p1} \alpha_{p1} + \tau_{friction of p1}$
$T_{1} r_{p1} - T_{2} r_{p1} = \frac{m_{p1} r_{p1}^2}{2} \frac{a_{1}}{r_{p1}} + \tau_{frictionofp1}$
$T_{1} - T_{2} = \frac{m_{p1} a_{1}}{2} + \frac{\tau_{frictionofp1}}{r_{p1}}$

Solve for $T_{1}$

$T_{1} = T_{2} + \frac{m_{p1} a_{1}}{2} + \frac{\tau_{frictionofp1}}{r_{p1}}$

Insert $T_{2}$

$T_{1} = m_{2} g + m_{2} a_{2} + \frac{m_{p2} a_{2}}{2} + \frac{\tau_{frictionofp2}}{r_{p2}} + \frac{m_{p1} a_{1}}{2} + \frac{\tau_{frictionofp1}}{r_{p1}}$

Now I also know that $T_{1} + T_{2} = \left(m_{1} + m_{p1}\right) g - \left(m_{1} + m_{p1}\right) a_{1}$

Solve for $m_{1}$

$m_{1} = \frac{T_{1} + T_{2}}{g - a_{1}} - m_{p1}$

Insert $T_{1}$ and $T_{2}$

$m_{1} = \frac{m_{2} g + m_{2} a_{2} + \frac{m_{p2} a_{2}}{2} + \frac{\tau_{frictionofp2}}{r_{p2}} + \frac{m_{p1} a_{1}}{2} + \frac{\tau_{frictionofp1}}{r_{p1}} + m_{2} g + m_{2} a_{2} + \frac{m_{p2} a_{2}}{2} + \frac{\tau_{frictionofp2}}{r_{p2}}}{g - a_{1}} - m_{p1}$

Simplify a bit

$m_{1} = \frac{2 m_{2} g + 2 m_{2} a_{2} + m_{p2} a_{2} + \frac{2 \tau_{frictionofp2}}{r_{p2}} + \frac{m_{p1} a_{1}}{2} + \frac{\tau_{frictionofp1}}{r_{p1}}}{g - a_{1}} - m_{p1}$

I also know that $a_{1} = \frac{a_{2}}{2}$ and can insert that in

$m_{1} = \frac{2 m_{2} g + 2 m_{2} a_{2} + m_{p2} a_{2} + \frac{2 \tau_{frictionofp2}}{r_{p2}} + \frac{m_{p1} \frac{a_{2}}{2}}{2} + \frac{\tau_{frictionofp1}}{r_{p1}}}{g - \frac{a_{2}}{2}} - m_{p1}$

Simplify some more

$m_{1} = \frac{4 m_{2} \left(g + a_{2}\right) + 2 m_{p2} a_{2} + \frac{4 \tau_{frictionofp2}}{r_{p2}} + \frac{m_{p1} a_{2}}{2} + \frac{2 \tau_{frictionofp1}}{r_{p1}}}{2g - a_{2}} - m_{p1}$

Is that right?

It seems to make sense, in that if I zero the mass and friction of the pulleys I get the same equation we had to solve for in another exercise where the pulleys had no mass or friction. And it makes sense that the friction of pulley 2 matters double versus the friction of pulley 1 because of the multiplying effect of pulley 1. But is it right that the inertia of pulley 2 matters 4 times as much as the inertia of pulley 1?

4. Nov 11, 2013

### haruspex

Yes. P2 rotates twice as fast as P1, so exerts twice the retardation. But it also has a 2:1 mechanical advantage. Another way to think of it is the P2 rotating twice as fast will gain 4 times the energy.
I agree with you final equation.

5. Nov 12, 2013

### johnschmidt

Ok, I think I understand. This is like where energy is a function of velocity squared, right?

Thanks haruspex!