# A pulley of mass m using conservation of energy

1. Mar 6, 2017

### physicsdude101

1. The problem statement, all variables and given/known data
Consider a massive pulley of mass m and radius r shown in the figure below, with two objects hanging off it having masses m and 2m, respectively. The pulley can be considered as a uniform disk, and the string is massless, does not stretch and does not slip.
By considering conservation of energy, determine the acceleration of the masses.

2. Relevant equations
$$I=\frac{1}{2}mr^2$$
$$K=\frac{1}{2}mv^2$$
$$K=\frac{1}{2}I\omega^2$$

3. The attempt at a solution
I think the total kinetic energy of the system is $$K=\frac{1}{2}mv^2+\frac{1}{2}(2m)v^2+\frac{1}{2}I\omega^2$$ but I'm not sure how to calculate the total potential energy of the system.

Last edited: Mar 6, 2017
2. Mar 6, 2017

### BvU

Don't see any equation involving energy in your relevant equations ?
Also don't see anything with acceleration ?
Do see unexplained variable $r$, though. Unknown ?

Last edited: Mar 6, 2017
3. Mar 6, 2017

### physicsdude101

I don't know how to get the acceleration without using Newton's Laws, which is what the question is asking me to do.

4. Mar 6, 2017

### BvU

DId you render the problem statement completely, litterally ? I read it asks you to do it with energy conservation.

5. Mar 6, 2017

### physicsdude101

Sorry that is what I meant to say. I am supposed to do it with conservation of energy rather than Newton's laws but I am unsure how to do so.

6. Mar 6, 2017

### BvU

I repeat posts 2 and 4.

7. Mar 6, 2017

### physicsdude101

I'm unsure of what you're asking.

8. Mar 6, 2017

### BvU

From post #2:
What is $r$ ?
Is it known or unknown ?
You now edited in $K$. What about potential energy ?

From post #4:
What is the complete problem statement ?

9. Mar 6, 2017

### physicsdude101

r is unknown. I'm not sure about potential energy because I don't know where the ground is defined. I'm also not sure about acceleration.

I copied the problem statement exactly as I saw it in my homework problems.

10. Mar 6, 2017

### BvU

Ok, means you are supposed to formulate the answer in terms of $m$ and $r$. Funny they don't even mention $r$ -- perhaps it divides out ultimately; we'll see.

You still need more equations (there are various relationships you haven't listed).
Ground can be anywhere. So pick something (you can even pick the ceiling if you want). More importantly: what coordinate system do you plan to define and use ?

11. Mar 7, 2017

### physicsdude101

I think I'll use Cartesian coordinate system with h as the initial height from the ground in both masses. Then the finish height would be 0 for mass 2m? Although I'm not sure about the final height for the second mass.

I suppose another relevant equation would be $$v^2-v_0^2=2a\Delta x$$

12. Mar 7, 2017

### BvU

That is indeed the energy conservation equation. You can take $v_0 = 0$.

m and 2m are connected by a rope (you implicitly use that already with the single variable $v$)
the rope doesn't slip on the pulley

To be honest, I still don't see how this can be answered using energy considerations only. Maybe you can get something containing acceleration by differentiating a form that contains $v^2$ .
If I were you I would embark on a solution attempt with a more conventional approach and then see if it can be bent to accommodate the instruction in the last line of the problem statement.

13. Mar 7, 2017

### physicsdude101

I think I have a solution. First define the two masses as being h metres off the ground initially. Then by the conservation of energy,
$$E_{initial}=E_{final}$$
$$\implies mgh+(2m)gh=\frac{1}{2}mv^2+\frac{1}{2}(2m)v^2+\frac{1}{2}I\omega^2+mg(2h)$$
$$\implies mgh=\frac{3}{2}mv^2+\frac{1}{4}mr^2\frac{v^2}{r^2}$$
$$\implies gh=\frac{7}{4}v^2$$
Now we use the kinematic equation: $$v^2-v_0^2=2a\Delta x$$ which gives
$$\frac{4gh}{7}-0^2=2ah$$
and finally, $$a=\frac{2g}{7}$$

14. Mar 7, 2017

### BvU

Consistent with $mg$ having to cause 3.5 $\times {1\over 2}mv^2$. Well done !

I like this exercise very much: both h and r divide out -- which I hoped for, but it wasn't obvious at the start -- and something simple remains.