Solid Disk Pulley + 2 Mass system, v derivation using 2 methods

[Dorian]

Homework Statement

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Homework Equations

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The Attempt at a Solution

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The issue for me starts with (but probably doesn't end with) replicating the velocity equation using the Conservation of Energy equations. Is this Second Law derived equation correct to begin with? I've tried and have arrived at a few different answers using the Conservation of Energy equations. One was V=(gh)^(1/2). Another was V=(2gh/3)^(1/2).

edit: Oh, shoot. You may have to zoom in for the relevant equations. My apologies :(

 

[kuruman]

Hi Dorian and welcome to PF.

Can you show your two different solutions using energy conservation? Your solution using dynamics has a problem. The first equation in (b) should result in $T_2=\frac{3}{2}mg\sin \theta-3ma$. You forgot to carry the sine through.

 

[Dorian]

Hi Dorian and welcome to PF.

Can you show your two different solutions using energy conservation? Your solution using dynamics has a problem. The first equation in (b) should result in $T_2=\frac{3}{2}mg\sin \theta-3ma$. You forgot to carry the sine through.

I didn't carry it because $T_2=3mg\sin \theta-3ma=3mg\sin(30)-3ma=\frac{3}{2}mg-3ma$

 

[kuruman]

OK, thanks for the clarification. Then your answer in (b) should be correct. What about energy conservation? Can you show how you got your different answers?

 

[Dorian]

OK, thanks for the clarification. Then your answer in (b) should be correct. What about energy conservation? Can you show how you got your different answers?

Sure thing! Thank you for the prompt responses, by the way. :)

I think the issue for me arises in the potential energy. For the right side of the equation, I have $\frac{1}{2}3mv^{2}+\tfrac{1}{2}mv^{2}+\frac{1}{2}(\frac{1}{2}2mR^{2})(\frac{v^{2}}{R^{2}})=\frac{5mv^{2}}{2}$. But the right side leaves me to believe I should have mgh on the left side. I don't understand how. 3mgh-2mgh, perhaps? I don't know.

 

[kuruman]

But the right side leaves me to believe I should have mgh on the left side. I don't understand how. 3mgh-2mgh, perhaps? I don't know.

Yes you need to add potential energy terms it doesn't matter on which side because the zero of potential energy is arbitrary. So let's say that the potential energy is zero at the point where they are released from rest. Then $U_i=0$ on the left side. For the right side, note that the hanging mass gains potential energy while the mass on the incline loses potential energy. Then $U_f=mgh_1-3mgh_2$. Now $h_2=h$ as shown in the figure, but $h_1 \neq h$. Put it together.

 

[Dorian]

Yes you need to add potential energy terms it doesn't matter on which side because the zero of potential energy is arbitrary. So let's say that the potential energy is zero at the point where they are released from rest. Then $U_i=0$ on the left side. For the right side, note that the hanging mass gains potential energy while the mass on the incline loses potential energy. Then $U_f=mgh_1-3mgh_2$. Now $h_2=h$ as shown in the figure, but $h_1 \neq h$. Put it together.

edit: I think I figured it out:
$h_{1}=\Delta s$
and
$sin(30^{\circ})=\frac{h}{\Delta s}\rightarrow {\Delta s}=\frac{h}{sin(30^{\circ})}=2h$

As such,
$PE_{0}=PE=3mgh=mg\Delta s\rightarrow 3mgh-mg\Delta s=3mgh-mg(2h)=3mgh-2mgh=mgh$

Now for Conservation of Energy (as simplified earlier):
$mgh=\frac{5}{2}mv^{2}\rightarrow v=\sqrt{\frac{2gh}{5}}$

Please let me know if this makes sense.

 

[kuruman]

Your treatment of potential energy doesn't make sense. Especially the equality $PE_0=PE$ What do these two stand for and why are they equal? Where is your zero of potential energy?
I suggest that you start by writing the energy conservation equation at two points: point A is the starting point from rest and point B is where mass $3m$ has dropped a vertical distance $h$ and mass $m$ has risen vertical distance $2h$. Assume that the potential energy at point A is $PE_A=0$. Since the system starts from rest, $KE_A =0$. By energy conservation, $KE_B+PE_B=0$ as well. At this point you need to find expressions for $KE_B$ and $PE_B$, substitute and solve for the speed $v$.

Now for Conservation of Energy (as simplified earlier):

As a matter of good habit I would recommend against using previous results unless you are 100% certain that they are correct to avoid propagating mistakes. Restart from the beginning, preferably using a different path.

 

[Dorian]

Your treatment of potential energy doesn't make sense. Especially the equality $PE_0=PE$ What do these two stand for and why are they equal? Where is your zero of potential energy?
I suggest that you start by writing the energy conservation equation at two points: point A is the starting point from rest and point B is where mass $3m$ has dropped a vertical distance $h$ and mass $m$ has risen vertical distance $2h$. Assume that the potential energy at point A is $PE_A=0$. Since the system starts from rest, $KE_A =0$. By energy conservation, $KE_B+PE_B=0$ as well. At this point you need to find expressions for $KE_B$ and $PE_B$, substitute and solve for the speed $v$.

As a matter of good habit I would recommend against using previous results unless you are 100% certain that they are correct to avoid propagating mistakes. Restart from the beginning, preferably using a different path.

They're from the Conservation of Energy equation, and what you stated is precisely what I did. I'm sorry you didn't interpret it the way I intended it to be interpreted. What I provided wasn't the complete product, merely me looking for a comment on if this part of my derivation made sense, not the entire picture.

$PE_{0}$ is the initial potential energy of the system. $PE$ is the final potential energy of the system. The same logic follows for initial kinetic energy $KE_{0}$ and final kinetic energy $KE$. I eschewed a few steps that I had written down already in the hopes that I wouldn't have needed to provide them since it was already discussed. I'm sorry, sincerely.

My reference height for $mass_{2}=3m$ is its ending point. My reference height for $mass_{1}=m$ is its starting point.

The entire formula for the system is

$PE_{0}+KE_{0}=PE+KE$,

which is

$\ PE_{m2}=PE_{m10}=KE_{m10}=KE_{m20}=0 \\ PE_{m20}=3mgh\ ,\ PE_{m1}=mg(2h)\ ,\ KE_{m1}=\frac{1}{2}mv^{2}\ ,\ KE_{m2}=\frac{1}{2}(3m)v^{2}\ ,\ KE_{rot}=\frac{1}{2}\mathbb{I} \omega^{2}=\frac{1}{2}(2mR^{2})(\frac{v^{2}}{R^{2}})$

$3mgh=mg(2h)+\frac{1}{2}(3m)v^{2}+\frac{1}{2}mv^{2}+\frac{1}{2}(2mR^{2})(\frac{v^{2}}{R^{2}})$

 

[kuruman]

Your last equation is correct, but I am not convinced that you know what you are doing as opposed to cooking up an equation that gives you the answer that you know is correct. Your expression for the total kinetic energy is correct. Now assume that zero potential energy is at the starting point. This means (in your notation) that $PE_{m10} = 0$ and $PE_{m20} = 0$ so that $PE_0 = PE_{m10}+PE_{m20}=0$ is the initial potential energy. The initial kinetic energy is zero so that the initial mechanical energy, $ME_0=PE_0+KE_0=0$, is also zero.

Energy conservation demands that the final mechanical energy must equal the initial mechanical energy, $ME=PE+KE=0$.
You have correctly found the final kinetic energy to be
$KE=\frac{1}{2}(3m)v^{2}+\frac{1}{2}mv^{2}+\frac{1}{2}(2mR^{2})(\frac{v^{2}}{R^{2}})$
What is the correct expression for the final potential energy $PE~$?

 

[Dorian]

but I am not convinced that you know what you are doing as opposed to cooking up an equation that gives you the answer that you know is correct.

Goodness. That was a little condescending and I've been nothing but respectful toward you. :/

In anyway event, I used different reference heights for the two masses. If I gave both $PE_{m10}$ and $PE_{m20}$ zero potential energy for both of their starting positions, then the final potential energy, $PE_{m1}$ and $PE_{m2}$ would be $mg(2h)$ and $-3mgh$, respectively.

So, $PE=mg(2h)-3mgh$

 

[kuruman]

Goodness. That was a little condescending and I've been nothing but respectful toward you. :/

In anyway event, I used different reference heights for the two masses. If I gave both $PE_{m10}$ and $PE_{m20}$ zero potential energy for both of their starting positions, then the final potential energy, $PE_{m1}$ and $PE_{m2}$ would be $mg(2h)$ and $-3mgh$, respectively.

So, $PE=mg(2h)-3mgh$

That's it, you got it! I meant no disrespect, and I am sorry if you thought I did.

 

[Dorian]

That's it, you got it!

Apology accepted. =) I am truly appreciative of your prompt responses and your help. Thank you, sincerely.