# Are the Force Equations for Rotational Motion Accurate?

• Bling Fizikst
Bling Fizikst
Homework Statement
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Relevant Equations
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Writing force equations for block ##m## : $$T-mr\omega^2=m\ddot{r}$$ and for block ##M## : $$Mg-T=M\ddot{r}$$ I think there are mistakes in my equations as they are leading to nowhere and morever i think force methods are really risky in this regard . It would be better to write the total energy of the system which i don't know how to . I tried to write $$\frac{1}{2}m\dot{r}^2 +\frac{1}{2}M\dot{r}^2 + \frac{L^2}{2mr^2} +U(r)=E$$ where ##L=mR^2\omega##

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Have you done any Lagrangian mechanics?

PeroK said:
Have you done any Lagrangian mechanics?
No

PeroK
Bling Fizikst said:
No
It should come out from conservation of AM and energy. You're aiming for an equation for ##\ddot r##. Which you should be able to get from:

Bling Fizikst said:
$$\frac{1}{2}m\dot{r}^2 +\frac{1}{2}M\dot{r}^2 + \frac{L^2}{2mr^2} +U(r)=E$$ where ##L=mR^2\omega##

... differentiating the energy equation should work.

PeroK said:
... differentiating the energy equation should work.
I did that but how do i deal with the potential energy of the system : ##U(r)## $$(m+M)\ddot{r}-\frac{L^2}{mr^3}+U'(r)=0$$

Bling Fizikst said:
I did that but how do i deal with the potential energy of the system : ##U(r)##
You should be able to express that in terms of ##r##.

PeroK said:
You should be able to express that in terms of ##r##.
Need help on this , really deadstuck . Writing forces and using ##\frac{-\partial U}{\partial r}=F## doesn't seem to help?

Bling Fizikst said:
I did that but how do i deal with the potential energy of the system : ##U(r)## $$(m+M)\ddot{r}-\frac{L^2}{mr^3}+U'(r)=0$$
What precisely is stopping you from writing down an expression for ##U(r)##?

Bling Fizikst said:
Need help on this , really deadstuck . Writing forces and using ##\frac{-\partial U}{\partial r}=F## doesn't seem to help?
How does the position of the mass M depend on ##r##? Take ##r = R## as the zero point.

Is it -Mgr

PeroK
Bling Fizikst said:
Is it -Mgr
What happens to ##M## as ##r## increases?

it goes up so +Mgr? please forgive me if i am acting dumb , this is my first time using the potential energy function in this way . It gives me the wrong ans anyways , so it is wrong

Bling Fizikst said:
it goes up so +Mgr? please forgive me if i am acting dumb , this is my first time using the potential energy function in this way
Yes. Note that ##Mg## is related to ##L## by the initial equilibrium.

$$(m+M)\ddot{r} -mr\omega^2- 2mr\omega^2=0$$ $$\implies \omega'=\sqrt{\frac{3m\omega^2}{m+M}}$$

Bling Fizikst said:
$$(m+M)\ddot{r} -mr\omega^2- 2mr\omega^2=0$$ $$\implies \omega'=\sqrt{\frac{3m\omega^2}{m+M}}$$
That's the right answer, but I'm not sure quite how you got there.

PeroK said:
That's the right answer, but I'm not sure quite how you got there.
##F = -kr## implies ##\omega = \sqrt{k/m}##.

vela said:
##F = -kr## implies ##\omega = \sqrt{k/m}##.
No need to consider only small oscillations?

Oh, you were asking about the steps preceding that last step. Yeah, the OP isn't exactly forthcoming with his reasoning.

Bling Fizikst said:
I did that but how do i deal with the potential energy of the system : ##U(r)## $$(m+M)\ddot{r}-\frac{L^2}{mr^3}+U'(r)=0$$
I wonder whether you replaced ##L^2## with ##mr^2w## here? To get from the above equation to this:
Bling Fizikst said:
$$(m+M)\ddot{r} -mr\omega^2- 2mr\omega^2=0$$
In any case, the solution only applies to small oscillations. You will need an approximation somewhere.

PeroK said:
I wonder whether you replaced ##L^2## with ##mr^2w## here? To get from the above equation to this:
Yes

Note that ##w## is a constant, like ##M, g## and ##R## and is the fixed angular velocity at the equilibrium. ##r## and ##\dot \theta## are the variables. We have:
$$L = mR^2w = mr^2\dot \theta$$Also, with ##U(r) = Mgr##, we have:
$$\frac d{dt} U(r) = Mg\dot r$$And
$$\frac d {dt}\dot r^2 = 2\dot r \ddot r$$

Bling Fizikst
Anyways , i found a really elegant solution in David Morin :

Bling Fizikst said:
Anyways , i found a really elegant solution in David Morin : View attachment 342284
The idea is that you do these problems yourself. Not just to find a solution online.

PeroK said:
The idea is that you do these problems yourself. Not just to find a solution online.
This was a new learning experience for me , so i don't really mind . But i do agree that one should try to solve problems on their own . At the end , it just matters that i learnt to solve it even if i had to look at the solution.
I just learnt a new technique and i am happy!

Also , i would love to thank you for patiently helping me throughout the problem

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