Are the Force Equations for Rotational Motion Accurate?

[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 $L=mR^2\omega$

 

[PeroK]

Have you done any Lagrangian mechanics?

 

[Bling Fizikst]

Have you done any Lagrangian mechanics?

No

 

[PeroK]

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:

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

 

[PeroK]

... differentiating the energy equation should work.

 

[Bling Fizikst]

... differentiating the energy equation should work.

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

 

[PeroK]

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 $r$.

 

[Bling Fizikst]

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?

 

[vela]

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 $U(r)$?

 

[PeroK]

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.

 

[Bling Fizikst]

Is it -Mgr

 

[PeroK]

Is it -Mgr

What happens to $M$ as $r$ increases?

 

[Bling Fizikst]

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

 

[PeroK]

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.

 

[Bling Fizikst]

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

 

[PeroK]

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

That's the right answer, but I'm not sure quite how you got there.

 

[vela]

That's the right answer, but I'm not sure quite how you got there.

$F = -kr$ implies $\omega = \sqrt{k/m}$.

 

[PeroK]

$F = -kr$ implies $\omega = \sqrt{k/m}$.

No need to consider only small oscillations?

 

[vela]

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

 

[PeroK]

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:

$$(m+M)\ddot{r} -mr\omega^2- 2mr\omega^2=0$$

In any case, the solution only applies to small oscillations. You will need an approximation somewhere.

 

[Bling Fizikst]

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

Yes

 

[PeroK]

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 :

 

[PeroK]

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.

 

[Bling Fizikst]

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