# Homework Help: Rigid body, physical pendulum

1. Nov 22, 2008

### fluidistic

1. The problem statement, all variables and given/known data
Hello,
I've tried all I could and get very confused about how to approach the problem.
Here it comes : A disk of mass M and radius R can rotates around an axis passing by the point A without friction situated on its circumference. The disk is in a vertical plane. We let it move freely when point B is at the same high than point A and point B is opposed to point A. (See figure to get clarification).

a)What is the tangential acceleration of point B just after the disk has been released?
b)What is its velocity when it pass by its lower point?
c)What is the acceleration of point B in function of the disk's position.

2. The attempt at a solution
First I drew the applied forces, which I think are the normal and the weight. The weight is applied in the center of mass of the disk that is in its center while the normal force is applied at point A and is constantly changing so I don't even know its magnitude nor its direction.
I believe that from a formula, $$\vec v_{B}=\vec{\alpha}\wedge \vec{r_{B}}+\vec{a}_{CM}$$.
I consider the origin of the system as being the point A. I want to know $$\vec{a}_{CM}$$.
From Newton, $$\frac{d\vec{P}}{dt}=m\vec{a}_{CM}=\vec{F}_e=Mgj+\vec{N}\Leftrightarrow \vec{a}_{CM}=gj+\frac{\vec N}{M}$$.
Thanks to the right hand rule, I got that $$\vec{\alpha}=-\alpha k$$.
So $$\vec v_{B}=-\alpha k \wedge Ri+gj+\frac{\vec N }{M}$$. The only unknowns remaining are the normal force and the angular acceleration. I've no clue how to find them.
I also tried other things, like writing down $$\frac{d\vec L}{dt}$$ or even $$\vec L= \vec L_o + \vec L_s$$ where $$\vec L_o$$ is $$L$$ orbital and $$\vec L_s$$ is $$L$$ spin. I calculated $$L_s$$ to be worth $$\frac{3MR^2\vec{\omega}}{4}$$ and I believe that $$L_o=\vec R_{CM} \wedge mv_{CM}$$ and I have much more unknowns including $$mv_{CM}$$ which seems at least as hard as to find $$v_{B}$$...
I must be missing many things. Can you help me please?

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2. Nov 23, 2008

### Staff: Mentor

Keep it simple. Start by using Newton's 2nd law applied to rotation to find the angular acceleration:
$$\tau = I \alpha$$

3. Nov 23, 2008

### fluidistic

Thank you very much for your help Doc Al.
Your $$\tau=I\alpha$$ is my $$\frac{d\vec L}{dt}$$ and I sadly didn't realize it was worth $$I\alpha$$.
I wrote $$\frac{d\vec L}{dt}=\vec P \wedge \vec r + \vec N \wedge \vec r = -Mg\frac{R}{2}k$$.
By luck I calculated $$L_s$$. Derivating it I obtain $$I\alpha$$ : $$\frac{3MR^2\vec{\alpha}}{4}$$. (I didn't realize $$L_o$$ would be equal to 0. That's because the motion is a pure rotation so there is no translation I think).
So $$\vec \tau = -\frac{3MR^2 \alpha}{4}k$$.
Right now I don't see how I can calculate $$\alpha$$. A little help would be appreciated
I'll try to do the next exercise alone.

4. Nov 23, 2008

### Staff: Mentor

How did you calculate the rotational inertia?
Figure out the torque.

5. Nov 23, 2008

### fluidistic

To calculate $$I$$ I used Steiner's theorem.
If I'm not wrong the moment of inertia of a disk rotating about an axis passing by its center is $$\frac{MR^2}{2}$$. Using Steiner we have that $$I_{A}=I_G+Md^2$$.
I made an error, you're right : $$Md^2=4MR^2$$.
So $$I=\frac{9MR^2}{2}$$.

Ok I will. A bit later though...
Thank you.

6. Nov 23, 2008

### Staff: Mentor

This is still incorrect. (Where are you getting the 4?)

7. Nov 23, 2008

### fluidistic

Sorry my fault, I took 2R instead of R the distance between the CM and A. I got confused with point B... nevermind.
So $$I=\frac{3MR^2}{2}$$.

8. Nov 23, 2008

### Staff: Mentor

Good. Now find the torque.

9. Nov 23, 2008

### fluidistic

$$\vec \tau = \vec r \wedge \vec F=Ri \wedge -Mgj=-RMgk$$ which is worth $$\frac{3MR^2 \vec \alpha }{2} \Leftrightarrow \vec \alpha=-\frac{2}{3R}k$$.
Later I'll answer to question a).

10. Nov 23, 2008

### Staff: Mentor

Looks OK, but you dropped off a g.

11. Nov 23, 2008

### fluidistic

Oops. Ok now I get it right.

As $$\vec a_T= \vec \alpha r$$ I get that $$\vec a_T=\frac{4g}{3}j$$ when the disk has just been released. Is it right? (I guess no because I should get $$\vec a_T=-\frac{4g}{3}j$$...)

12. Nov 23, 2008

### Staff: Mentor

Good. Don't get hung up on the sign of the acceleration, that just depends on your convention. It's rotating clockwise, so the tangential acceleration is down and thus negative.

13. Nov 23, 2008

### fluidistic

Thanks for all your help Doc Al. I also think that in the last formula, it is $$\vec g$$ so it is -gj.
Sorry for all the time I took to solve the problem and all the errors. I might say it's in part due to the fact that temperature here is very hot today, more than 35°C and humidity is high so it's uncomfortable to think well.

Last edited: Nov 23, 2008
14. Nov 23, 2008

### fluidistic

For the b) I have a question.
My attempt would be to find out $$\omega (t)$$ which I think is simply $$\int \alpha dt$$.
So that I can use the relation $$\vec v_B=\vec r_B \wedge \vec \omega_B$$. But I doubt it will help since I don't know at what time it will reach its lower position.
Maybe considering the energy of the body? It is a pure rotation so $$E=\frac{I\omega ^2}{2}$$ but it makes no sense at all... $$E_{initial}=E_{final}$$ but $$\omega$$ changes and $$I$$ is a constant so it is not possible. I don't understand why I can't apply the law of conservation of energy. Can you (or someone else) help me?

15. Nov 23, 2008

### Staff: Mentor

That method is the hard way. Note that α is not a constant.
Mechanical energy is conserved. Why do you think you can't apply it?

16. Nov 23, 2008

### fluidistic

If it is conserved then $$E=\frac{I\omega ^2}{2}$$ is not true because as $$I$$ is a constant, it implies that $$\omega$$ is also a constant, but as you said $$\alpha$$ is not constant and thus $$\omega$$ also varies.
So as $$E$$ is a constant in this problem, it must equals something else than $$\frac{I\omega ^2}{2}$$ but as the motion is a pure rotation I must say I'm missing something.
Thanks once again. You've been very useful to me.

17. Nov 23, 2008

### Staff: Mentor

That's just the kinetic energy, not the total energy. It's the total mechanical energy that's conserved. (Don't forget gravity!)

18. Nov 23, 2008

### fluidistic

Ahaha, I missed the obvious.
One last question and I get the answer to part b)
From the conservation of the mechanical energy I got that $$\omega _f=\sqrt{\omega _i^2+ \frac{4MgR}{I}}$$. How do I get $$\omega _i$$?
Is it from $$\alpha _i$$ that I already got? If so, I'm lost.

19. Nov 23, 2008

### Staff: Mentor

Where did the 4 come from?
Assume the disk is released from rest.

20. Nov 23, 2008

### fluidistic

Oh... yes of course. Second time I make a similar error. The angular acceleration can be different from 0 rad/s^2 while the angular velocity is worth 0 rad/s.

The difference of height between point B final position versus point B initial position is a diameter that is twice the radius (2R). So I took the origin of the system as being point B final position.
Thus $$E_i=\frac{I\omega _i^2}{2}+Mg2R$$.
$$E_f=\frac{I\omega _f^2}{2}$$.
Hence $$\frac{I\omega _i^2}{2}+Mg2R=\frac{I\omega _f^2}{2} \Leftrightarrow I\omega _f=I \omega _i^2+4MgR$$. From it I reached what I wrote.
(I believe that point B reaches the bottom of the picture I uploaded, thus the difference of height of point B is 2R.)