Energy of one cylinder rolling down another cylinder

1. Jan 26, 2009

C. Darwin

1. The problem statement, all variables and given/known data
A cylinder with mass $m$ and radius $a$ rolls down a fixed cylinder with radius $b$. Let $\theta$ be the angle between the plane containing the axis of each cylinder and the vertical. Find the normal component of the reaction force exerted on the rolling cylinder.

2. Relevant equations

3. The attempt at a solution

$$T = \frac{1}{2} m v^2 + \frac{1}{2}I\omega^2 = \frac{1}{2} m (R\dot{\theta})^2 + \frac{1}{2}(\frac{1}{2}ma^2)(\frac{b}{a}\dot{\theta})^2 = m[\frac{1}{2}(a+b)^2 + \frac{1}{4}b^2]\dot{\theta}^2$$

$$V = mgh = mgRcos(\theta) = mg(a+b)cos(\theta)$$

$$E = T + V \Rightarrow mg(a+b) = m[\frac{1}{2}(a+b)^2+\frac{1}{4}b^2}]\dot{\theta}^2 + mg(a+b)cos(\theta)$$

$$\Rightarrow \dot{\theta}^2 = \frac{g(a+b)}{\frac{1}{2}(a+b)^2 + \frac{1}{4}b^2}(1-cos(\theta))$$

Now here's where I'm not so sure (although my statements of energy may be off):

$$F_N = mgcos(\theta) - m\frac{v^2}{r} = mgcos(\theta) - mr\dot{\theta}^2 =mgcos(\theta) - m(a+b)\dot{\theta}^2$$

$$F_N = mg[cos(\theta) - \frac{(a+b)^2}{\frac{1}{2}(a+b)^2 + \frac{1}{4}b^2}(1-cos(\theta))]$$

I'm lost at this point, I think my statements for T and V are correct because I arrived at the same $$\ddot{\theta}$$ from both the energy analysis and the Langrangian.

According to the book, the answer is: $$F_N = mg(7cos(\theta) - 4)/3$$

Last edited: Jan 27, 2009
2. Jan 26, 2009

chrisk

Your energy equation is not quite correct. You have to account for two rotational energies:

The rotational energy of the center of mass of the rolling cylinder (it's traveling in a circle of radius a+b = r) and the rotational energy of the rolling cylinder. Let the angle phi be the angular displacement of the rolling cylinder and theta the angular displacement of the center of mass of the rolling cylinder.

$$T=\frac{1}{2}\mbox{m}\dot{r}^2\mbox{ + }\frac{1}{2}\mbox{m}r^2\dot{\theta}^2\mbox{ + }\frac{1}{2}\mbox{I}\dot{\phi}^2$$

3. Jan 26, 2009

C. Darwin

chrisk,

I'm not sure I follow you. The large cylinder is fixed, only the one with radius a can move. The first term in the kinetic energy is for the translation motion of the moving cylinder (constrained to move along the fixed cylinder). The second term in the kinetic energy is for the rotational motion of the moving cylinder.

As far as I can see, $$\dot{r}^2$$ and $$r^2\dot{\theta}^2$$ are the same thing. (I tried to add the proper boldness for vectors vs scalars but it didn't seem to render properly... I think the context is clear.

Last edited: Jan 26, 2009
4. Jan 27, 2009

xboy

Darwin, as the cylinder is rotating on a curved surface (another cylinder) the normal to the point of contact is not always in the same direction. So when you consider how much the cylinder has rotated, you can't just take he angle it makes with the normal. You have to take into account the rotation of the normal also. That will enter in the second term in the energy equation.

5. Jan 27, 2009

C. Darwin

The second term of the kinetic energy equation is $$\frac{1}{2}I\omega^2 = \frac{1}{4}b^2\dot{\theta}^2$$. The takes into account the change in the point of contact because $$a\theta = b\phi \Rightarrow a\dot{\theta} = b\dot{\phi} \Rightarrow \dot{\theta}^2 = (\frac{b}{a}\dot{\phi})^2$$ where $$\phi$$ is the angle of revolution of the moving cylinder.

6. Jan 27, 2009

chrisk

Darwin,

View the second problem of this site. It may help.

http://electron6.phys.utk.edu/phys594/archives/mechanics/Lagrangian/lagrangian3.htm [Broken]

Last edited by a moderator: May 3, 2017
7. Jan 27, 2009

C. Darwin

Ok, I solved my problem. I had previously thought that $$\dot{\theta}=\frac{b}{a}\omega$$ but actually it should have been $$\dot{\theta}=\frac{a+b}{a}\omega$$. This lead to a lot of simplification...

PS - chrisk
Thanks for the link, but I think that analysis is way overly complicated. The system is presented as having three degrees of freedom when there is actually only one. Also, he carries out his work with $$\dot{\rho}$$ throughout, but in actuality, $$\dot{\rho}=0$$ because the two cylinders stay touching.

Last edited: Jan 27, 2009
8. Jan 27, 2009

C. Darwin

To the PF community:

Is it customary for me to work out the correct solution to the problem and/or edit the information that was incorrect in my OP? Also, being that this was from a specific problem in a specific textbook, is there a proper way for me to indicate that? I think I've seen forums that index hw-type questions by book...

9. Jan 28, 2009

xboy

Now I have confused myself. Can you please explain to me why it should be so?

10. Jan 30, 2009

C. Darwin

Eh, actually I had a little typo last time, what I should have said was:

I had previously thought that $$b\dot{\theta} = a \omega$$ but it should actually be $$(a+b)\dot{\theta} = a \omega$$. So $$\omega^2 = (\frac{a+b}{a}\dot{\theta})^2$$

That being said, I still don't quite understand it. If you think about the problem as two gears. Shouldn't the radius times the degrees one turns = the radius times the degrees the other turns?

11. Jan 30, 2009

xboy

I think the angle here is being measured with respect to an axis fixed on the body. If that is so, this axis (which may be the normal to the point of contact) is also rotating. So you add the two angles. This explanation seems to work, but I am not sure at all if it is the correct explanation.
I could do with some help here.

12. Jan 31, 2009

C. Darwin

And since the problem spoke nothing of it, I'm certain this is an inertial reference frame.

13. Feb 1, 2009

xboy

In Spiegel's Theoretical Mechanics, this sort of problem is worked out by adding the angles. There$$b\dot{\theta} = a \omega$$

and the total angle of revolution of the smaller sphere is :$$\theta + \omega$$

And the result comes out the same.