# Energy of one cylinder rolling down another cylinder

• C. Darwin
In summary, the homework statement states that a cylinder with mass m and radius a rolls down a fixed cylinder with radius b. The normal component of the reaction force exerted on the rolling cylinder is found to be mgcos(\theta) - mr\dot{\theta}^2.
C. Darwin

## Homework Statement

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.

## 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: $$F_N = mg(7cos(\theta) - 4)/3$$

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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.

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

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.

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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.

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.

Darwin,

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

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

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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.

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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...

C. Darwin said:
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...
.

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

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

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?

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.

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

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.

## 1. What is the principle behind the energy of one cylinder rolling down another cylinder?

The energy of one cylinder rolling down another cylinder is based on the principle of conservation of energy. This means that the total amount of energy in a closed system remains constant, and energy can be transferred from one form to another but cannot be created or destroyed.

## 2. How does the shape and size of the cylinders affect the energy of one rolling down the other?

The shape and size of the cylinders can affect the energy of one rolling down the other in several ways. For instance, a larger and heavier cylinder will have a greater potential energy at the top of the ramp, and a smaller and lighter cylinder will have a lower potential energy. The shape of the cylinders can also affect the amount of friction and resistance during the rolling motion, which can impact the energy transfer.

## 3. How does the surface of the cylinders impact the energy of one rolling down the other?

The surface of the cylinders can play a significant role in the energy of one rolling down the other. A rough surface will create more friction and resistance, resulting in a decrease in the overall energy transfer. On the other hand, a smooth surface will have less friction and allow for a more efficient transfer of energy between the two cylinders.

## 4. Can the energy of one cylinder rolling down another cylinder be converted into other forms of energy?

Yes, the energy of one cylinder rolling down another can be converted into other forms of energy. For example, as the cylinder rolls down the ramp, its potential energy is converted into kinetic energy. This kinetic energy can then be transferred to other objects, such as a ball at the bottom of the ramp, through collisions or other interactions.

## 5. Are there any real-life applications of the energy of one cylinder rolling down another cylinder?

Yes, there are several real-life applications of the energy of one cylinder rolling down another cylinder. One example is the use of water mills, where the energy of water flowing down a ramp is used to turn a wheel and grind grains. Roller coasters also utilize the principle of energy transfer between cylinders to create thrilling rides. Additionally, the concept is used in various industrial processes, such as conveyor belts and assembly lines.

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