# A rocking hemisphere

1. Feb 8, 2013

### Karol

1. The problem statement, all variables and given/known data

The hemisphere is of radius R.
Prove that the period equals that of a mathematical pendulum of length $4R/3$
I have a solution, but i don't understand.

2. Relevant equations

Center of gravity:
$$a=\frac{3}{8}R$$
Moment of inertia round the center of mass:
$$I_C=\frac{83}{320}mR^2$$
Kinetic energy consists of: T1, linear velocity of center of mass and energy of rotation round center of mass, T2:
$$T_1=\frac{1}{2}m \left(\dot{x}^2+\dot{z}^2\right)$$
$$T_2=\frac{1}{2}I_C \dot{\phi}^2$$
Potential energy: V=mgz

3. The attempt at a solution

There is no sliding, so, the x coordinate of the center of mass is:
$$x_C=\phi R$$
The connection between $\phi$ and x and z:
$$x=R \left(\phi -\frac{3}{8}\sin \phi \right)$$
$$z=R \left(1-\frac{3}{8}\cos\phi \right)$$
Small angles approximation:
$$x=R \frac{3}{8} \phi R$$
$$z=R \left( \frac{5}{8} + \frac{3}{16} \phi^2 \right)$$
Now i start not to understand.
It is written that:
$$V=\frac{1}{2} \frac{3}{8} mgR \phi^2$$
It has the units of energy, but which?
And:
$$T_1=\frac{1}{2} \frac{25}{64} mR^2 \dot{\phi}^2$$
If i compute the members of the kinetic energy, it is only the first member in it.
And the period is:
$$\omega^2=\frac{\frac{3}{8} mgR}{\left( \frac{25}{64} + \frac{83}{320} \right) mR^2}$$
The denominator is the moment of inertia round the contact point with the floor.
It resembles the equation for kinetic energy:
$$E=\frac{1}{2} I \omega^2$$
But in our solution ω isn't the angular velocity, but the period.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Feb 8, 2013

### TSny

The expression for $x$ is not correct. You have two factors of R (probably just a typo), but in addition it doesn't appear that you made the small angle approximation correctly for $x$. The expression for $z$ looks ok.
Remember that you can always add or drop a constant when defining the potential energy
Did you include the $\dot{z}$ contribution to $T_1$?

3. Feb 8, 2013

### Karol

What do you mean two factors? what is a typo? i am not an american.
There is a mistake in the header of x. the equation x=R$\phi$ is for the center of the hemisphere, maybe that was your intention.

4. Feb 8, 2013

### TSny

Sorry. "typo" is an abbreviation for "typographical error". You wrote $x = R\frac{3}{8} \phi R$. The left hand side of the equation has dimensions of length while the right hand side has dimensions of length squared. So, that can't be correct. Also, the 3/8 factor is incorrect. Can you show the details of how you made the small angle approximation for x?

5. Feb 8, 2013

### Karol

Right, a small mistake, the approximation is:
$$x=\frac{3}{8}R \phi$$

6. Feb 8, 2013

### TSny

I think it's possible to argue that the contribution of the $\dot{z}^2$ term to the kinetic energy is negligible for small oscillations.
That's not the period, it's the square of the angular frequency. But there is a simple relationship between $\omega$ and the period.
I don't understand these last remarks.

7. Feb 8, 2013

### TSny

No, the 3/8 factor is incorrect. Can you show how you got it?

8. Feb 8, 2013

### Karol

Right, the approximation should be:
$$x=\frac{5}{8}R\phi$$
My last remark says exactly what you say, that it's the square of the angular velocity, not the period.
What is the simple relation between the period and ω?
ω that we found is constant, how can we relate it to the period?

9. Feb 8, 2013

### TSny

In the first equation $\omega$ is the angular frequency of the simple harmonic motion. In the second equation for energy, $\omega$ is something different. It's the angular velocity $\dot{\phi}$.

10. Feb 8, 2013

### TSny

Your equation for $\omega^2$ gives the square of the angular frequency of the simple harmonic motion, it does not give the square of the angular velocity of the hemisphere.
Angular frequency $\omega$ equals $2\pi f$ where $f$ is the frequency. And $f = 1/T$ where $T$ is the period.

11. Feb 9, 2013

### Karol

I meant frequency when i wrote period, i don't know the terms.
So, in the equation:
$$\omega^2=\frac{\frac{3}{8} mgR}{\left( \frac{25}{64} + \frac{83}{320} \right) mR^2}$$
ω is the frequency?
I don't know this equation, what is it?
It looked like the equation for energy, but you say it isn't, so, what is the root, the general form for this equation?

12. Feb 9, 2013

### TSny

How did you arrive at this equation in the first place? Its meaning should be clear from its derivation.

13. Feb 9, 2013

### Karol

I didn't arrive to it, i received it as part of the solution.
So, is ω the frequency, in this equation? and you say it isn't:
$$E=\frac{1}{2} I \omega^2$$
So, what is it, then? it has exactly the same form

14. Feb 9, 2013

### TSny

For small oscillations of the hemisphere, the angle of displacement from equilibrium, $\phi$, will vary with time as $\phi(t) = \phi_{max}\; cos(\omega t)$ if you choose t = 0 at maximum displacement. Here $\phi_{max}$ is the amplitude of the oscillations and $\omega$ is a constant called the angular frequency. $\omega$ is related to the period as I gave before.

The energy of the rocking hemisphere is the sum of the kinetic and potential energies:

$E = \frac{1}{2} I \dot{\phi}^2 + \frac{1}{2}B\phi^2$

where $I$ is the moment of inertia about the equilibrium point of contact and $B$ is a constant that you can find in your original post.

Since E is a constant, you may evaluate it at the instant the hemisphere is passing through the equilibrium position $\phi = 0$. Then $\dot{\phi}$ will have its maximum value which you can show is $\dot{\phi}_{max} = \omega \phi_{max}$. Then you find

$E = \frac{1}{2} I \omega^2 \phi_{max}^2$

15. Feb 9, 2013

### Karol

The potential energy when $\phi=\phi_{max}$ is:
$$mgz_{max}=mgR\left(1-\frac{3}{8}\cos \phi_{max}\right)=mgR \left(\frac{5}{8}+\frac{\phi_{max}^2}{2}\right)$$
Equating the energies at this point and at the lowest point:
$$mgz_{max}=\frac{1}{2} I \omega^2 \phi_{max}^2$$
$$\Rightarrow mgR \left( \frac{5}{8} + \frac{\phi_{max}^2}{2} \right)=\frac{1}{2} \left( \frac{13}{20} mR^2 \cdot \omega^2 \cdot \phi_{max}^2 \right)$$
And the $\phi_{max}$ on the two sides don't reduce.

16. Feb 9, 2013

### TSny

Looks like you dropped the 3/8 factor in the term.

You should include the potential energy at the lowest point on the right side.

Last edited: Feb 9, 2013
17. Feb 9, 2013

### Karol

$$mgR \left(\frac{5}{8}+\frac{3\phi_{max}^2}{16}\right)=\frac{13}{40} mR^2 \cdot \omega^2 \cdot \phi_{max}^2 + mg\frac{5}{8}R$$
Still don't reduce

18. Feb 9, 2013

### TSny

Now you can solve for ω2. What do you get?

19. Feb 9, 2013

### Karol

We are getting closer.
You started from the assumption that for small oscillations of the hemisphere, the angle of displacement from equilibrium, $\phi$, will vary with time as $\phi(t) = \phi_{max}\; cos(\omega t)$
Why? i have to prove this is a harmonic motion.
I get:
$$\omega^2=\frac{15g}{26R}$$
I should get:
$$\omega^2=\frac{3g}{4R}$$
But i will check calculations again

20. Feb 9, 2013

### TSny

To prove it, you can derive the differential equation for $\phi(t)$ and show that the solution is simple harmonic motion. To get the differential equation you can write out the expression for the total energy E as a function of $\phi$ and $\dot{\phi}$ and then take the time derivative of the expression keeping in mind that E is a constant.
I think your answer is the correct one. I don't think the expression in the statement of the problem is correct. Hope I haven't overlooked anything, but I got what you got. I will go back over it.