# Prove Period of a Rocking Hemisphere of Radius R is Equal to 4R/3

• Karol
In summary, the problem is to prove that the period of a hemisphere with radius R is equal to that of a mathematical pendulum with length 4R/3. The solution involves using the center of gravity and moment of inertia around the center of mass to calculate the kinetic energy and potential energy. The small angles approximation is used to find the coordinates of the center of mass and the expression for the potential energy is adjusted by a constant. The final equation for the period, which is the square of the angular frequency, resembles the equation for kinetic energy but has a different meaning. The simple relation between angular frequency and period is also mentioned.
Karol

## Homework Statement

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.

## Homework 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

## 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.

#### Attachments

• Hemisphere.jpg
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Karol said:
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:
$$x=R \frac{3}{8} \phi R$$
$$z=R \left( \frac{5}{8} + \frac{3}{16} \phi^2 \right)$$
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.
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?
Remember that you can always add or drop a constant when defining the potential energy
$$T_1=\frac{1}{2} \frac{25}{64} mR^2 \dot{\phi}^2$$
Did you include the ##\dot{z}## contribution to ##T_1##?

TSny said:
The expression for ##x## is not correct. You have two factors of R (probably just a typo)
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.

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?

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

Karol said:
$$T_1=\frac{1}{2} \frac{25}{64} mR^2 \dot{\phi}^2$$
If i compute the members of the kinetic energy, it is only the first member in it.
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.
And the period is:
$$\omega^2=\frac{\frac{3}{8} mgR}{\left( \frac{25}{64} + \frac{83}{320} \right) mR^2}$$
That's not the period, it's the square of the angular frequency. But there is a simple relationship between ##\omega## and the period.
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.
I don't understand these last remarks.

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

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

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?

Karol said:
$$\omega^2=\frac{\frac{3}{8} mgR}{\left( \frac{25}{64} + \frac{83}{320} \right) mR^2}$$
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.

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}##.

Karol said:
My last remark says exactly what you say, that it's the square of the angular velocity, not the period.
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.
What is the simple relation between the period and ω?
ω that we found is constant, how can we relate it to the period?

Angular frequency ##\omega## equals ##2\pi f## where ##f## is the frequency. And ##f = 1/T## where ##T## is the period.

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?

Karol said:
$$\omega^2=\frac{\frac{3}{8} mgR}{\left( \frac{25}{64} + \frac{83}{320} \right) mR^2}$$
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?

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

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

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##

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.

Karol said:
The potential energy when $\phi=\phi_{max}$ is:
$$mgz_{max}=mgR\left(1-\frac{3}{8}\cos \phi_{max}\right)=mgR \left(\frac{5}{8}+\frac{\phi_{max}^2}{2}\right)$$
Looks like you dropped the 3/8 factor in the term.

Equating the energies at this point and at the lowest point:
$$mgz_{max}=\frac{1}{2} I \omega^2 \phi_{max}^2$$
You should include the potential energy at the lowest point on the right side.

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

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

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

Karol said:
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.
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 get:
$$\omega^2=\frac{15g}{26R}$$
I should get:
$$\omega^2=\frac{3g}{4R}$$
But i will check calculations again

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.

Karol said:
i have to prove this is a harmonic motion.

Another approach is to start with the equation E = KE + PE written in terms of ##\phi## and ##\dot{\phi}## and separate variables so that ##\phi## and ##d\phi## occur on one side of the equation and ##dt## on the other. Then integrate both sides to determine ##\phi## as a function of time.

Last edited:
TSny said:
Another approach is to start with the equation E = KE + PE written in terms of ##\phi## and ##\dot{\phi}## and separate variables so that ##\phi## and ##d\phi## occur on one side of the equation and ##dt## on the other. Then integrate both sides to determine ##\phi## as a function of time.

$$E_{tot}=mgR \left( \frac{5}{8}+\frac{3\phi^2}{16}\right)+\frac{13}{40}mR^2\left(\frac{d\phi}{dt}\right)^2$$

And i can't seem to separate the dt to the left and ##\phi## and ##d\phi## to the right.
Maybe, in this case, this method doesn't work

Karol said:
$$E_{tot}=mgR \left( \frac{5}{8}+\frac{3\phi^2}{16}\right)+\frac{13}{40}mR^2\left(\frac{d\phi}{dt}\right)^2$$

And i can't seem to separate the dt to the left and ##\phi## and ##d\phi## to the right.
Maybe, in this case, this method doesn't work

You an write this as ##A = B\phi^2+C\dot{\phi}^2## where you can figure out expressions for A, B, and C.

Solve for ##\dot{\phi} = f(\phi)## where you'll see what you get for the function ##f(\phi)##

Separate variables ##d\phi/f(\phi) = dt##

Integrate both sides and consult an integral table if you need to for the left hand side.

$$A = B\phi^2+C\dot{\phi}^2 \Rightarrow \sqrt{\frac{A-B\phi^2}{C}}=\dot{\phi}$$
$$\int dt=\int \frac{d\phi}{\sqrt{\frac{A-B\phi^2}{C}}}$$
$$\Rightarrow \sqrt{\frac{C}{B}} \arcsin \left( \phi \sqrt{\frac{B}{A}}\right)=t$$
$$\Rightarrow \phi=\sqrt{\frac{A}{B}}\sin\left(\sqrt{\frac{B}{C}t}\right)$$

And it has the form of harmonic motion.

What does this sequence: \; do in Latex? you used it, but i don't see any effect

Good. So, what do you get for ##\omega## in terms of the original parameters of g and R?

Each \; just adds a little space

TSny said:
Then ##\dot{\phi}## will have its maximum value which you can show is ##\dot{\phi}_{max} = \omega \phi_{max}##.
$$\phi(t) = \phi_{max}\; cos(\omega t) \Rightarrow \dot{\phi}=-\omega \phi_{max} \; \sin(\omega t)$$
The lowest point is at a quarter of the period, at:
$$t=\frac{T}{4}=\frac{\pi}{2\omega} \Rightarrow \dot{\phi}=-\omega \phi_{max}$$

Yes, that will give ##\dot{\phi}## when the system first passes through the equilibrium position.

If you just want the period, then use the fact that when ##t = T## the cosine function must have completed one cycle. So, the argument of the cosine function will be ##2\pi##. Hence, ##\omega \: T = 2\pi##.

Last edited:

## 1. What is the period of a rocking hemisphere of radius R?

The period of a rocking hemisphere of radius R is equal to 4R/3. This means that it takes 4R/3 units of time for the hemisphere to complete one full back-and-forth rocking motion.

## 2. How is the period of a rocking hemisphere calculated?

The period of a rocking hemisphere is calculated using the formula T=2π√(R/g), where T is the period, R is the radius, and g is the acceleration due to gravity. Plugging in the value of R, the formula becomes T=2π√(4R/3g). Simplifying further, we get T=4R/3√(g), which is equal to 4R/3 since the square root of g is a constant value.

## 3. What is the significance of the period of a rocking hemisphere?

The period of a rocking hemisphere is significant because it represents the time it takes for the hemisphere to complete one full back-and-forth rocking motion. It is an important factor in understanding the dynamics of a rocking hemisphere and can be used to calculate other properties such as the frequency and amplitude of the rocking motion.

## 4. How does the radius of the hemisphere affect its period?

The radius of the hemisphere directly affects its period, as seen in the formula T=2π√(R/g). A larger radius will result in a longer period, while a smaller radius will result in a shorter period. This is because a larger radius means a larger distance for the hemisphere to travel in one back-and-forth rocking motion, thus taking more time to complete.

## 5. Can the period of a rocking hemisphere be changed?

Yes, the period of a rocking hemisphere can be changed by altering the radius or the acceleration due to gravity. Changing the radius will directly impact the period, while changing the acceleration due to gravity will indirectly impact the period through the formula T=2π√(R/g). Other factors such as the mass and shape of the hemisphere may also affect the period.

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