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

[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.

 

[TSny]

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 $T_1$?

 

[Karol]

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.

 

[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?

 

[Karol]

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

 

[TSny]

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.

 

[TSny]

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?

 

[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?

 

[TSny]

\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 $\dot{\phi}$.

 

[TSny]

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.

 

[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?

 

[TSny]

\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.

 

[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

 

[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

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

 

[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.

 

[TSny]

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.

 

[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

 

[TSny]

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

 

[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

 

[TSny]

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.

 

[TSny]

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.

 

[Karol]

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

 

[TSny]

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.

 

[Karol]

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

 

[TSny]

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

Each \; just adds a little space

 

[Karol]

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}

 

[TSny]

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, $\omega \: T = 2\pi$.