# Oscillation of a solid hemisphere

1. Jan 28, 2014

### Saitama

1. The problem statement, all variables and given/known data
I don't have the exact wordings of the problem statement. I hope the following is enough to understand the problem.

A solid hemisphere is kept on a plane horizontal frictionless surface. The hemisphere is made to tumble (or toss, I am not sure about the correct word). Find the time period of oscillation.

The mass of hemisphere is M and radius is R.

2. Relevant equations

3. The attempt at a solution
I have described the situation in the attachment.

The CM of hemisphere is at a distance 3R/8 from the centre of base of hemisphere. To find the time period, I am thinking of writing down the expression for energy at any instant and then set its time derivative to zero.

The orange coloured sketch is the new position of hemisphere. For the energy equation, I need the angular velocity, velocity of CM and height of CM from the ground.

For angular velocity, I write $d\theta/dt$. I can also find the height of CM but I have trouble finding the velocity of CM or specifically, the horizontal component of velocity. How do I find the horizontal displacement ($x$) of CM? I have found out rest of the variables in terms of $\theta$ and I am stuck on finding the horizontal displacement from hours. (I hope the sketch is easily decipherable).

Any help is appreciated. Thanks!

#### Attached Files:

• ###### hemisphere tossing.png
File size:
6.3 KB
Views:
414
2. Jan 28, 2014

### haruspex

"base" might be a bit ambiguous here. I guess you mean 3R/8 from centre of whole sphere.
Can you see straightaway what the displacement of the point of contact is as a function of theta? That gives you the displacement of the centre of the sphere. Then you just need a small correction to come back to x.

3. Jan 28, 2014

### consciousness

See the figure to get x as function of theta.
Since the hemisphere doesn't lose contact with the ground, the distance between the "vertex" of the hemisphere and the point of contact is x i.e. distance traveled.

Edit: Just noticed that your x is distance traveled by CM. To correct for this add the small distance between CM and vertical axis passing through point of contact with ground. My figure is wrong.

Edit 2: Figure now corrected

#### Attached Files:

• ###### hemisphere tossing.png
File size:
6.8 KB
Views:
349
Last edited: Jan 28, 2014
4. Jan 28, 2014

### ehild

Is the CM displaced at all if the surface is frictionless?

ehild

5. Jan 28, 2014

### haruspex

Well spotted.

6. Jan 28, 2014

### Saitama

Hi ehild, haruspex and consciousness! :)

Yes, you are right, I am sorry about that.

If I modify the question this way, let the surface provides enough friction to allow pure rolling, would it be correct?

Ah yes, thanks consciousness!

So we have $x+y'=R\theta$ (I will use $y$ to denote the vertical displacement of CM) where $y'=(3R/8)\sin\theta$, hence $x=R\theta-(3R/8)\sin\theta$.

The vertical displacement of CM is given by $y=(3R/8)(1-\cos\theta)$. Hence

$$\frac{dx}{dt}=\left(R-\frac{3R}{8}\cos\theta\right)\left(\frac{d\theta}{dt}\right)$$
$$\frac{dy}{dt}=\frac{3R}{8}\sin\theta \left(\frac{d\theta}{dt}\right)$$

The energy at any instant of time,
$$E=\frac{1}{2}M\left(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt} \right)^2 \right)+\frac{1}{2}I\left(\frac{d\theta}{dt}\right)^2+mgh(\theta)$$

where $I$ is the moment of inertia of hemisphere about CM, I found it to be:
$$I=\frac{83}{320}MR^2$$
and
$$h(\theta)=R-\frac{3R}{8}\cos\theta$$
I plugged in the expressions for dx/dt and dy/dt and got:
$$E=\frac{1}{2}M\left(\frac{73}{64}R^2-\frac{3R^2}{4}\cos\theta\right)\left(\frac{d\theta}{dt}\right)^2+\frac{1}{2}\frac{83MR^2}{320}\left(\frac{d\theta}{dt}\right)^2+Mg\left(R-\frac{3R}{8}\cos\theta\right)$$

This is getting a bit dirty so I would like to know if I am correct so far. :)

7. Jan 28, 2014

### consciousness

I get a different value for moment of inertia.

Moment of inertia of a solid hemisphere about the axis passing through the center of its base lets say I.
$$I=\frac{1}{5}MR^2$$
We want to find ICOM moment of inertia about axis passing the COM of the body parallel to the plane of the base lets say x. By parallel axis theorem,
$$I=x+Mr^2$$
$$x=\frac{1}{5}MR^2-\frac{9}{64}MR^2$$
$$x=\frac{19}{320}MR^2$$
Don't see any error apart from this

Last edited: Jan 28, 2014
8. Jan 28, 2014

### haruspex

2/5. Pranav-Arora's 83/320 is correct.

9. Jan 28, 2014

### haruspex

Well, which is the actual question? Both make sense.

10. Jan 28, 2014

### haruspex

Looks right to me. You can simplify a little by taking the zero potential at height 5R/8 instead of ground. Collecting up the $MR^2\frac{d\theta}{dt}^2$ terms yields a much simpler coefficient.

11. Jan 28, 2014

### consciousness

2/5 is for a solid sphere. For a hemisphere divide it by two.

12. Jan 28, 2014

### haruspex

It's half the mass. Formula is still 2Mr2/5.

13. Jan 29, 2014

### consciousness

You are right I messed up

14. Jan 29, 2014

### Saitama

Yes, you are right, it gets really simple with that. I got:
$$E=\frac{1}{2}MR^2\left(\frac{7}{5}-\frac{3}{4}\cos\theta\right)\left(\frac{d\theta}{dt}\right)^2+mgR\left(1-\frac{3}{8}\cos\theta\right)$$
Differentiating wrt time,
$$\frac{dE}{dt}=\frac{1}{2}MR^2\left(\frac{3}{4}\sin\theta\dot{\theta}^3+\left(\frac{7}{5}-\frac{3}{4}\cos\theta\right)2\dot{\theta}\ddot{\theta}\right)+\frac{3}{8}MgR\sin\theta \dot{\theta}=0$$
I used the approximation that $\sin\theta \approx \theta$ and $\cos\theta \approx 1$ and got:
$$\frac{3}{4}\theta\dot{\theta}^2+\frac{13}{10}\ddot{\theta}+\frac{3g}{8R}\theta=0$$
But it doesn't seem that it would be an oscillatory motion. I fed the above equation to wolfram alpha but got nothing. :(

I don't see how the frictionless surface makes sense. If the CM doesn't displace from its position, how is it supposed to oscillate?

15. Jan 29, 2014

### consciousness

This proves that the motion isn't simple harmonic (even for small displacements). To get a solution we need to solve that DE. If its any consolation, you have solved the physics part. Only the mathematics is left :tongue:

For a frictionless surface, consider what happens when you displace the hemisphere angularly and allow it to oscillate. The gravitational force will act downwards on the COM and the normal reaction will act upwards (below the center of the base). I think it will be an SHM (easier than this question)

16. Jan 29, 2014

### Saitama

Isn't this the same case we are discussing currently in this thread?

17. Jan 29, 2014

### ehild

Keep the terms linear in theta. (Ignore the first term).

ehild

18. Jan 29, 2014

### Saitama

But that's $\dot{\theta}^2$, not $\theta^2$.

19. Jan 29, 2014

### haruspex

Next I would approximate the cos(θ) terms as 1-θ2/2. The first one you can just approximate as 1 since it has a factor $\dot \theta^2$, which will be small. The 'base' energy is 5mgR/8, which gets rid of the the 1 in the approximation for the second cos:
$E=\frac{1}{2}MR^2\left(\frac{7}{5}-\frac{3}{4}\right)\left(\frac{d\theta}{dt}\right)^2+mgR\left(\frac{3}{16}\theta^2\right)$
But you're right, it's not SHM. Solution would involve inverse trig functions. So.. onto the frictionless version?

With no friction, the CoM will just bob up and down instead of (mostly) rocking from side to side. The point of contact will still move from side to side, but not as far.

20. Jan 29, 2014

### voko

$\dot x^2 + c^2 x^2 = c^2 A^2 \Rightarrow x = A \sin \left( ct + k \right)$. Not SHM?