# Ball rolling down slope (with wedge-shaped groove)

1. Mar 23, 2014

### BenjyPhysics

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

The question is B7 here: http://www-teach.phy.cam.ac.uk/dms/dms_getFile.php?node=7787

I managed to derive the acceleration required in the first part, but the second part is giving me trouble.

2. Relevant equations

3. The attempt at a solution

I have calculated the moment of inertia of the ball about the axis it is instantaneously rotating around (I don't quite know what is meant by an 'instantaneous axis' - but anyway), I arrived at...

$${I_{AB}} = \frac{2}{5}m{a^2} + m{a^2}{\sin ^2}(\frac{1}{2}\phi ) = m{a^2}(\frac{2}{5} + {\sin ^2}(\frac{1}{2}\phi ))$$
1. The problem statement, all variables and given/known data

Now, I have an equation for the linear motion of the centre of mass of the ball, and it is... $$mg\sin \theta - 2F = m\frac{{dv}}{{dt}}$$ where F is the frictional force acting in the direction of the slope, providing the moment (two points of contact so I wrote it as 2F)

The rotational motion is confusing me, I tried to write...

$$G = I\frac{{d\omega }}{{dt}} = \frac{I}{a}\frac{{dv}}{{dt}} = ma(\frac{2}{5} + {\sin ^2}(\frac{1}{2}\phi ))\frac{{dv}}{{dt}}$$

but how do I find G, the total external moment, and which axis do I take the moment about - (because if I were to take the moment about AB I'd have cos terms in my result)? Which I am guessing is supplied by the two frictional forces? Am I thinking about this all wrong?

2. Mar 23, 2014

### BvU

Hello Benjy and welcome to PF.
Nice word, invigilator (my first language isn't english).
PF guidelines don't like it if you skip over 1 by posting a big fat pdf and also don't like skipping the relevant equations, but you can make up for that in your next post(s)...

Compliments for the $\TeX$ use. Looks neat!.

Instantaneous axis is the line between stationary points. Simple, isn't it?
Stationary points are useful because they help you connect linear motion to rotation.

Your $\omega$ doesn't look quite right to me. Can you explain how it comes about ?

I also don't understand why G should have a cos term if your I doesn't have that either...

3. Mar 23, 2014

### BenjyPhysics

I found omega like so $$v = a\omega \Rightarrow \omega = \frac{v}{a} \Rightarrow \frac{{d\omega }}{{dt}} = \frac{1}{a}\frac{{dv}}{{dt}}$$ That^ is actually the result I used in the first derivation - but I'm not sure if it applies in the second scenario because the ball is rotating about a different axis :/ Yet if this isn't correct, how do I find a suitable expression for omega?

Anyway, I was thinking of calculating the moment (about the instantaneous axis AB) provided by each friction force (acting on the two stationary points) as $$Fa\cos (\frac{\phi }{2})$$ using $$a\cos (\frac{\phi }{2})$$ as the perpendicular distance from each frictional force to the instantaneous rotation axis.

So I end up with... $$2Fa\cos (\frac{\phi }{2}) = ma(\frac{2}{5} + {\sin ^2}(\frac{1}{2}\phi ))\frac{{dv}}{{dt}}$$ which I would like to plug into my first equation (describing the linear motion) but the pesky cosine term is there!

(Oh and yes apologies about the pdf and the lack of relevant equations - I shall indeed try to redeem myself in my next post :D )

Last edited: Mar 23, 2014
4. Mar 23, 2014

### BenjyPhysics

Oops I forgot to quote you

5. Mar 23, 2014

### BvU

In the second scenario it will have to rotate faster to get the same speed. Check how far it proceeds in one revolution. The factor is the same for distance, speed and acceleration.

6. Mar 23, 2014

### BenjyPhysics

I'm not sure how to 'check how far it proceeds in one revolution' Could you please tell me what the angular velocity in the second case should be - so that I can understand exactly how you worked it out?

7. Mar 23, 2014

### BvU

In the second diagram, you have correctly derived that the distance of AB to the center of the ball is $r\sin\phi/2$. Can you imagine the trajectory of the contact point on the ball when it makes one revolution? The righthand picture is an "end view"; consider the "side view". Given that there is no slip at the contact point, the length of the trajectory on the ball must be the same as the distance travelled in the groove...

8. Mar 24, 2014

### BenjyPhysics

I'm not even sure I can imagine the ball rotating about the instantaneous axis AB, and I have no idea how to visualize the trajectory of either of the contact points of the ball :( Obviously it will be a circle but I'm not sure how to figure out the radius of said circle

9. Mar 24, 2014

### BvU

Well, the separation between the contact points is constant, right? If you cut the ball vertically through the contact points (lines CA and CB), you slice off two domes with the same radius (you have it already), and you are left with two flat circular areas on the ball, whose circumference is the distance that the ball travels during one revolution !

Extreme case: if A and B are diametrically opposite on the ball, no progress at all for one revolution!

Still not clear? find a ball and two wooden beams ! (the V shape isn't necessary, only the distance between the beams (hence the contact points))

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