A spinning ball hits a wall and bounces back question

[greytomato]

Hi, my question is this:

A solid ball with mass M, radius R, speed U0, angular speed W0 hits a wall at the angle theta0. after the hit, the ball goes away with an angle theta1, angular speed W1 and speed U1.

i'm asked to find the angular momentum for the point of hit, before and after the hit.
i'm also asked to find the neccesary conditions of U0, W0, and theta0 so that the path of the ball after the hit would be 90 degrees from the wall.

I've been trying for some time now to crack this question but no luck...
any tips, ideas would be extremely helpful!

 

[StatusX]

You need more information. Is the collision elastic, does the ball slip? You can't use conservation laws when you have an immovable wall, because the wall can absorb momentum, angular momentum, and energy at will.

 

[greytomato]

The collision is elastic, the wall cannot move, the ball doesn't slip.

 

[BerryBoy]

I think that it is meant the collision is elastic with the words hard and immovable.

Right, I think that the horizontal velocity of the ball will remain unchanged (because it is an elastic collision) but the energy lost in the rotational energy will be transferred to the kinetic energy of the ball (through means of the friction on the wall). This change in energy can be used to calculate the new vertical component of velocity.

As for the 90 degree change, I see a dot product of U0 & U1 needs to be used.

Can you go from here?
I don't know what you've done so far.

Sam

 

[greytomato]

i too found that the component of speed (horizontal) which is perpendicular to the wall is the same (only the sign changes).
so this leaves only the parallel speed (vertical) and the spin speed.
i'm pretty sure i need to use angular momentum only I'm not sure how to do this...

 

[BerryBoy]

I show me what you've tried, and I can help you.

Hint:
Moment inertia of a sphere -
$$I = \frac{2}{5} mr^2$$

Regards,
Sam

 

[greytomato]

the angular momentum should remain the same after hitting the wall?

i have 2 equations i can use for that:
i can calculate the angular momentum for a constant point, so i choose point P as the hitting point
1) H= R *cross* MV
2) H = IW + R *cross* MV

so I'm guessing you mean i should use equation #2?

 

[BerryBoy]

OK, the angular momentum will change because it will be retarded by friction on impact with the wall.

Consider this equation:

$$\frac{1}{2} m \underline{u_0}^2 + \frac{1}{2}I {\omega_0}^2 = \frac{1}{2} m \underline{u_1}^2 + \frac{1}{2}I {\omega_1}^2$$

With this you can split up u into:

$$\underline{u}^2 = {u_x}^2 + {u_y}^2$$

Then you can substitue the value for moment of inertia.

Can you go from here?

Sam

 

[BerryBoy]

the angular momentum should remain the same after hitting the wall?

i have 2 equations i can use for that:
i can calculate the angular momentum for a constant point, so i choose point P as the hitting point
1) H= R *cross* MV
2) H = IW + R *cross* MV

so I'm guessing you mean i should use equation #2?

Remember that in, $\underline{H} = \underline{r} \times m \underline{v}$ (where H is angular momentum, I usually use L). v is the surface speed of the object, not the object's linear speed.

Regards,
Sam

 

[greytomato]

by surface speed you mean v = wr?
the vercor r you wrote, is that the vector from the point of center mass of the ball to the point of hit?

 

[BerryBoy]

Exactly, but we do not want to know this speed. You want to know the object's linear speed.

Consider the equation I wrote above. Then rewrite it with $\underline{v}$ in terms of its x and y components. With this you can write $\underline{v}_1$ in terms of $\underline{v}_0$.

If I am confusing you, please tell me and I'll try and get a diagram up.

Regards,
Sam

 

[greytomato]

i think it's all becoming a salad for me, a diagram would be very helpful...

 

[BerryBoy]

OK, here's the diagram I promised:

http://www.berrys.plus.com/wall.gif

I have used $u$ as the initial velocity and $v$ as the final velocity. The most important thing to notice here is that, due to the fact the collision is elastic and there is no horizontal friction, the horizontal velocity remains unchanged.

Hope this helps.

Sam

 

[greytomato]

this is perfectly clear! thanks very much indeed :)
so this means that the system's energy is conserved although the collision occurs because it's an elastic collision? hope i got this right.

 

[BerryBoy]

Sounds good, glad to help.

Regards,
Sam