Calculating Angular Velocity of a Disk after Inelastic Collision

[fatcat39]

Homework Statement

A ball with mass m and velocity v goes on a path tangent to a disk with radius R and a mass of M. There is a completely inelastic collision at the outer rim of the disk, and the entire system begins to spin around the axis. What was the final angular speed of the system?

Homework Equations

Conservation of Angular Momentum: I w = I$$_{F}$$ w$$_{F}$$
Conservation of Momentum: m$$_{a}$$v$$_{a}$$ + m$$_{b}$$v$$_{b}$$ = m'$$_{a}$$v'$$_{a}$$ + m'$$_{b}$$v'$$_{b}$$

Speed: w^2 = w + 2a$$\vartheta$$

The Attempt at a Solution

Since it's inelastic, I'm not really sure if i should be using angular or regular momentum. If someone could just point me in the right direction, I'm sure i coul figure out the rest of it.

 

[Astronuc]

Is the disk initially at rest.

Initially the ball has linear momentum, but then becomes attached (fixed) to the disk, which is the point of the inelastic collision (as opposed to elastic in which the ball would simply collide and scatter off the disk).

The ball and disk each contribute the total moment of inertia.

 

[fatcat39]

Yes, the disk is initially at rest.

So can I relate linear and rotational momentum to each other?

 

[fatcat39]

but it has to be momentum right? not kinetic energy, because of the totally inelastic part.

 

[Astronuc]

Yes, the linear momentum is transferred into the angular momentum of the combined disk and mass.

 

[fatcat39]

L1 = L2
angular = linear
Iw = rmv(perpendicular)
MR^2w = rmv
(M+m)R^2w = rmv
So: W = (rmv)/(MR^2+mR^2)Is this correct?

 

[Astronuc]

MR^2w = rmv

Almost correct.

Linear momentum (mv) has units of mass*velocity (in SI - kg-m/s). Angular momentum, I*$\omega$, has units of moment of inertia (~ MR²) and angular velocity (s^-1) or in SI kg-m²/s.

When the small mass m traveling with velocity v strikes the disc at R, it contributed to the angular momentum of the system. That angular momentum is mvR.

Now the small mass becomes embedded in the larger mass, but m is not distributed in M, but limited to a point at R from the center.

Let $\omega$ be the angular velocity = V/R after the collision.

Then the angular momentum must be due the I_disc*V/R and the mVR of the smaller mass, with V < v, so

L = 1/2 MR²*V/R + mVR.



See http://hyperphysics.phy-astr.gsu.edu/hbase/tdisc.html#dis

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html - third plate - Common Moments of Inertia

 

[fatcat39]

Sorry - where did the 1/2 come from?

and is it then mvr = (1/2 MR^2*V/R + mVR)w ??

Sorry, you lost me with the "angular momentum must be due...V<v, so" part. *confused*