Moment of Inertia and Collisions

[the keck]

Homework Statement

A uniform symmetric ellipsoid (Mass M) has a large semi axis c and small semi axis a. A particle of mass m<<M is moving along a straight line parallel to the x-axis. Its y-coordinate is a/2 and its z-coordinate it c/2. After an inelastic collision, it sticks to the ellipsoid and then it (The ellipsoid) starts to move and rotate. One can assume the moment of inertia of the composite system equals to that of the ellipsoid.

Find its linear velocity after collision
Find the angular momentum of the system
Find the precession of the ellipsoid c-axis around direction of the angular momentum

Homework Equations

Moment of Inertia about z-axis is I=(2/5)*M*a^2
Moment of Inertia about other axes is I=(1/5)*M*(a^2+c^2)
KE=0.5*I*w^2 where w is angular velocity
L= I*w

The Attempt at a Solution

I suspect that you need the linear velocity to determine both the angular momentum and the precession. But my linear velocity cancels out when I equate KE=0.5*(m+M)*v^2=0.5*(0.4Ma^2)*(v/a)^2 [v/a=w]

Any help would be nice! Thanks a lot

Regards,
The Keck

 

[chaoseverlasting]

By conservation of linear momentum, mv(i)+M(0)=(m+M)v'. Here v' is velocity vector of the center of mass of the system.

By conservation of angular momentum, m(v i)x(a/2 j+ c/2 k)=Iw, where I=I cm about the x axis. From this you can find the angular velocity of the system. $$w=\frac{mv}{I_{cm}}(\frac{a}{2}k-\frac{c}{2}j)$$.

I don't know what you mean by precision but if you explain it, I may be able to help you.

 

[the keck]

So that would mean you actually need the initial velocity of the particle m, cause the question seems to indicate you don't need it. (I a bit confused about what you are doing...are you finding w and v independently and then use the fact that v=a*w or something?)

I'll check up on that the precession means, cause personally I'm not sure on it myself.

Thanks a lot

Regards,
The Keck

 

[the keck]

Okay...I checked up on my lecturer who wrote this question, and he said that he had forgotten to put in an initial velocity v(i). But I'm still not sure on what you are doing with the angular velocity. Are you saying that the angular momentum is mv(i) x (a/2j + c/2k)?

Regards,
The Keck

 

[the keck]

Okay...I checked up on my lecturer who wrote this question, and he said that he had forgotten to put in an initial velocity v(i). But I'm still not sure on what you are doing with the angular velocity. Are you saying that the angular momentum is mv(i) x (a/2j + c/2k)?

Regards,
The Keck