Rod rotates when hit by point mass/ Cons. of angular momentum

[Epictetus]

A thin uniform stick of length 4m and mass 3 kg is hung vertically from a support 1m from its end. The stick is initially at rest. A point mass of 7 kg traveling at 6.11 m/s is moving and impacts the stick at the bottom-most point. Determine how high the stick and point mass will rotate before coming to rest momentarily.

I applied the conservation of angular momentum theorum

L = r (moment arm) mv
= (3)(7)(6.11) = 128.31

I = (summation) m r(moment arm)^2 = 3 (1)^2 + 3 (3)^2 = 30 kg m^2
(where 1 and 3 are the distances from the pivot point to the center of mass and pivot point to the point of impact, respectively)

I` = I(rod) + 7(3)^2 = 93 Kg m^2

L = I omega
128.31 = 93 omega
omega = 1.38 rad/sec

Kinetic energy is then transferred to potenial:

1/2 I` omega ` ^2 = mgh
1/2 (93) (1.38) ^2 = mgh
88.51 = g[7 (3cos theta) + 3 cos theta]
9.022 = 24 cos theta
theta = 67.92 deg.

Are the moment arms that I used for the (I = (summation) m r(moment arm)^2) formula correct? Assuming I did my calculations correct, how do I determine the height from the information that I got?

 

[Doc Al]

A thin uniform stick of length 4m and mass 3 kg is hung vertically from a support 1m from its end. The stick is initially at rest. A point mass of 7 kg traveling at 6.11 m/s is moving and impacts the stick at the bottom-most point. Determine how high the stick and point mass will rotate before coming to rest momentarily.

I assume that the point mass attaches itself to the stick? (An inelastic collision?)

I applied the conservation of angular momentum theorum

L = r (moment arm) mv
= (3)(7)(6.11) = 128.31

OK.

I = (summation) m r(moment arm)^2 = 3 (1)^2 + 3 (3)^2 = 30 kg m^2
(where 1 and 3 are the distances from the pivot point to the center of mass and pivot point to the point of impact, respectively)

I assume this is supposed to be the moment of inertia of the stick? If so, you'll have to redo it. What's the moment of inertia of a stick about one end? or about its center? (Look it up!) Then use the parallel axis theorem (Or just treat it as two short sticks added together) to find the moment of inertia about the pivot point.

...

Are the moment arms that I used for the (I = (summation) m r(moment arm)^2) formula correct?

No. See comments above.

Assuming I did my calculations correct, how do I determine the height from the information that I got?

Use conservation of angular momentum to find the angular speed after the collision. Then find the rotational KE of the system after the collision. Then apply conservation of mechanical energy. (Hint: Find the change in height of the center of mass, then convert that to an angle.)