# Rotational Momentum Problem Help

• xnitexlitex

#### xnitexlitex

A ballistic pendulum consists of an arm of mass M and length L = 0.55 m. One end of the arm is pivoted so that the arm rotates freely in a vertical plane. Initially, the arm is motionless and hangs vertically from the pivot point. A projectile of the same mass M hits the lower end of the arm with a horizontal velocity of V = 1.8 m/s. The projectile remains stuck to the free end of the arm during their subsequent motion. Find the maximum angle to which the arm and attached mass will swing in each case.

a) The arm is treated as an ideal pendulum, with all of its mass concentrated as a point mass at the free end.

b) The arm is treated as a thin rigid rod, with its mass evenly distributed along its length.

Relevant equations:
mvL = (Irod + Iprojf
Ei = Ef
There might be a simpler equation, though.

Here's how I tried to solve it:
First, I solved for ωf by substituting the moments of inertia. I ended up with this:

ωf = mvL / [(1/3)mL2 + mL2] , which became:
3mv/(mL+3m).

Then, I used energy conservation, resulting in this equation:
1/2(Irod + Iprojf2 + mg(L/2) = mg(L-L cos θ) + mg{L/2 + [(L/2) - (L/2 cos θ)]}

I canceled the masses, yet I still got it wrong. I can't seem to find a way to simplify the equation! All the above is for part B. If I can get part B, I can get A, since it's the same, but with one less moment of inertia to worry about. All help is appreciated.

I'm sorry if this post isn't perfect. This is my first post here.

1st stage: initially both projectile and pendulum are at rest.
2nd stage: the projectile hits and gets embedded in the pendulum.
3rd stage: both move as one body to a final angle.

I do not think that energy is conserved from stage (1) to stage (3) as shown above because in the impact at stage (2) energy is 'lost'.

1st stage: initially both projectile and pendulum are at rest.
2nd stage: the projectile hits and gets embedded in the pendulum.
3rd stage: both move as one body to a final angle.

I do not think that energy is conserved from stage (1) to stage (3) as shown above because in the impact at stage (2) energy is 'lost'.

Okay, but using momentum conservation doesn't get me the angle. Do I need Newton's Second Law?

Momentum conservation will give the speed of the combined system just after impact. THEN conservation of energy can be used and this will give the angle required.