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.(adsbygoogle = window.adsbygoogle || []).push({});

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 = (I_{rod}+ I_{proj})ω_{f}

E_{i}= E_{f}

There might be a simpler equation, though.

Here's how I tried to solve it:

First, I solved for ω_{f}by substituting the moments of inertia. I ended up with this:

ω_{f}= mvL / [(1/3)mL^{2}+ mL^{2}] , which became:

3mv/(mL+3m).

Then, I used energy conservation, resulting in this equation:

1/2(I_{rod}+ I_{proj})ω_{f}^{2}+ mg(L/2) = mg(L-L cos θ) + mg{L/2 + [(L/2) - (L/2 cos θ)]}

I cancelled 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.

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# Rotational Momentum Problem Help!

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