2008 Fnet=ma Exam: Bullet and Bob

[compwiz3000]

A bullet of mass m_1 strikes a pendulum of mass m_2 suspended from a pivot by a string of length L with a horizontal velocity v_0. The collision is perfectly inelastic and the bullet sticks to the bob. Find the minimum velocity v_0 such that the bob (with the bullet inside) completes a circular vertical loop.

What did I do wrong:
Using conservation of momentum, we have
m_1 v_0 = \left(m_1+m_2\right) v_n.
Then
v_n=\frac{m_1 v_0}{m_1+m_2}
and
0=v_n^2+2a dx.
Then
v_n^2=-2 \int_0^\pi -g \sin \theta L d \theta?
I think that's where I screwed up?
If I follow my wrong steps, I end up with
v_0=\left(m_1+m_2\right)2 \sqrt{Lg}/m_1,
but that's not the answer.

 

[adriank]

This seems to be a bit of a trick question. The string is not rigid; it cannot provide an outward force at the top, so at the top of the loop the pendulum better have enough speed to not fall down. Your solution is correct if the string was replaced by a rigid rod.

For brevity I'll write M = m₁ + m₂. The speed required at the top of the loop is that for uniform circular motion with acceleration g: v_{top}^2/L = g, so at this point its kinetic energy is M v_{top}^2/2 = MgL/2. At the minimum initial speed, the kinetic energy the pendulum and bullet have must equal the difference in potential energy at the top and bottom, plus this kinetic energy:

\frac12 M v_n^2 = 2MgL + \frac12 MgL = \frac52 MgL
v_n = \sqrt{5gL}
v_0 = \frac{M}{m_1} v_n = \frac{m_1 + m_2}{m_1} \sqrt{5gL}.

 

[compwiz3000]

Thanks.

Also, in my original solution, I wasn't sure if it was correct to say
v_n^2+2a\, dx = 0
Can I just put that dx in there? I kind of just guessed that I could do that.
Can somebody more rigorously do my solution for a rigid rod?

Also, can you elaborate on this?
"At the minimum initial speed, the kinetic energy the pendulum and bullet have must equal the difference in potential energy at the top and bottom, plus this kinetic energy"
I didn't understand what you were referring to in the equations.

EDIT:Wait, 2MgL is just potential energy right?

 

[adriank]

Yeah sorry, 2MgL is the change in potential energy and MgL/2 is the kinetic energy the thing must have at the top.

As for your initial question, what is a?

For the rigid rod, you can drop the MgL/2 term in the energy equation, and your original answer will be correct:
v_0 = 2 \frac{m_1 + m_2}{m_1} \sqrt{gL}.

 

[compwiz3000]

So is my calculus way rigorous or correct? I tried to use v_f^2=v_0^2+2ax in my first attempt...is that wrong?

And for the rigid rod, I can drop the term because I can just have almost 0 kinetic energy at the top, right?

 

[adriank]

Yeah, that won't work since your acceleration isn't constant. It's really easy if you just use energy; that way you don't have to deal with possibly incorrect integrals and all that. You're correct about the 0 kinetic energy.

 

[compwiz3000]

Thanks for the energy solution. I do prefer that solution. I just didn't think of how to do it that way.

Could you also show me a way to do it with integration?