Posted by **jim** on Sunday, March 22, 2009 at 11:57pm.

As shown below, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? (Use L for , g for gravity, and M and m as appropriate.)

I got the answer (4M/m)sqrt(gL) but the problem is that I did it a while ago and I completely forgot why this is the answer. Can anyone explain it to me?

- physics -
**drwls**, Monday, March 23, 2009 at 12:07am
If you derived it once, you should be able to derive it again.

The speed V of the pendulum bob after the bullet has passed through is given by

MV = m(v - v/2) = mv/2

V = mv/(2M)

To swing all the way to the top,

(1/2) M V^2 = M g (2L)

V^2 = 4 L g = m^2/4M^2 * v^2

v = sqrt (16 L g M^2/m^2)

= 4 (M/m) sqrt (Lg)

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