Homework Statement

Abullet of mass m strikes a pendulum of mass M suspended from a pivot by a string of length L with a horizontal velocity v0. The collision is perfectly INelastic and the bullet sticks to the bob. Find the minimum velocity v0 such that the bob (with the bullet inside) completes a sircular vertical loop

p=mv

KE=1/2 mv^2

PE=mgh

The Attempt at a Solution

mvo=(m+M)vf

mg(2L)=1/2 m vf^2 --------- This is the Vf needed to comlete one whole loop (also i think this is where im missing something)

vf=2 sqrt(Lg)

and so the answer i got was vo= ((m+M)2 sqrt(Lg)) / m

but the ACTUAL asnwer is vo= ((m+M) sqrt(5Lg)) / m

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gneill
Mentor
What equations are relevant? Where's your attempt at a solution?

By the way, your post title is pretty unspecific about the actual problem content.

mg(2L)=1/2 m vf^2 --------- This is the Vf needed to comlete one whole loop (also i think this is where im missing something)
Indeed. That is NOT the velocity needed to complete one loop, that is the minimum velocity required for the bob (and bullet) to just reach the top of the circular loop. For the bob and bullet to complete a circular loop, it cannot have zero velocity at the top of the loop! There is a minimum velocity that it must possess - think back to your concepts of circular motion.

Lol I shouldve drew a diagram.

so up top its F+mg=mv^2/r and the F (tension) cancels out since we're trying to find minimum speed. so v=sqrt(gr) and if we go back to my original work equation

it should be 1/2 m vf^2 = mgh + 1/2 m vtop ^2

then you find the Vf and then plug that into the momentum equation

Thanks so much guys!