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Homework Help Overview

The problem involves a bullet striking a pendulum in a perfectly inelastic collision, where the goal is to determine the minimum initial velocity required for the pendulum-bullet system to complete a circular vertical loop.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relevant equations, including momentum and energy conservation. There is an exploration of the conditions necessary for the pendulum to complete a loop, with emphasis on the minimum velocity at the top of the loop.

Discussion Status

Some participants have provided insights into the necessary conditions for completing the loop, questioning the original poster's calculations and suggesting that the velocity at the top of the loop must be greater than zero. There is an acknowledgment of the need for a diagram to clarify the problem setup.

Contextual Notes

Participants note that the original poster's title lacks specificity regarding the problem's content, which may affect clarity in the discussion.

physics517
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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


Homework Equations



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 I am 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
 
Last edited:
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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.
 
physics517 said:
mg(2L)=1/2 m vf^2 --------- This is the Vf needed to comlete one whole loop (also i think this is where I am 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 possesses - 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!
 

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