- #1

- 10

- 0

## Homework Statement

A mass m with velocity v

_{0}makes an inelastic collision with second mass M that is suspended by a string of length L. The velocity v

_{0}is perpendicular to the vertical string. After the collision the combined masses, m+M, rotate in a vertical plane around the point of suspension of the string. Find the minimum value v

_{0}such that the string remains always under tension for a complete rotation.

## Homework Equations

∑F

_{r}=mv

^{2}/r

T=1/2mv

^{2}

U=mgh

## The Attempt at a Solution

The condition for the an inflexible string to remain under tension is that it never goes slack, i.e. length=L for the entire rotation.

Since the collision is inelastic, momentum is conserved while kinetic energy is not. So,

mv

_{0}=(m+M)v' where v' is the magnitude of the x-directed velocity of the combined masses (m+M) after the collision. so v'=m/(m+M)v

_{0}.

__At the bottom of the vertical circle (In Polar Coordinates),__

letting m'=(m+M)

∑F

_{r}= T

_{b}-m'g=m'v'

^{2}/L (*)

__At the top,__

∑F

_{r}=T

_{t}+m'g=m'v''

^{2}/L (**)

__Using Energy,__

1/2m'v'

_{2}=1/2m'v''

^{2}+m'g(2L) (***)

v'

^{2}=v''

^{2}+4gL

v''

^{2}=v'

^{2}-4gL

I am going to pause here to see if there is anything I am missing.