# Physics

A ball of mass m=4.6, at one end of a string of length L=6.6, rotates in a vertical circle just fast enough to prevent the string from going slack at the top of the circle. Assuming mechanical energy is conserved, the speed of the ball at the bottom of the circle is:

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1. at the top:
mv^2/r=mg

v^2=rg
KE at top: 1/2 m v^2=1/2 m rg
PE at top mag(2r)
KE at bottome+1/2 m rg+mg(2r)
1/2 mvb^2=1/2 mrg+2rmg

vb^2=5rg
veloecitybaottom=sqrt(5rg

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