I see what I did wrong. I was using the PE at the top not from the pivot point. The EQ for the speed at the top of the pivot is still from TE=PE+KE TE=9.8*1.13*mass PE=9.8*.43*mass the .43 is from 1.14-.915 which gives the radius=.215 r*2=.43 This gives me 6.86 which equals KE=1/2mv^2...
Homework Statement
The string in the Figure is L = 113.0 cm long and the distance d to the fixed peg P is 91.5 cm. When the ball is released from rest in the position shown, it will swing along the dashed arc. How fast will it be going when it reaches the lowest point in its swing?
B-How...