Physics homework - how to solve?

Question

A ball is attached to a horizontal cord of length L whose other end is fixed. (a) If the ball is released what will be its speed at hte lowest point of its path? (b) A peg is located a distance h directly below the point of attachment of the cord. If h=0.8.L, what will be the speed of the ball when it reaches the top of its circular path about the peg?

Answer: (a) sq.rt. 2gL (b) sq.rt.1.2gL

Answer 1

Throw your homework onto the fire...go out and find the one that you love.

Answer 2

tul b is right with part a, I'll try to help with b.

The peg changes the path the ball follows in part a. From part a, when the cord makes contact with the peg, its speed is sq.rt. 2gL and its kinetic energy is
KE = 1/2mv^2 = 1/2m2gL.
It wraps around the peg and, when the ball is at the top of its circular path about the peg, some of that kinetic energy is converted back to potential.

Call the kinetic part of the ball's energy above the peg KEp, and the potential part when at that point PEp. The potential energy
PEp = mg*.4L

(I had trouble here - the ball is .2L below the peg at the bottom and swings to .2L above the peg. Delta height = .4L.)

So
KE = 1/2m2gL = KEp + PEp = KEp + mg*.4L
So
1/2m2gL = KEp + mg*.4L
KEp = 1/2m*Vpt^2 = 1/2m2gL - mg*.4L
where Vpt is speed at the top of the path around the peg.

Simplify
Vpt^2 = 2gL - .8gL = 1.2gL
Vpt = sqrt 1.2gL

Answer 3

Using the Law of Conservation of Energy, and Conservation of Momentum should be able to provide you the solution have.

Answer 4

a)Potential Energy, PE at start=mgL
Kinetic Energy , KE at lowest point=1/2mv^2

Based on principle of conservation of energy,

PE=KE

mgL=1/2mv^2

(m cancels out)

v^2=2gL
v=sq. rt 2gL

b)PE=mg*0.8L
KE=1/2mv^2

PE=KE

mg*0.8L=1/2mv^2

(m cancels out)

v^2=1.6gL
v=sq.rt 1.6gL

Maybe I didn't understand the question. My answer is different from the one you gave.