A ball is attached to a horizontal cord of length L whose other end is fixed, as

Question

A ball is attached to a horizontal cord of length L whose other end is fixed, as shown below.

(a) If the ball is released, what will be its speed at the lowest point of its path? (Use the following as necessary: g and L.)

(b) A peg is located a distance h directly below the point of attachment of the cord. If h = 0.86 L, what will be the speed of the ball when it reaches the top of its circular path about the peg? (Use the following as necessary: g and L.)

A ball is attached to a horizontal cord of length L whose other end is fixed, as shown below. (a) If the ball is released, what will be its speed at the lowest point of its path? (Use the following as necessary: g and L.) (b) A peg is located a distance h directly below the point of attachment of the cord. If h = 0.86 L, what will be the speed of the ball when it reaches the top of its circular path about the peg? (Use the following as necessary: g and L.)

Explanation / Answer

V^2 = 2gL is the square of the velocity when the ball has dropped L distance. You are correct with part 1.

Thus the kinetic energy at the bottom is KE = mgL and at the top of the peg swing total energy TE = ke + pe = 1/2 mv^2 + 2mg(L - h) = mgL = KE from the conservation of energy. ke is the kinetic energy of the ball at the top of its swing around the peg and pe = mg2(L - h) is its potential energy at that point. L - h = r the rotation radius around the peg.

Therefore v^2 + 4g(L - h) = 2gL; so that v^2 = 2g(L - .86L) = 2g(.14L) = .28gL The answer to part 2 is v = sqrt(.28gL)

We assumed the ball swings around the peg with a rotation axis r = .14L as the peg is .86L from the point where the string is fixed. That is, when the ball is at the bottom, the string entangles with the peg that is .36L above the ball when its at the bottom of its drop from the horizontal.

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