A ball is attached to a horizontal cord of length L = 1.38m whose other end is f

Question

A ball is attached to a horizontal cord of length L = 1.38m whose other end is fixed, as seen in figure below.

2. A peg is located a distance h directly below the point of attachment of the cord. If h = 0.880 L, what will be the speed of the ball when it reaches the top of its circular path about the peg?

A ball is attached to a horizontal cord of length L = 1.38m whose other end is fixed, as seen in figure below. If the ball is released, what will be its speed at the lowest point of its path? 2. A peg is located a distance h directly below the point of attachment of the cord. If h = 0.880 L, what will be the speed of the ball when it reaches the top of its circular path about the peg?

Explanation / Answer

1. using work - energy method,

mgL = mv^2 /2

v = sqrt(2gL) = sqrt(2 x 9.8 x 1.38) =5.2 m/s

2. after that ball will go h = 2 x (1 - 0.880)L = 0.24L high fromlowest position,

again using work-energy method,

mv^2 /2 = mg(0.24L) + mv'^2 /2

5.2^2 /2 = 9.8 x 0.24 x 1.38 + v'^2 /2

V' = 4.50 m/s

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