Use conservation of energy arguments to find the speed of the ball when it reaches the top of the circular path about the peg. your answer should be expressed in terms of l, g and d.

Using law of conservation of energy, "Energy can neither be created nor be destroyed but it can be transformed from one form to another".

That is, sum of Kinetic energy and Potential energy at initial point is equal to the sum of Kinetic energy and Potential energy at final point.

We can say that,

(K.E)1 + (P.E.)1 = (K.E)2 + (P.E)2

1/2 m1 * v1² + m1*g*h1 = 1/2 m2 * v2² + m2* g*h2

For motion in circular path,

Initial velocity v1 is at h1 =0

Final velocity v2 is at h2= d

For rotational motion, inertia is denoted by I.

1/2 I1 * v1² = 1/2 I2* v2² + I2 gd

1/2 I2*v2² = 1/2 I1*v1²- I2 gd

v2²= I1/I2 * v1²- 2 gd

v2= √(I1/12* v1²- 2gd)

The above equation gives the speed of the ball when it reaches the top of circular path in term of I, g and d.

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