Question

A 10.0 kg ball is attached to a massless rope that is 1.30 m long. The rope is pushed to 45 degrees from the vertical and released from rest. Ignore the rotation of the ball as the rope swings, and this does not appreciably affect the dymanics. What is the potential energy of the ball with reference to its lowest position? B) How fast will the ball be moving at the bottom of the swing? and B) How long does it take for the ball to swing back to its original position?

Answer 1

A) h = L –Lcos45 = 1.3 - 1.3cos45 = 0.3808 m

PE = mgh = 10*9.8*0.3808 = 37.31 J

  

B) Use law of conservation of energy,

K.E_f + P.E_f = K.E_i + P.E_i

Initially ball is start from rest so its initial kinetic energy is zero i.e K.E_i = 0

At bottom of swing final potential energy is become zero   so P.E_f= 0

Therefore above equation become,

                                                  K.E_f = P.E_i   

1/2*m*v² = mgy

v²= 2gy

v = 2.73 m/s

c) In this part asked to find time taken by the ball to swing back to its original position

To come back to original position, ball completed one rotation. Generally the time required to complete one rotation is called the period of object. It is calculated by using following equation.

T = 2π*sqrt(L/g)   

T = 2*3.14*sqrt(1.3/9.8)

T = 2.28 s