Aball of mass m = 4.6 kg, at one end of a string of length l= 6.6 m, rotates in a vertical circle just fast enough to prevent the string from going slack at the top of the circle. assuming mechanical energy is conserved, calculate the speed of the ball at the bottom of the circle.

 v₂ = 17.98 m/s

Explanation:

given,

mass of ball = m = 4.6 Kg

length of string = L = 6.6 m

force acting toward the center is equal to the force exerted by centripetal acceleration

now, calculating the speed of ball at the bottom of the circlr

work done by the gravity = change in kinetic energy

 v₂ = 17.98 m/s

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s