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Aleksandr-060686 [28]
3 years ago
11

A ball 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 t

he 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.
Physics
1 answer:
VashaNatasha [74]3 years ago
8 0

Answer:

 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

m g = \dfrac{mv_1^2}{r}

v_1 = \sqrt{gr}

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

work done by the gravity = change in kinetic energy

- m g (2R) = \dfrac{1}{2}m(v_1^2-v_2^2)

-4 gR =v_1^2-v_2^2

-4 gR =g R-v_2^2

v_2^2 = 5 g R

v_2=\sqrt{5\times 9.8 \times 6.6}

 v₂ = 17.98 m/s

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