A stone at the end of a sling is whirled in a vertical circle of radius 1.20 m a

Question

A stone at the end of a sling is whirled in a vertical circle of radius 1.20 m at a constant speed v_0 = 2.20 m/s as in Figure P4.57. The center of the string is 1.50 m above the What is the range of the stone if it is released when the sling is inclined at 30.0degree with the horizontal at A? What is the range of the stone if it is released when the sling is inclined at 30.0degree with the horizontal at B? What is the acceleration of the stone just before it is released at A? Magnitude (absolute value) toward the center of the circle upward downward in the direction of v_0 What is the acceleration of the stone just after it is released at A? Magnitude (absolute value)

Explanation / Answer

Given,

R = m ; V0 = 2.2 m/s ; h = 1.5 m

Vx = V0 cos(90-30) = 2.2 x cos60 = 1.1 m/s

Vy = V0 Sin(90-30) = 2.2 x sin60 = 1.91 m/s

a) we know that, ay = Vy / t

t = Vy / ay = 1.91/9.8 = 0.195 s

di = 0.3 + 1.2 + sin30 x 1.2 = 2.1 m

dtop = -1/2 g t2 + Vx t + di

dtop = - 0.5 x 9.8 x 0.195 x 0.195 + 1.1 x 0.195 + 2.1 = 2.13 m

dbottom = dtop = 2.13

1/2 a t22 = 2.13 => t2 = 0.66 sec

total time will be t = t1 + t2 = 0.195 + 0.66 = 0.86 sec

from the defination of speed, Vx = X / t => X = Vx x t = 1.1 x 0.86 = 0.95 m

Hence, X = 0.95 m

b)Let this be X'

di = 1/2 g t2 + Vo x cos30 x t

4.9 t2 + 1.91 t - 2.1 = 0

t = 0.49 s

Again using, Vx = X' / t => X' = 1.1 x 0.49 = 0.54 m

Hence, X' = 0.54 m

c)acceleration will be:

a = v2/R = 2.2 x 2.2 / 1.2 = 4.03 m/s2

Hence, a =  4.03 m/s2

directed, towards the center of the circle.

d)when it is released, the only force that acts on the stone is the force of gravity, so its acceleration would be

a = g = 9.8 m/s2 acting downwards.

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