A pendulum consists of a 2.0 kg stone swinging on a 4.0 m string of negligible m

Question

A pendulum consists of a 2.0 kg stone swinging on a 4.0 m string of negligible mass. The stone has a speed of 9.0 m/s when it passes its lowest point.

(A) what is the speed when the string is at an angle of 60 degrees from the vertical?

(B) what is the greatest angle ( with the vertical) that the string will reach during the stones motion?

(C) just as the stone passes its lowest point, the string is cut (making the stone a horizontally launched projectile with initial speed of 9.0 m/s). The stone hits the ground 4.50 m below. Calculate the speed of the stone just before it hits the ground.

Explanation / Answer

a) when the string is 60 deg to the vertical, its height above its lowest point is

L-L cos 60 = L(1-cos 60) = 1/2 L where L is the length of the string, so the stone is 2m above its lowest point

at the bottom of the swing, the stone has 1/2 m v^2 kinetic energy (v=9m/s)

when it is 60 deg from vertical, the total energy is 1/2 m V^2 + m gh

where V is the speed at the new height and h = 2 m

therefore

1/2 m (9m/s)^2 = 1/2 m V^2 + m g (2m)

solve for V=6.465m/s

b) if the stone converts all its KE to PE, its height above the lowest level will be given by

1/2 m (9m/s)^2 = m gh or h = (9m/s)^2 / 2 g = 4.13m

recall from above that height above lowest point = L(1-cos(theta))

so here 4.13 m = 4m (1-cos(theta))

theta = 91.9 deg

c) total mechanical energy = PE + KE =1/2 *m*v^2 + mgh= 1/2 *2*(9^2) +2*g*4.5

if PE =0 at the bottom, then the total mechanical energy = KE at bottom =1/2 *m*v^2= 1/2 x 2kg x (Vm/s)^2

so,,1/2 x 2kg x (Vm/s)^2 = 1/2 *2*(9^2) +2*g*4.5

solving, Velocity before it hits the ground = V = 13 m/s

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