Question

a warrior swings a slingshot in a horizontal circle above his head at a constant speed. The sling is 1.5 m long, and the stone has a mass of 50 g. The tension in the string is 3.3 N. When he releases the sling, what will the stone's speed be?

Answer 1

r = 1.5 m , m = 0.05 kg
Tension = Centrifugal force
T= mrw² , w = angular velocity
3.3 N = 0.05 x 1.5 x w²
w = 6.63 rps
Linear Velocity = rw = 1.5 x 6.63 = 9.945 m/s

Answer 2

The speed of the stone when he releases the sling is equal to 9.95 m/s.

Given the following data:

• Length of sling (radius) = 1.5 meter

• Mass of stone = 50 grams to kg = 0.05 kg

• Tension = 3.3 Newton

To determine the speed of the stone when he releases the sling:

First of all, we would solve for the angular velocity of the stone by applying

the law of conservation of momentum.

Tension in the string = Centrifugal force of the stone

$Tension = mrw^2$

$\omega = \sqrt{\frac{Tension}{mr} } \\\\\omega = \sqrt{\frac{3.3}{0.05 \times 1.5} }\\\\\omega = \sqrt{44}$

Angular velocity = 6.63 rad/s

Now, we can determine the speed of the stone when he releases the sling:

$V = r\omega\\\\V=1.5 \times 6.63$

Speed, V = 9.95 m/s

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