My question is about the rotatioanal motion

Question

A some is attached to the string.The stone is rotating in a circle, the center is C and the radius is 85cm, at constant speed in a verticle plane, the stone has mass of 50g and it takes 0.12s to complete one revolution.Determine for this stone its angular speed, and the centripetal acceleration?

2.Now the stone is rotating in a horizontal circle, determine the tension in the string under this condition?

3.With the help of a free body diagram ,explain what will happen to the tension in the string when the string is vertical with the stone below C and above C?

Answer 1

Let

r = 85 cm = 0.85 m be the radius of rotation,

m = 50g = 0.05 kg be the mass of the stone,

T = 0.12 s be the period of rotation,

g = 9.81 m/s² be gravitational acceleration.

1.

The stone's tangential speed is the ration between the length of the circle

and the time it takes to perform one rotation.

v = 2πr/T

The angular speed is

w = v/r = 2π/T = 52.36 s^-1

The centripetal acceleration is

a = v²/r = 4π²r/T² = 2330.3 m/s²

2.

The tension, which is the centripetal force is

F = ma = mv²/r = 116.5 N

3.

When the stone is below C then the tension in the string must support the

weight of the stone in addition to providing the centripetal force.

So the tension will be

ma + mg = 116.5 + 0.05×9.81 = 117 N

When the stone is above C then the weight of the stone contributes to the

the centripetal force.

So the tension will be

ma - mg = 116.5 - 0.05×9.81 = 116 N