# A small 0.5 kg mass is attached to a string of length 1 meter and is swung around in a circle making 6 revolutions in 12 seconds. 1. What is the period of revolution? 2. What is the centripetal...

A small 0.5 kg mass is attached to a string of length 1 meter and is swung around in a circle making 6 revolutions in 12 seconds.

1. What is the period of revolution?

2. What is the centripetal acceleration?

3. What force physically causes the centripetal acceleration?

4. Find the value of this force?

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### 1 Answer

1. Period of revolution is the time taken by one complete rotation.

`T = t/N = 12/6 =2 seconds`

2. Angular speed is the total angle over the time. Hence

`omega = ("angle")/(time) = (2*pi*N)/t = (2*pi)/T = pi (rad)/s`

Centripetal acceleration depends on the radius of the movement,

`a_(cp) = omega^2*R = pi^2*1=9.87 (rad)/s^2`

3. The centripetal acceleration is directed always towards the center of the circle. This means that the force that gives this cetripetal acceleration is the tension in the wire `T` .

4. `T= F_(cp)= m*a_(cp) = 0.5*9.87 = 9.34 N`

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