Question

The same experiment is performed. This time the stoppers mads is known m= 0.05 kg, the radius of the string is 0.4 m and the period is 0.25s. what is the weight of the hanging mass?

Answer 1

As we know that weight of the hanging mass will be same as tension in the string

so we will have

`T = Mg`

now also we know that stopper will revolve in the circle with constant speed and the tension force will be the centripetal force on it

`T = m\omega^2 R`

now we will have

`Mg = m\omega^2 R`

`M(9.8) = 0.05((2\pi)/(0.25))^2(0.4)`

`M(9.8) = 12.63`

`M = 1.3 Kg`

so mass will be 1.2 kg