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

$Mg = m\omega^2 R$

$M(9.8) = 0.05(\frac{2\pi}{0.25})^2(0.4)$

$M(9.8) = 12.63$

$M = 1.3 Kg$

so mass will be 1.2 kg