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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?

Respuesta :

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

so we will have

[tex]T = Mg[/tex]

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

[tex]T = m\omega^2 R[/tex]

now we will have

[tex]Mg = m\omega^2 R[/tex]

[tex]M(9.8) = 0.05(\frac{2\pi}{0.25})^2(0.4)[/tex]

[tex]M(9.8) = 12.63[/tex]

[tex]M = 1.3 Kg[/tex]

so mass will be 1.2 kg

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