Answer to Question #105722 in Classical Mechanics for Alyssa

Question

A flywheel mounted on a shaft of radius 2 cm has a moment of inertia 1.6*10^-3 kg m^2. A string of length 2 m and with a mass of 1 kg is wound on the shaft and is let to fall down. The mass takes 10 s to get off the shaft's pin. Find the work done per revolution.

Answer 1

As per the given question,

Radius of the shaft= 2cm

Moment of inertia of flywheel$=1.6\times10^-3 kg m^2.$

Length of the string = 2m

Mass of the string = 2kg

Time taken to get off the string =10s

Now potential energy lost by string

=$mg\dfrac{l}{2}=2\times 9.8\times 1=19.6J$

total energy required to 19.6J

total length of the string 2m,

so number of revolution, =$\dfrac{2}{2\pi r}=\dfrac{2}{2\times 3.14\times 2\times 10^{-2}}=15.92$

So, energy energy per revolution, = $\dfrac{19.6}{15.92}=1.23J$