Question #105722
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.
1
Expert's answer
2020-03-17T09:59:20-0400

As per the given question,

Radius of the shaft= 2cm

Moment of inertia of flywheel=1.6×103kgm2.=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

=mgl2=2×9.8×1=19.6Jmg\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, =22πr=22×3.14×2×102=15.92\dfrac{2}{2\pi r}=\dfrac{2}{2\times 3.14\times 2\times 10^{-2}}=15.92

So, energy energy per revolution, = 19.615.92=1.23J\dfrac{19.6}{15.92}=1.23J


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Assignment Expert
19.03.20, 17:26

Dear Alyssa, this is the energy lost for the center of mass of the string.

Alyssa
18.03.20, 18:31

Thank you for the answer! But could you please explain why the energy lost by string is 1/2mgl and not just mgl?

LATEST TUTORIALS
APPROVED BY CLIENTS