A string is wound round the rim of a mounted flywheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord. Neglecting friction and mass of the string, the angular acceleration of the wheel is

Options

A. 50 rad `s^(-2)`
B. 25 rad `s^(-2)`
C. 12.5 rad `s^(-2)`
D. 6.25 rad `s^(-2)`

Answer

C

Explanation

Here M = 20 kg
R = 20 cm = `(1)/(5)` m
Moment of inertia of flywheel about its axis is `I = (1)/(2) MR^(2)`
= `(1)/(2) xx 20 kg xx ((1)/(5) m)^(2) = 0.4 kg m^(2)`

As `tau = I alpha` (where `alpha` is the angular acceleration)
`therefore alpha = (tau)/(I) = (FR)/(I) = (25 xx (1)/(5))/(0.4) = (5 N m)/(0.4) = (5 N m)/(0.4 kg m^(2)) = 12. 5 "rad" s^(-2)`