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 in `rad//s^(2)` is

Options

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

Answer

C

Explanation

m= 20 kg, r= 20cm, F= 25N
`tau = I alpha = ((1)/(2) MR^(2)) alpha`
`tau= F r = 25 xx 0.2= 5 N-m`
`alpha= (tau)/(I)= (5)/((1)/(2) xx 20 xx (0.2)^(2))= 12.5 "rad/s"^(-2)`