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=20 "cm", F=25 N`
`tau=I alpha=((1)/(2)MR^(2))alpha`
`tau=Fr=25xx0.2=5` N-m
`alpha=(tau)/(I)=(5)/((1)/(2)xx20xx(0.2)^(2))=12.5"rad/s"^(-2)`