A uniform circular disc of radius `50 cm` at rest is free to turn about an axis, which is perpendicular to the plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of `2.0 rad//s^(2)`. Its net acceleration in `m//s^(2)` at the end of `2.0 s` is approximately

Options

A. `8.0`
B. `7.0`
C. `6.0`
D. `3.0`

Explanation

`omega = omega_(0) + alpha t = 0 + 2 xx 2 = 4 rad//s`
`a_(c ) = omega^(2) R = (4)^(2) xx (0.5) = 8 m//s^(2)`
`a_(t) = R alpha = 0.5 xx 2 = 1 m//s^(2)`
`a = sqrt(a_(c )^(2) + a_(t)^(2)) = sqrt((8)^(2) + (1)^(2)) ~= 8 m//s^(2)`