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`

Answer

A

Explanation

(a) Particle at periphery will have both radial and tangential acceleration
`a_t = R prop = 0.5 xx 2 = 1 m//s^2`
`omega = omega_0 + prop t`
`omega = 0 + 2 xx 2 = 4 rad//sec`
`a_c = omega^2 R = (4)^2 xx 0.5 = 16 xx 0.5 = 8 m//s^2`
`a_(total) = sqrt(a_p^2 + a_c^2) = sqrt(1^2 - 8^2) ~~ 8 m//s^2`.