A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its 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. `7.0`
B. `6.0`
C. `3.0`
D. `8.0`

Explanation

Angular acceleration, `alpha = 2 "rad"//s^(2)`
diameter of the disc, R = 50 cm = 0.5 m
Angular speed of the disc after 2 s,
`omega = omega_(0) + alpha` t
= `(0 + 2 xx 2) " "[ because omega_(0) = 0]`
At that instant, radial acceleration,
`a_(r ) = R omega^(2) = 0.5 xx (4)^(2) = 8 m//s^(2)`
and tangential acceleration,
`a_(t) = R alpha = 0.5 xx 2 = 1 m//s^(2)`
`therefore` Resultant acceleration,
a = `sqrt(a_(r )^(2) + a_(t)^(2)) = sqrt( 8^(2) + 1^(2)) approx 8 m//s^(2)`
The option D is correct.