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

Answer

C

Explanation

Given , r = 50 cm = 0.5 m `alpha = 2.0` rad `s^(-2) , omega_(0) = 0`
At the end of 2 s ,
Tangential acceleration , `a_(t) = r alpha = 0.5 xx 2 = 1 ms^(-2)`
Radial acceleration , `a_(r) = omega_(2) r = (omega_(0) + alpha t)^(2) r`
`= (0 + 2 xx 2)^(2) xx 0.5 = 8 m s^(-2)`
`therefore` Net acceleration ,
`a = sqrt(a_(t)^(2) + a_(r)^(2)) = sqrt(1^(2) + 8^(2)) = sqrt(65) = 8 m s^(-2)`