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

Options

A. 7
B. 6
C. 3
D. 8

Answer

D

Explanation

a uniform circular disc of radius 50 cm at rest is free to turn about an axis having perpendicular to its plane and passes through its centre. This situation can be
`:.` Angular acceleration `alpha=2rad s^(-2)` (given)
Angular speed, `omega=alphat=4 rad s^(-2)`
`because` Centripetal acceleration ,`a_c=omega^2r=(4)^2xx0.5=16xx0.5`
`a_c= 8 ms^(-2)`
`because` Linear acceleration at the end of 2s `a_t=alphar=2xx0.5implies a_t=1 ms^(-2)`
Therefore, the net acceleration at the end of 2s is given by
`a=sqrt(a_c^2+a_1^2)`
`a=sqrt((8)^2+(1)^2)=sqrt65 implies a ~~ 8 ms^(-2)`