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. 7
B. 6
C. 3
D. 8

Answer

D

Explanation

According to given question, 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. The situation can be shown by the figure given below :
`:.` Angular acceleration, `alpha=2 "rad s"^(2)` (given)
Angular speed, `omega=alphat=4 "rad s"^(-1)`
`because` Centripetal acceleration, `a_(c)=omega^(2)r=(4)^(2)xx0.5=16xx0.5`
`a_(c)=8 ms^(-2)`
`because` Linear acceleration at the end of 2 s
`a_(t)=alphar=2xx0.5 rArra_(t)=1ms^(-2)`
Therefore, the net acceleration at the end of 2.0 s is given by
`a=sqrt(a_(c)^(2)+a_(t)^(2))`
`a=sqrt((8)^(2)+(1)^(2))=sqrt(65)rArra~~8 ms^(-2)`.