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 `ms^(-2)` at the end of 2.0 s is approximately .............

Options

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

Answer

D

Explanation

`alpha=2` rad `s^(-2), omega=alphat=2xx2=4` rad `s^(-1)`,
`r=0.5m`
`a_(c)=romega^(2)=0.5xx(4)^(2)=8m//s^(2)`
`a_(T)=ralpha=0.5xx2`
`=1" rad/s"^(2)`
`therefore a=sqrt(a_(c)^(2)+a_(T)^(2))`
`=sqrt(8^(2)+1^(2))=sqrt(64+1)=sqrt(65)=8.06`
`therefore a~~8ms^(-2)`