A pulley os radius 2m is rotated about its axis by a force `F= (20 t- 5t^2)` newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is `10kg m^2` the number of rotaitons made by the pulley before its direction of motion is reversed, is:

Options

A. more than 3 but less than 6
B. more than 6 but less than 9
C. more than 9
D. less than 3

Answer

A

Explanation

We have, Torque `=r xx F= l alpha`
So, `2(20t-5t^(2))=10alpha`
`alpha= 4t-t^(2)`
But `alpha = (d omega)/(dt)`, so
`(d omega)/(dt)= 4t-t^(2)`
`d omega = (4t- t^(2))dt`
Integrating, `omega= 2t^(2)- (t^(3))/(3)`
`omega` will be zero at t= 6s
So, `omega (d theta)/(dt) =2t^(2)- (t^(3))/(3)`
or `d theta= (2t^(2)- (t^(3))/(3)) dt`
Again, integrating both sides
`theta= (2t^(3))/(3)- (t^(4))/(12)`
Since, t=6 s
so, `theta= (2 xx 6^(3))/(3)- (6^(4))/(12)=36`
Number of turns, `n= (theta)/(2pi) =(36)/(2pi)= 5.73`
So, option (a) is right.