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

Answer

A

Explanation

R=2m
To reverse the direction, `inttaud""theta=0` (work done is zero)
As the force is acting tangentially
`:.tau=RFsin90^(@)=2(20t-5t^(2))`
`=40t-10t^(2)`
`a=(tau)/(I)=(40t-10t^2)/(10)=4t-t^(2)`
`omega=int_(0)^(t)alphadt=(2t^(2)-(t^(2))/(3))`
If `omega=0,2t^(2)-(t^(2))/(3)=0,t=6sec.`
Angular displacement, `theta=intomegadt`
`=int_(0)^(6)(2t^(2)-(t^(2))/(3))dt`
`[(2t^(3))/(3)-(t^(4))/(12)]_(0)^(6)=216[(2)/(3)-(1)/(2)]=36` rad.
No. of revolutions `=(36)/(2pi)` which is less than 6