A pulley has 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 rotations 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

Torque = `r xx F = 1 alpha`
`:. 2(20t - 5t^(2)) = 10 alpha or alpha = 4t - t^(2)`
or `(d omega)/(dt) = 4t - t^(2)`
or `d omega = (4t - t^(2)) dt`
On integration, we get `omega = 2t^(2) - (t^(3))/(3)` ..(i)
`omega` will be zero at `t = 6 s`
`omega = (d theta)/(dt) = 2t^(2) - (t^(3))/(3)`
`d theta = (2t^(2) - (t^(3))/(3)) dt`
On intengration, we get `theta = (2 t^(3))/(3) - (t^(4))/(12)`
When `t = 6s, theta = (2 xx 6^(3))/(3) - (6^(4))/(12) = 36`
No. of turns, `n = (theta)/(2pi) = (36)/(2pi) lt 6`
Thus, option (a) is correct.