Question

A pulley of radius 2 m is rotated about its axis by a force F =20t-5t² newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kgm² , the number of rotations made by the pulley before its direction of motion it reversed,

Options:

(a) more than 6 but less than 9

(b) more than 9

(c) less than 3

(d) more than 3 but less than 6

Answer 1

Torque is given by τ = FR

Or α = FR/I

Given

F =20t – 5t2 R = 2m,

I = 10 kgm2

α = $\frac{(20t-5t^2)\times 2}{10}$

α = 4t – t 2

 

$\omega ={\int }_t^2\alpha dt=2t^2-t^3/3$

at $\omega =0=2t^2-t^3/3$

$t^3=6t^2$

$t=6$

$\theta ={\int }_0^6\omega dt={\int }_0^6(2t^2-t^3/3)dt$

$\theta =\frac{36}{2\pi }<6$

Hence, your answer should be (d) more than 3 but less than 6