Question

A pulley of radius 2m is rotated about its axis by a force F = ($20t-5t^2$)N (Where , t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 Kg-m2, then the number of rotations made by the pulley before its direction of motion is reserved, is:

• 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 1

The correct option is A more than 3 but less than 6

to reverse the direction

$\int \tau d\theta =0$

$\tau =(20t-5t^2)\times 2=40t-10t^2$

$\alpha =\frac{\tau }{I}=\frac{40t-10t^2}{10}=4t-t^2$

$\omega ={\int }_0^t\alpha dt=2t^2-\frac{t^3}{3}$

at $\omega =0⟹2t^2-\frac{t^3}{3}=0⟹t^3=6t^2⟹t=6$

$\theta ={\int }_0^6\omega dt={\int }_0^6(2t^2-\frac{t^3}{3})dt=[\frac{2t^3}{3}-\frac{t^4}{12}{]}_0^6=36rad=\frac{36}{2\pi }<6$