Question

A pulley of radius $2m$ is rotating about its axis by a force $F=(20t-5t^2)N$ (where, '$t$' is measured in second) 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 if reversed, is:

• A. More than $3$ but less than $6$
• B. greater 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 $6$

Given that,

Force $f=(20t-5t^2)N$

let, Torque $=\tau$

radius r = 2m

moment of Inertia I $=10kg-m^2$

We know,

$\begin{array}{cc}& \tau =f\times r\\ & \tau =I\alpha \\ \therefore & F\times r=I\cdot \alpha \\ & =(20t-5t^2)\times 2=10\times \alpha \\ & \Rightarrow 4t-t^2=\alpha \end{array}$

$now$

$d=\frac{dw}{dt}$

or ${\int }_0^{\infty }dw={\int }_0^t(qt-t^2)dt$

$or=2t^2-\frac{t^3}{3}$

If $w=0,t=0$ and $t=6$

$\begin{array}{cc}\text{again,}w& =\frac{d\theta }{dt}\\ \text{or}d\theta & =(2t^2-\frac{t^3}{3})dt\\ \text{or}{\int }_0^{\theta }& d\theta ={\int }_0^6(2t^2-\frac{t^3}{3})dt\\ or& \\ & \theta =36\text{radiun}\end{array}$

no. of revolution $\frac{36}{2\pi }$ revolutions $<6$

Ans : Option-A