Question

In a continuous printing process, paper roll is wrapped on a cylindrical core which is free to rotate about a fixed horizontal axis. The paper is drawn into the process at a constant speed $v$ , as shown in the figure. If $r$ is the radius of paper on the roll at any given time and $b$ is the thickness of the paper, then the angular acceleration of the roll at this instant is

• A. $\frac{v^{2}b}{2{\pi }^2{r}^3}$
• B. zero because $v$ is constant
• C. $\frac{v^{2}b}{2\pi r^3}$
• D. cannot be calculated from given information

Answer 1

Answer: (C)

In a continuous printing process suppose the total length of paper is $l$ .
If $r_1$ is the inner radius of roll then, $\pi r^2-\pi r_1^2=l\times b$ ...(i)
And at this moment, $v=r\omega$ ...(ii)

Differentiating equation (i) w.r.t. time,
$\Rightarrow \pi \times 2r\frac{dr}{dt}=\frac{dl}{dt}\cdot b$
$\Rightarrow \frac{dr}{dt}=\frac{v.b}{2\pi r}$ ...(iii)
Differentiating equation (ii) w.r.t time
$=>\frac{dv}{dt}=r\left(\frac{d\omega }{dt}\right)+\omega \left(\frac{dr}{dt}\right)$
$\text{⇒}0=r\alpha +\omega \frac{dr}{dt}$ $\left(\because \frac{\text{dr}}{\text{dt}}\text{is}-\text{ve}\right)$
$\text{⇒}\frac{dr}{dt}=\frac{r\alpha }{\omega }$
$\text{⇒}\alpha =\frac{\omega }{r}\left(\frac{dr}{dt}\right)$ ...(iv)
Using (iii) and (iv), $\frac{vb.\omega }{2\pi r^2}=\frac{v^{2}b}{2\pi r^3}=\alpha$