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