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 the 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 the given information

Answer 1

The correct option is C $\frac{v^{2}b}{2\pi r^3}$

In a continuous printing process, suppose the total length of paper on the roll is $l$. If $r_1$ is the inner radius of the roll, then
$\pi r^2-\pi r_1^2=l\times b...\left(1\right)$
And at this moment, $v=r\omega ...\left(2\right)$
[where $v$ is the speed at which paper is drawn into the process]


Differentiating equation $\left(1\right)$ w.r.t time,
$\pi \times 2r\frac{dr}{dt}-0=b\times \frac{dl}{dt}$
or $\frac{dr}{dt}=\frac{-vb}{2\pi r}....\left(3\right)$
[$\because \frac{dl}{dt}=-v$ i.e length of the paper on the roll is decreasing]
Differentiating equation $\left(2\right)$ w.r.t time
$\frac{dv}{dt}=r\left(\frac{d\omega }{dt}\right)+\omega \left(\frac{dr}{dt}\right)$
$\Rightarrow 0=r\alpha +\omega \frac{dr}{dt}$
$[\because v=\text{constant}]$
$\Rightarrow \frac{dr}{dt}=\frac{-r\alpha }{\omega }$
$\Rightarrow \alpha =\frac{-\omega }{r}\left(\frac{dr}{dt}\right)....\left(4\right)$

Using $\left(3\right)$ and $\left(4\right)$
$\alpha =\frac{-\omega }{r}\times \left(\frac{-vb}{2\pi r}\right)$
$=\frac{v^{2}b}{2\pi r^3}$ [$\because v=wr$]