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Systems of Particles and Rotational Motion — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsSystems of Particles and Rotational Motion5 Marks
Question
A solid disc and a ring, both of radius 10cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The co-efficient of kinetic friction is $\mu_\text{k}=0.2$
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Answer

Given,
Radii of the ring and the disc, r = 5cm = 0.05m
Initial angular speed, $\omega_0=8\pi\text{rads}^{-1}$
Coefficient of kinetic friction, $\mu_\text{k}=0.2$
Initial velocity of both the objects, u = 0a
Motion of the two objects is caused by force of friction. According Newton’s second,
force of friction, $\text{f}=\text{ma}$
$\mu_\text{k}\text{mg}=\text{ma}$
Where,
a = Acceleration produced in the disc and the ring
m = Mass
$\therefore\ \text{a}=\mu_{\text{k}}\text{g}\ ...(\text{i})$
Using the first equation of motion,
$\text{v}=\text{u}+\text{at}$
$=0+\mu_\text{k}\text{gt}$
$=\mu_\text{k}\text{gt}\ ...(\text{ii})$
The frictional force applies a torque in perpendicularly outward direction and reduces the initial angular speed.
Torque, $\text{T}=-\text{I}\alpha$
Where, $\alpha=$ Angular acceleration
$\mu_\text{k}\text{mgr}=-\text{I}\alpha$
$\therefore\alpha=\frac{-\mu_\text{k}\text{mgr}}{\text{I}}\ ...(\text{iii})$
According to the first equation of rotational motion, we have,
$\omega=\omega_0+\alpha\text{t}$
$=\omega_0+\Big(\frac{-\mu_\text{k}\text{mgr}}{\text{I}}\Big)\text{t}\ ...(\text{iv})$
Rolling starts when linear velocity, $\text{v}=\text{r}\omega$
$\therefore\ \text{v}=\text{r}\Big(\omega_0-\frac{\mu_\text{k}\text{mgrt}}{\text{I}}\Big)\ ...(\text{v})$
Using equation (ii) and equation (v), we have,
$\mu_\text{k}\text{gt}=\text{r}\Big(\omega_0-\frac{\mu_\text{k}\text{mgrt}}{\text{I}}\Big)$
$=\text{r}\omega_0-\frac{\mu_\text{k}\text{mgr}^2\text{t}}{\text{I}}\ ....(\text{vi})$
For the ring,
$\text{I}=\text{mr}^2$
$\therefore\ \mu_\text{k}\text{gt}=\text{r}\omega_0-\frac{\mu_\text{k}\text{mgr}^2\text{t}}{\text{mr}^2}$
$=\text{r}\omega_0-\mu_\text{k}\text{gt}$
$2\mu_\text{k}\text{gt}=\text{r}\omega_0$
$\therefore\ \text{t}=\frac{\text{r}\omega_0}{2\mu_\text{k}\text{g}}$
$=\frac{(0.05\times8\times3.14)}{(2\times0.2\times9.8)}=0.32\text{s}\ ...(\text{vii})$
For the disc, $\text{I}=\Big(\frac{1}{2}\Big)\text{mr}^2$
$\therefore\ \mu_\text{k}\text{gt}=\text{r}\omega_0-\frac{\mu_\text{k}\text{mgr}^2\text{t}}{\big(\frac{1}{2}\big)\text{mr}^2}$
$=\text{r}\omega_0-2\mu_\text{k}\text{gt}$
$3\mu_\text{k}\text{gt}=\text{r}\omega_0$
$\therefore\ \text{t}=\frac{\text{r}\omega_0}{3\mu_\text{k}\text{g}}$
$=\frac{(0.05)\times8\times3.14}{(3\times0.2\times9.8)}=0.213\text{s}\ ...(\text{viii})$
Since $\text{t}_\text{D}>\text{t}_\text{R},$ the disc will start rolling before the ring.

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