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 $\mu_\text{k}=0.2$
Exercise
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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.
$\mu_\text{k}\text{mg}=\text{ma}$ 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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