Question
Asolid disc and a ring, both of radius 10cm are placed on a horizontal table simultaneously, with initial angular speed equal to $10\pi\  rad\  s^{-1}?.$ Which of the two will start to roll earlier? The coefficient of kinetic friction is

Answer

As motion starts due to friction, F = ma $\Rightarrow\mu_{\text{k}}\text{mg}=\text{ma}$
$\Rightarrow\text{a}=\mu_{\text{k}}\text{g}$ Torque due to friction $\tau=\text{I}\alpha$ $\Rightarrow\mu_{\text{k}}\text{mgR}=\text{I}\alpha$
$\Rightarrow\alpha=\frac{\mu_{\text{k}}\text{mgR}}{\text{I}}$ Rolling begins, when $\text{v}=\omega\text{R}$But $\text{v}=0+\alpha\text{t}$
$\Rightarrow\text{v}=\mu_{\text{k}}\text{gt}$
and $\omega=\omega_0+\alpha\text{t}$
$\omega=\omega_0-\frac{\mu_{\text{k}}\text{mgR}}{\text{I}}\text{t}$
$\Rightarrow\frac{\text{v}}{\text{R}}=\omega_0\frac{\mu_{\text{k}}\text{mgR}}{\text{I}}\text{t}$
$\Rightarrow\text{t}=\frac{\text{R}\omega_0}{\mu_{\text{k}}\text{g}\Big(\frac{\text{I}+\text{mR}^2}{\text{I}}\Big)}$ For disc, $\text{I}=\text{mR}^2$
$\Rightarrow\text{t}=0.53\text{s}$ For ring, $\text{I}=\text{mR}^2$ $\Rightarrow\text{t}=0.80\text{s}$Thus, disc begins to roll earlier.

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