A uniform disc of radius R, is resting on a table on its rim.The coefficient of friction between disc and table is $\mu.$Now the disc is pulled with a force F as shown in the figure. What is the maximum value of F for which the disc rolls without slipping?
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Let 'a', $\alpha$ be the linear and angular acceleration respectively. For linear motion, F - f = Ma ....(i)
Where M is mass of disc. Force of friction (f) applies torque about centre O. But torque due to F is along 'O'. $\therefore$ Torque to disc $\tau=\text{I}_\text{D}\alpha$ $\because$ M.I. of disc is $\text{I}_\text{D}=\frac{1}{2}\text{MR}^2$ $\text{f.R}=\frac{1}{2}\text{MR}^2\cdot\frac{\text{a}}{\text{R}}\ \because\ \text{a}=\text{Ra}$ $\text{f.R}=\frac{1}{2}\text{MRa}$ $\text{Ma}=2\text{f}\ ....(\text{ii})$ $\text{F}-\text{f}=2\text{f}$ $3\text{f}=\text{F}$ Or $\text{f}=\frac{\text{F}}{3}\ \because\ \text{N}=\text{Mg}$ $\mu.\text{M}=\frac{\text{F}}{3}$ $\text{F}=3\mu\text{ Mg}$ is the maximum force applied on disc to roll on surface without slipping.
Where M is mass of disc. Force of friction (f) applies torque about centre O. But torque due to F is along 'O'. $\therefore$ Torque to disc $\tau=\text{I}_\text{D}\alpha$ $\because$ M.I. of disc is $\text{I}_\text{D}=\frac{1}{2}\text{MR}^2$ $\text{f.R}=\frac{1}{2}\text{MR}^2\cdot\frac{\text{a}}{\text{R}}\ \because\ \text{a}=\text{Ra}$ $\text{f.R}=\frac{1}{2}\text{MRa}$ $\text{Ma}=2\text{f}\ ....(\text{ii})$ $\text{F}-\text{f}=2\text{f}$ $3\text{f}=\text{F}$ Or $\text{f}=\frac{\text{F}}{3}\ \because\ \text{N}=\text{Mg}$ $\mu.\text{M}=\frac{\text{F}}{3}$ $\text{F}=3\mu\text{ Mg}$ is the maximum force applied on disc to roll on surface without slipping.
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