Obtain an expression for linear acceleration of a cylinder rolling down an inclined plane and hence find the condition for the cylinder to roll down the inclined plane without slipping.
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When a cylinder rolls down on an inclined plane, then forces involved are,
From free body diagram, $\text{N}-\text{mg}\cos\theta=0$ $\text{or }\text{N}-\text{mg}\cos\theta$ Also, if a = acceleration of centre of mass down the plane, then $\text{F}_{\text{net}}=\text{ma}=\text{mg}\sin\theta-\text{f}\dots(1)$ As friction produces torque necessary for rotation, $\tau=\text{I}\alpha=\text{f R}$ $\Rightarrow\text{f}=\frac{\text{I}\alpha}{\text{R}}=\frac{\text{Ia}}{\text{R}^2}\Big[\therefore\alpha=\frac{\text{a}}{\text{R}}\Big]$ Substituting for f in Eq. (i), we get $\text{ma}=\text{mg}\sin\theta-\frac{\text{Ia}}{\text{R}^2}$ $\Rightarrow\text{a}=\text{g}\sin\theta-\frac{\text{Ia}}{\text{mR}^2}$ For cylinder, $\text{I}=\frac{1}{2}\text{mR}^2$ $\therefore\text{a}=\text{g}\sin\theta-\frac{\text{a}}{2}$ $\Rightarrow\frac{3\text{a}}{2}=\text{g}\sin\theta$ $\Rightarrow\text{a}=\frac{2\text{g}\sin\theta}{3}$ and from Eq. (i), the value of friction is $\text{f}=\text{mg}\sin\theta-\text{ma}$ $=\text{mg}\sin\theta-\frac{2}{3}\text{mg}\sin\theta$ $=\frac{1}{3}\text{mg}\sin\theta$ $\text{If }\mu_{\text{s}}=$ coefficient of static friction, then $\mu_{\text{s}}=\frac{\text{f}}{\text{N}}\text{ or }\mu_{\text{s}}=\frac{1}{3}\tan\theta$ $\therefore$ For perfect rolling, $\mu_{\text{s}}\geq\frac{1}{3}\tan\theta$
- Weight mg.
- Normal reaction N.
- Friction f.
From free body diagram, $\text{N}-\text{mg}\cos\theta=0$ $\text{or }\text{N}-\text{mg}\cos\theta$ Also, if a = acceleration of centre of mass down the plane, then $\text{F}_{\text{net}}=\text{ma}=\text{mg}\sin\theta-\text{f}\dots(1)$ As friction produces torque necessary for rotation, $\tau=\text{I}\alpha=\text{f R}$ $\Rightarrow\text{f}=\frac{\text{I}\alpha}{\text{R}}=\frac{\text{Ia}}{\text{R}^2}\Big[\therefore\alpha=\frac{\text{a}}{\text{R}}\Big]$ Substituting for f in Eq. (i), we get $\text{ma}=\text{mg}\sin\theta-\frac{\text{Ia}}{\text{R}^2}$ $\Rightarrow\text{a}=\text{g}\sin\theta-\frac{\text{Ia}}{\text{mR}^2}$ For cylinder, $\text{I}=\frac{1}{2}\text{mR}^2$ $\therefore\text{a}=\text{g}\sin\theta-\frac{\text{a}}{2}$ $\Rightarrow\frac{3\text{a}}{2}=\text{g}\sin\theta$ $\Rightarrow\text{a}=\frac{2\text{g}\sin\theta}{3}$ and from Eq. (i), the value of friction is $\text{f}=\text{mg}\sin\theta-\text{ma}$ $=\text{mg}\sin\theta-\frac{2}{3}\text{mg}\sin\theta$ $=\frac{1}{3}\text{mg}\sin\theta$ $\text{If }\mu_{\text{s}}=$ coefficient of static friction, then $\mu_{\text{s}}=\frac{\text{f}}{\text{N}}\text{ or }\mu_{\text{s}}=\frac{1}{3}\tan\theta$ $\therefore$ For perfect rolling, $\mu_{\text{s}}\geq\frac{1}{3}\tan\theta$
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