Discuss the rolling motion of a cylinder on an inclined plane.
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Consider a spherical body of mass M and radius r, rolling down on an inclined plane without slipping. Let the angle of inclination of the plane is $\theta$ with the horizontal (See fig.)It $\omega$ be the angular velocity of the body, then the linear velocity of the body is,
$\text{v} = \text{r} \omega\dots(1)$
The various forces acting on the body are:
- The weight (Mg) of the body in the vertically downward direction.
- Normal reaction (R) of the surface of the plane on the body which acts vertically upward.
- The force of friction (F) which acts opposite to the direction of motion of the body.
- $\text{Mg}\cos\theta$ is the horizontal component, which is equal and opposite to the normal reaction $\text{i.e.,}\text{ R}=\text{Mg}\cos\theta\dots(2)$
- $\text{Mg}\sin\theta$ is the vertical component, which acts in the direction of the motion of the body
The external torque acting on the body is produced by the force of friction (F) and its lever arm is r (the radius of the spherical body) [The lines of action of Mg and R pass through the centre of mass of body and hence do not contribute anything to the torque about the centre of mass]
$\therefore\tau=\text{Fr}\dots(4)$
If $\alpha$ be the angular acceleration produced in the body whose moment of inertia is I, then $\tau=\text{I}\alpha\dots(5)$
From eqns (4) and (5), we get
$\text{Fr}=\text{I}\alpha\text{ or }\text{F}=\text{I}\frac{\alpha}{\text{r}}\dots(6)$
Using eqn. (6) in eqn. (3), we get
$\therefore\text{Ma}=\text{Mg}\sin\theta-\text{I}\frac{\alpha}{\text{r}}$
But $\text{a}=\text{ra}\text{ or }\alpha=\frac{\text{a}}{\text{r}}$
$\therefore\text{Ma}=\text{Mg}\sin\theta-\text{I}\frac{\text{a}}{\text{r}^2}$
$\Rightarrow\text{a}=\text{g}\sin\theta-\frac{\text{I}_{\text{a}}}{\text{Mr}^2}$
$\text{or }\text{a}\Big(1+\frac{\text{I}}{\text{Mr}^2}\Big)=\text{g}\sin\theta$
$\text{ or }=\frac{\text{g}\sin\theta}{1+\Big(\frac{\text{I}}{\text{Mr}^2}\Big)}$
Now $\alpha=\frac{\text{a}}{\text{r}}=\frac{\text{g}\sin\theta}{\text{r}\Big[1+\frac{\text{I}}{\text{Mr}^2}\Big]}$
Substituting this value in eqn. (6), we get
$\text{F}=\frac{\text{I}}{\text{r}}\frac{\text{g}\sin\theta}{\Big[1+\frac{\text{I}}{\text{Mr}^2}\Big]}=\frac{\text{Ig}\sin\theta}{\text{r}^2\Big[1+\frac{\text{I}}{\text{Mr}^2}\Big]}$
This is the force of friction, required by the body to roll down an inclined plane without slipping.
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