Cylinder rolling down an incline without slipping

Tuesday, 10 November 2009 09:15 Bruno Lebon

After attempting simple examples of motions, we will now turn to a constrained problem: a cylinder of radius $r$ , mass $m$ and moment of inertia $I$ about its axis, rolling on an incline of length $Z$ without slipping. $\theta$ is the angular displacement of the cylinder and the angular velocity is given by $\omega$ .

The Lagrangian for this system is

$L = \frac{1}{2}m\dot{x}^{2} + \frac{1}{2}I\omega^{2} - mg\left(Z - x + r\right)\sin\alpha$

The two generalised coordinates $x$ and $\theta$ are related by

$f = x - r\theta = 0$

For this constrained system, we obtain the following equations of motion:

$m\ddot{x} - mg\sin\alpha = \lambda$

$I\ddot{\theta} = -\lambda r$

Differentiating $f$ twice with respect to time gives the relationship

$\ddot{x} = r\ddot{\theta}$

Therefore,

$-\frac{mr^{2}}{I}\lambda - mg\sin\alpha = \lambda$

Noting that the moment of inertia of a cylinder about its axis is given by

$I = \frac{1}{2}mr^{2}$

$\lambda = \frac{mg\sin\alpha}{3}$

Therefore the translational acceleration of the cylinder becomes

$\ddot{x} = \frac{2}{3}g\sin\alpha$

Note that this relationship applies only when the cylinder rolls without slipping. If the cylinder starts slipping, the relationship $f = 0$ does not hold true anymore.