Worked example 8.7: Rotating cylinder

Question: A uniform cylinder of radius $b=0.25 {\rm m}$ is given an angular speed of $\omega_0=35 {\rm rad./s}$ about an axis, parallel to its length, which passes through its centre. The cylinder is gently lowered onto a horizontal frictional surface, and released. The coefficient of friction of the surface is $\mu=0.15$. How long does it take before the cylinder starts to roll without slipping? What distance does the cylinder travel between its release point and the point at which it commences to roll without slipping?

Answer: Let $v$ be the velocity of the cylinder's centre of mass, $\omega$ the cylinder's angular velocity, $f$ the frictional force exerted by the surface on the cylinder, $M$ the cylinder's mass, and $I$ the cylinder's moment of inertia. The cylinder's translational equation of motion is written

$$M \dot{v} = f.$$

Note that the friction force acts to accelerate the cylinder's translational motion. Likewise, the cylinder's rotational equation of motion takes the form

$$I \dot{\omega} = - f b,$$

since the perpendicular distance between the line of action of $f$ and the axis of rotation is the radius, $b$, of the cylinder. Note that the friction force acts to decelerate the cylinder's rotational motion. If the cylinder is slipping with respect to the surface, then the friction force, $f$, is equal to the coefficient of friction, $\mu$, times the normal reaction, $M g$, at the surface:

$$f = \mu M g.$$

Finally, the moment of inertia of the cylinder is

$$I = \frac{1}{2} M b^2.$$

The above equations can be solved to give

$$\dot{v} = \mu g,$$

$$b \dot{\omega} = - 2 \mu g.$$

Given that $v=0$ (i.e., the cylinder is initially at rest) and $\omega=\omega_0$ at time $t=0$, the above expressions can be integrated to give

$$v = \mu g t,$$

$$b \omega = b \omega_0 - 2 \mu g t,$$

which yields

$$v - b \omega = -(b \omega_0 - 3 \mu g t).$$

Now, the cylinder stops slipping as soon as the ``no slip'' condition,

$$v = b \omega,$$

is satisfied. This occurs when

$$t = \frac{b \omega_0}{3 \mu g}=\frac{0.25\times 35}{3\times 0.15\times 9.81} = 1.98 {\rm s}.$$

Whilst it is slipping, the cylinder travels a distance

$$x = \frac{1}{2} \mu g t^2 = 0.5\times 0.15 \times 9.81\times 1.98^2=2.88 {\rm m}.$$