# [SOLVED] Difficult rotational mechanics question?

• Apr 30th 2009, 08:52 PM
fardeen_gen
[SOLVED] Difficult rotational mechanics question?
A uniform circular cylinder of mass $m$ and radius $r$ is given an initial angular velocity $\omega_{0}$ and no initial translational velocity. It is placed in contact with a plane inclined at an angle $\alpha$ to the horizontal. If there is a coefficient of friction $\mu$ for sliding between the cylinder and plane, find the distance the cylinder moves up before sliding stops. Also, calculate the maximum distance it travels up the plane. Assume $\mu > \tan\alpha$

Spoiler:
$d_{1} = \frac{r^2\omega_{0}^2(\mu \cos \alpha - \sin \alpha)}{2g(3\mu\cos \alpha - \sin \alpha)^2}$
$d_{max} = \frac{r^2\omega_{0}^2(\mu \cos \alpha - \sin \alpha)}{4g\sin \alpha(3\mu\cos \alpha - \sin \alpha)}$

• May 15th 2009, 08:20 AM
the_doc
This is actually quite simple to solve and here is how you do it:

Spoiler:

To begin with here's a diagram of the initial state of the cylinder:

For the duration of time when there is still sliding taking place then the friction force is at maximum and therefore the acceleration is a constant and so is the torque on the cylinder. This means the following equations of motion hold:

$\omega = \omega_0 + \ddot{\theta}t$ and $v = a t$

where $v$ and $\omega$ are the velocity and angular velocity at the point when sliding stops respectively. Also
$v = r \omega$.
Hence, $v$ can be found in terms of $\ddot{\theta}$ and $a$ as
$v = \frac{a}{a-r \ddot{\theta}} r \omega_0$ .

This equation for the velocity can then be used with the equation of motion relating velocity to distance to find $d_1$, i.e.
$v^2 = u^2 +2aS$ so then
$\color[rgb]{0,0,1}d_1 = \frac{a}{(a-r\ddot{\theta})^2} \frac{r^2 \omega_{0}^{2}}{2}$.

Now all you need to do is find $\ddot{\theta}$ and $a$ and substitute in the above equation. These two can now be found by resolving forces in the diagram.
Friction force is given by

$F_\mu = \mu R$

Resolving perpendicular to the plane we have:

$R = m g \cos \alpha$.

Resolving along the plane

$F_\mu - mg \sin \alpha = m a$ .

From these three equations we find that

$\color[rgb]{0,0,1} a = (\mu \cos \alpha - \sin \alpha) g$.

Finally the torque is given by:

$I \ddot{\theta} = -r F_\mu$ where $\color[rgb]{0,0,1}I = \frac{1}{2} m r^2$.

Hence

$\color[rgb]{0,0,1} \ddot{\theta} = - \frac{2 \mu g \cos \alpha}{r}$ .

So that gives you $d_1$ easily enough!

For the second part you can just use the conservation of energy to work out the further distance travelled, $d_2$, before stopping. So equating kinetic energy lost to the potential energy gained gives:
$\color[rgb]{0,0,1}mg \, d_2 \, \sin \alpha = \frac{1}{2} I \omega^2 +\frac{1}{2} m v^2$

then

$d_{max} = d_1 +d_2$ .