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:
and
where
and
are the velocity and angular velocity at the point when sliding stops respectively. Also
.
Hence,
can be found in terms of
and
as
.
This equation for the velocity can then be used with the equation of motion relating velocity to distance to find
, i.e.
so then
![\color[rgb]{0,0,1}d_1 = \frac{a}{(a-r\ddot{\theta})^2} \frac{r^2 \omega_{0}^{2}}{2}](http://latex.codecogs.com/png.latex?\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
and
and substitute in the above equation. These two can now be found by resolving forces in the diagram.
Friction force is given by
Resolving perpendicular to the plane we have:
.
Resolving along the plane
.
From these three equations we find that
![\color[rgb]{0,0,1} a = (\mu \cos \alpha - \sin \alpha) g](http://latex.codecogs.com/png.latex?\color[rgb]{0,0,1} a = (\mu \cos \alpha - \sin \alpha) g)
.
Finally the torque is given by:
where
![\color[rgb]{0,0,1}I = \frac{1}{2} m r^2](http://latex.codecogs.com/png.latex?\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}](http://latex.codecogs.com/png.latex?\color[rgb]{0,0,1} \ddot{\theta} = - \frac{2 \mu g \cos \alpha}{r})
.
So that gives you
easily enough!
For the second part you can just use the conservation of energy to work out the further distance travelled,
, 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](http://latex.codecogs.com/png.latex?\color[rgb]{0,0,1}mg \, d_2 \, \sin \alpha = \frac{1}{2} I \omega^2 +\frac{1}{2} m v^2)
then
.
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and radius
is given an initial angular velocity
and no initial translational velocity. It is placed in contact with a plane inclined at an angle
to the horizontal. If there is a coefficient of friction
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
