I'm trying to solve the following problem, but i have no idea how to begin.

A wheel of mass $M$, radius of gyration $k$, spins smoothly on a fixed horizontal axle of radius a which passes through a hole of slightly larger radius at the hub of the wheel. The coefficient of friction between the bearing surfaces is $\mu$. If the wheel is initially spinning with angular velocity $w_{0}$, find the time and the number of turns that it takes to stop.

Any help?

  • $\begingroup$ Think about what the normal force due to gravity will be at each point on the axle. $\endgroup$ – George G Apr 2 '14 at 14:03

The kinetic enegy at $t=0$ is equal to the friction work done when the wheel has stopped:
that is
$1/2J \omega_0^2=F_fs=mg \mu 2 \pi r n $ (with $J=mk^2$)
where $r$ is the radius of the bearing bore an n the number of revolutions. Solving for the revolutions gives:
$n= \frac{J \omega_0^2}{4 \pi m g \mu r}$
The angular acceleration due to friction is constant: $\dot{\omega}=\tau /J=\frac {F_f r}{J} =\frac{mg \mu r}{J}=const$
and therfore the the time to stop is
$ \omega =\omega_0 -\dot{\omega}t \rightarrow t=\omega_0/\dot{\omega}=\frac{J \omega_0}{ m g \mu r}$

  • $\begingroup$ $J$ is the moment of inertia? if so, why it is $J = M.k^2$? you have a disc and not a simple particle. And about this formula $\dot{\omega}=M/J=\frac {F_f r}{J} =\frac{mg \mu r}{J}=const$ you could explain why $M =F_f.r$ ? $\endgroup$ – Mike Apr 2 '14 at 16:56
  • $\begingroup$ @ExampleMo Yes $J$ is the moment of inertia. I used $m$ for mass and $M$ for torque, which might be confusing, sorry. Regarding gyradius I looked it up in en.wikipedia.org/wiki/Radius_of_gyration (applications in mechanics) $\endgroup$ – Mirc Breitschuh Apr 2 '14 at 19:53
  • $\begingroup$ you made a mistake @mirc-breitschuh. You have $\dot{\omega} = \frac{\tau}{J}$ where $\tau $ is the torque $\endgroup$ – Mike Apr 2 '14 at 22:04
  • $\begingroup$ @ExampleMo Thank you for editing. I put a minus sign in the last line to make it consistent. $\endgroup$ – Mirc Breitschuh Apr 5 '14 at 11:34

Not the answer you're looking for? Browse other questions tagged or ask your own question.