The question is if a bug is moving from center to periphery of a rotating disk or lets say vice versa; there is a tangential friction acting on the bug which is producing some torque. The small doubt that I have here is that what if such a tangential friction never existed?? Would the bug simply lose contact and fly of tangentially or would there be some other kind of motion?

In a previous question of mine (Which force exerted a torque?), the explanation of the concept by one of the members, provoked me to think about this question.To summarize, the change in angular momentum of the disk was due to this tangential friction provided by bug. But, how can we conclude with strong statements that tangential friction does exist barring the fact that angular momentum changes??

When we walk on road, we apply friction in only one direction. But in a rotating frame, is it always necessary for the bug to apply both radial and tangential frictions(I know radial friction is absolutely necessary to provide centripetal force for rotation, and this fact isn't my doubt)?? What is the utility of tangential friction in moving forward (that is perpendicular to line of action of tangential friction) ??

It would be nice if someone explained the physical feel in perspective of the bug because that is what I seek for. By rotational dynamics as said earlier, it is easy to conclude that we need tangential friction to cause change in torque, but I wish to feel what the bug feels and what the bug has to do - its compulsions.

  • $\begingroup$ You should not think of tangential and radial friction as two different things, rather as one friction against the net force. The bug reacts to the net force at any given time. $\endgroup$ Commented Jan 15, 2021 at 16:19
  • $\begingroup$ Okay but then what is this net force here then?? Normal and mg cancel each other out, and there is possibly no other force which friction will act against.. $\endgroup$
    – Sam
    Commented Jan 15, 2021 at 16:32
  • $\begingroup$ Radial from centrifugal force, and when the bug's radius changes while rotating with the disk there is tangential force from conservation of angular momentum. $\endgroup$ Commented Jan 15, 2021 at 17:33
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    $\begingroup$ Looking over the previous anwer. I found that the $\omega (r)$ is not quite correct. Because during the moving outward of the bug, these two system, the bug and the disc, have relative motion. That didn't promise that the two system are still have a same angular velocity. My conlcusion is that they shouldn't have same $\omega$ during the period of relative motion. $\endgroup$
    – ytlu
    Commented Jan 16, 2021 at 14:20

1 Answer 1


In the rotation frame of the disc, the bug (mass $m$) position $\vec{r}'(t)$ relative to the disc. Note that $|\vec{r}'| = | \vec{r} | = r$, the disc frame differs from the prime system only in the angular velocity. The bug moving outward suffers the following forces:

  1. The kinetic frictional force, $ -\mu_k m g \hat{v}'$;
  2. the centrifugal force, $m \omega^2 \vec{r}'$;
  3. the Coriolis force, $-2 m \vec{\omega} \times \vec{v}'$;
  4. the variant $\omega$, $-m\frac{d\vec{\omega}}{dt} \times \vec{r}'$.

The equation for motion of $m$ \begin{equation} m \frac{d\vec{v'}}{dt} = -\mu_k m g \hat{v}' + m \omega^2 \vec{r}' - 2 m \vec{\omega} \times \vec{v}' - m\frac{d\vec{\omega}}{dt} \times \vec{r}'. \tag{1} \end{equation} Due to the Coriolis force, the trajectory of bug is not along the radial direction, but is a curve bended to the right (for counter-clockwise rotation) shown in Figure as the blue line.

 Trajectory of bug on the disc.

The change of angular momentum $\vec{r}' \times $ equ.(1), \begin{equation} m \frac{d ( \vec{r}' \times \vec{v}')}{dt} = -\mu_k m g \vec{r}' \times \hat{v}' - 2 m \vec{r}' \times \{ \vec{\omega} \times \vec{v}' \} - m \vec{r}' \times \{ \frac{d\vec{\omega}}{dt} \times \vec{r}' \}. \tag{2} \end{equation} The triple cross product can be expanded: $$ \vec{r}' \times \{ \vec{\omega} \times \vec{v}' \} = \vec{\omega} \{ \vec{r}' \cdot \vec{v}' \} - \vec{v}' \{ \vec{r}' \cdot \vec{\omega} \} = \vec{\omega} \{ \vec{r}' \cdot \vec{v}' \} $$ since $\vec{r}'$ and $\vec{\omega}$ are mutually vertical, and similarly, $$ \vec{r}' \times \{ \frac{d\vec{\omega}}{dt} \times \vec{r}' \} = \frac{d\vec{\omega}}{dt} r^2 - \vec{r}' \{ \vec{r}' \cdot \frac{d\vec{\omega}}{dt} \} = \frac{d\vec{\omega}}{dt} r^2 $$ Substitue these two into the equation of motion: \begin{equation} m \frac{d ( \vec{r}' \times \vec{v}') }{dt} = - m \mu_k g \vec{r}' \times \hat{v}' - 2 m \vec{\omega} \{ \vec{r}' \cdot \vec{v}' \} - m \frac{d\vec{\omega}}{dt} r^2. \tag{3} \end{equation} Note that $ m ( \vec{r}' \times \vec{v}') + m r^2 \omega =\vec{ L}_m$, the angular momentum of $m$ in Lab frame. \

The frictional force is also change the rotation of the disc (mass $M$, inertial moment $I_M$): \begin{equation} I_M \frac{d \vec{\omega} } {dt} = m \mu_k g \vec{r}' \times \hat{v}' \tag{4} \end{equation} Substitue Eq.(4) into Eq.(3): \begin{equation} I_M \frac{d \vec{\omega}}{dt} + m r^2 \frac{d \vec{\omega} }{dt} = - m \frac{d ( \vec{r}' \times \vec{v}')}{dt} - 2 m \vec{\omega} \{ \vec{r}' \cdot \vec{v}' \} \end{equation}

The left-hand-side of the above equation is the angular momentum assuming both disc and bug rotate in a same angular velocity $\omega$. Taking the Coriolis force into account, the trajectory is not in the radial direction, and therefore, the two term in the LHS is not constant of motion.


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