1
$\begingroup$

A wheel of radius $R$ spins about its center with a centripetal acceleration of $v^2/R$. I get that the acceleration at all points on the rim of the wheel point towards the center of the wheel. But, what happens when the acceleration of the wheel is slowing down? Let's say the wheel is spinning clockwise.

So, at the top, in the case where the wheel is not spinning, there is only one acceleration vector (pointing down towards the center of the wheel). When the wheel starts to slow down, I would guess that the centripetal acceleration remains at $v^2/R$ but now there's another component - the acceleration vector that points to the left, tangent to the circle, and perpendicular to the centripetal acceleration. So... ←↓

Adding these up gives me the new acceleration, right? How does this relate to the new period of the wheel though? Since the wheel is slowing down, I cannot just use $T = 2 \times \pi \times R / v$... Or can I?

Improve this question
$\endgroup$
2
  • $\begingroup$ How is the period a useful concept for a monotonically increasing or decreasing tangential speed? $\endgroup$
    – Bill N
    Commented Feb 6, 2015 at 16:31
  • 1
    $\begingroup$ "...where the wheel is not spinning." Do you mean where the wheel is not changing speed? $\endgroup$
    – Bill N
    Commented Feb 6, 2015 at 16:35

1 Answer 1

Reset to default
0
$\begingroup$

If the wheel is undergoing some sort of angular acceleration, then the magnitude of its velocity - its speed - would be expressed as a function of time, $v\equiv v(t)$.

So what does this mean for the period of rotation? That means the period also becomes a function of time. However, since the period still represents the time required for the wheel to rotate through $2\pi$ radians, the equation won't change.

$$T(t)=\frac{2\pi R}{v(t)}$$

An alternate way of thinking about this is that the period is defined as the inverse of the frequency, $$T=\frac{1}{f}$$ And we can also define the angular frequency as $\omega=2\pi f$ to find that $$T=\frac{2\pi}{\omega}$$

However, it's clear that the angular frequency must be the same as the angular velocity, which is $\omega=\frac{v}{R}$, thus we get the original equation for period as a function of speed and when $v$ changes with time, the equation for period must remain the same but become a function of time as well.

If you want to know what the new period is at any time, you must figure out the function $v(t)$ and substitute that into the equation for $T(t)$.

For example, if the speed is changing such that $v(t)=v_0-at$, then the equation for the period would become $$T(t)=\frac{2\pi R}{v_0-at}$$

Improve this answer
$\endgroup$

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