Deriving to find speed (angular motion)

[Erenjaeger]

Homework Statement

Going through a lecture recording from a lecture I missed the lecturer says "to get the relationship between translational speeds and angular velocities we will take a differential" what is that relationship she is talking about? does she mean converting between the two? The derivative was,
v=ds/dt = d(rθ)/dt
v = r d(θ)/dt ← the r was pulled out the front here because this was assuming an unchanging radius.
therefore v = rω
i understand that rω comes about because the d(θ)/dt was defined as the angular velocity, whereas i thought it was just Δθ/Δt that was angular velocity? not the derivative of θ with respect to t, which is was d(θ)/dt is right?
Aslo would someone be able to show me how I would apply this to a question?

Homework Equations

v=ds/dt = d(rθ)/dt
v = r d(θ)/dt
v = rω

The Attempt at a Solution

Would applying that last part, v=rω does that just mean to find the velocity (as a scalar) it is just the product of the radius and angular velocity ??

 

[PeroK]

What you are asking is the relationship between speed and angular velocity for circular motion.

Angular velocity, like linear velocity, can be analysed using small finite changes in time or by continuos changes in time. In the first case, you have a small, finite time interval $\Delta t$ and a small, finite change of angle $\Delta \theta$. The (average) angular velocity during this interval is:
$\omega = \Delta \theta / \Delta t$.

You could think of this as the measurement taken during an experiment.

Theoretically, however, you can also imagine the angle to be changing continuously with time, in which case $\theta$ is a function of time and you can get an instantaneous angular velocity by differentiating this function:

$\omega = \frac{d\theta}{dt}$

In both cases if $r$ is constant, then the linear speed is given by:

$v = r\omega$

 

[Erenjaeger]

What you are asking is the relationship between speed and angular velocity for circular motion.

Angular velocity, like linear velocity, can be analysed using small finite changes in time or by continuos changes in time. In the first case, you have a small, finite time interval $\Delta t$ and a small, finite change of angle $\Delta \theta$. The (average) angular velocity during this interval is:
$\omega = \Delta \theta / \Delta t$.

You could think of this as the measurement taken during an experiment.

Theoretically, however, you can also imagine the angle to be changing continuously with time, in which case $\theta$ is a function of time and you can get an instantaneous angular velocity by differentiating this function:

$\omega = \frac{d\theta}{dt}$

In both cases if $r$ is constant, then the linear speed is given by:

$v = r\omega$

oh right so the derivative is just for instantaneous angular velocity and then delta theta over delta time is the average, makes sense.
for v=rω does it matter how you work out the value ω or do you specifically have to either use average angular or instantaneous velocity ??
thanks.

 

[PeroK]

oh right so the derivative is just for instantaneous angular velocity and then delta theta over delta time is the average, makes sense.
for v=rω does it matter how you work out the value ω or do you specifically have to either use average angular or instantaneous velocity ??
thanks.

In one case you get the average speed over the finite time interval. In the other case, you get an instantaneous speed as a function of time.