0
$\begingroup$

A particle is moving in a plane with constant radial velocity 12 m/s and constant angular velocity 2 rad/s. When a particle is at a distance r=8 m from origin what is the magnitude of instantaneous velocity?

I am familiar with angular velocity(ω). But what is meant by radial velocity? Is it the velocity of a body along the radius vector or that along the tangent drawn to the path traversed by the object which is usually perpendicular to radius vector? How can these three terms be related?

$\endgroup$
4
  • 1
    $\begingroup$ I would draw an image with some arrows (length corresponding to velocity) to get an idea on how to compute it. $\endgroup$
    – mic
    Nov 7 '19 at 15:50
  • $\begingroup$ Please send those pictures $\endgroup$
    – mig001
    Nov 7 '19 at 15:57
  • 3
    $\begingroup$ By radial velocity I assume they mean the velocity along the radius vector. $\endgroup$
    – NDewolf
    Nov 7 '19 at 16:09
  • $\begingroup$ I have removed the part asking for a solution, as that is not the intent of this site. $\endgroup$ Nov 7 '19 at 19:27
2
$\begingroup$

The line element in polar coordinates is $\vec{ds}=dr\hat{r}+rd\theta\hat{\theta}$.

Divide by dt, you get $\frac{\vec{ds}}{dt}=\frac{dr}{dt}\hat{r}+r\frac{d\theta}{dt}\hat{\theta}$

We are given that $\frac{dr}{dt}=12m/s$ and $\frac{d\theta}{dt}=2$rad/s. We multiply by 8 m to get $r\frac{d\theta}{dt}=16$m/s.

The instantaneous velocity is the vector sum of the two. The magnitude of the instantaneous velocity is the square root of the sum of the square of the components, 20 m/s.

$\endgroup$
0
$\begingroup$

Yes,the radial velocity is parallel to the radius. If you are moving from an inner region to an outer region of a rotating merry-go-round you have a radial velocity. The tangential velocity would be R*w. You combine these two components to get the velocity vector.

$\endgroup$
2
  • 1
    $\begingroup$ Radial velocity isn't just parallel to the radius, a radial velocity vector must point from the object to the center of rotation (or vice versa). A velocity vector that's parallel to the radius but not pointing through the center of rotation has an angular component to it. Being parallel to the radius is a necessary condition for a velocity vector to be a radial velocity vector, but it's not sufficient by itself. $\endgroup$ Nov 7 '19 at 19:57
  • $\begingroup$ If you measure the radius from the center of rotation to the object in question, then a velocity component which points toward the center is parallel to the radius. $\endgroup$
    – R.W. Bird
    Nov 8 '19 at 14:53
0
$\begingroup$

As R. Bird pointed out, you are asked to make a vectorial sum of components of the velocity vector. For movement on a plane, every vector has two components and, in this case, this two are the Radial and angular ones, who are perpendicular (better said orthogonal) to each other.

If draw your coordinate system -and imagine yourself at the origin of it-, the radial velocity component $v_r$ corresponds to how much the particle moves farther directly from where you are, while the angular velocity $\omega$ measures how many times the particles goes in circles around you; the bigger the radius $R$ of the circle it goes, the more velocity it has, so the component of tangential velocity $v_t$ is related to the angular velocity by $v_t=R \omega$.

So, how do you add this velocities who have directions $\hat{r},\hat{\theta}$ to get the magnitude of this instantaneous velocity vector is something you should find out easily.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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