blazinfury Posted March 3, 2013 Posted March 3, 2013 When an object undergoes centripetal motion, it's centripetal velocity is always changing direction and thus it's centripetal acceleration is non-zero but a constant value. My quest is about tangential velocity and acceleration. As an object spins, it's tangential velocity is always changing due to direction and in some cases magnitude (ex. roller coaster). So this would mean that tangential acceleration is also non-zero but a constant value, correct?
elfmotat Posted March 3, 2013 Posted March 3, 2013 The tangential velocity is changing due to the centripetal acceleration. The acceleration is always perpendicular to velocity, which means that the speed of the object remains constant. If you add a nonzero tangential acceleration, the speed of the rotating object will increase.
swansont Posted March 3, 2013 Posted March 3, 2013 It may help to view the system in cylindrical or spherical coordinates. The position is described with a radial value and an angle. The centripetal acceleration is along r, so it's a constant. Tangential speed is [math]r\dot{\theta}[/math], and there is no tangential acceleration if [math]\dot{\theta}[/math] is constant
blazinfury Posted March 3, 2013 Author Posted March 3, 2013 @Shaken, That makes sense, but my only qualm with that is that tangential speed can change and so can tangential direction. For instance, if you have a rollercoaster, the coaster slows down at the top and speeds up at the bottom. Isn't that indicative of tangential velocity changes due to both speed and direction changes?
swansont Posted March 3, 2013 Posted March 3, 2013 @Shaken, That makes sense, but my only qualm with that is that tangential speed can change and so can tangential direction. For instance, if you have a rollercoaster, the coaster slows down at the top and speeds up at the bottom. Isn't that indicative of tangential velocity changes due to both speed and direction changes? (user names are on the top left) A loop-the-loop has both a radial and tangential force, so you're combining the effects.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now