Angular displacement, Angular velocity

[Subliminal]

Angular displacement times r = |Linear displacement| (Intuitively I understand this but i can't put it into words, I know r*theata = L and if L = 0 then L after t seconds is the size of linear displacement t seconds after the object is released from the centripetal force)

Angular displacement over time = |Linear velocity| (as the size of the velocity is constant and acceleration only changes the direction of the object)

 

v = r*w, I would like to know what this means, when looking at a situation, what is the significance of it, can I express angular velocity in terms of angular velocity?

 

Is, that the tangential velocity of an orbiting object is equal to distance traveled by the object over the time taken?

 

One more thing, I learned that tangential acceleration is related to angular acceleration in the following way.

 

a = r*alpha

 

I'm not sure what this means as I thought that linear (tangential) velocity had to be constant for v = r*w to be true, so how can tangential acceleration (which would break the current circular path)

 

Also expressing centripetal acceleration in terms of angular speed, I know how to do it, but i don't know what it means, it's significance.

 

 

Also this looks quite close to the derivation of Kepler's third law, i just did this while messing around with equations, it probably means nothing, who knows.

 

a_c = v²/r

w = v/r (linear - angular velocity rearranged)

a_c= (v/r)²/r

a_c= v²/r³

Edited November 30, 2013 by Subliminal

[Endercreeper01]

v=r*w means that the velocity is the radius times the angular velocity. The angular velocity is a measure of how much the angle is changing (in radians) per unit time, and it is 2pi*f, where f is the frequency. It can also be written as the derivative of the angle with respect to time, or da/dt, where a is the angle and t is time

And yes, the tangental velocity would be s/t, where s is distance, but when you have accelerations, you would use the derivative of distance with respect to time, or ds/dt for velocity.

[Subliminal]

v=r*w means that the velocity is the radius times the angular velocity. The angular velocity is a measure of how much the angle is changing (in radians) per unit time, and it is 2pi*f, where f is the frequency. It can also be written as the derivative of the angle with respect to time, or da/dt, where a is the angle and t is time

And yes, the tangential velocity would be s/t, where s is distance, but when you have accelerations, you would use the derivative of distance with respect to time, or ds/dt for velocity.

 

 

Is if the path of the object is a circle I can use the circle as some sort of vector diagram and calculate the magnitude of the velocity using angular velocity as some sort of ratio and the radius as a scaling instrument? s = r theta, v = r w, a = r alpha, that sort of thing? I understand intuitively why linear displacement can be found this way, but i only understand mathematically why the other quantities are found.

I suppose the further away a point is from the point of rotation the further it travels for a given change in theta, giving it a larger velocity than points which are closer, similarly a change in rate of ration would result in a bigger change in acceleration the magnitude of which would be determined by the points distance from the point of rotation.

As for tangential acceleration, does this simply mean a decreases in the magnitude of T?

[studiot]

 

Angular displacement times r = |Linear displacement|

 

Is, that the tangential velocity of an orbiting object is equal to distance traveled by the object over the time taken?

 

Yes but remember that the object is travelling in a curved path, so the distance travelled is along that curved path. It is not a straight line distance.

 

If the radius is constant the path is an arc of a circle.

 

If the angular velocity is constant, the angular acceleration is zero (just like its linear counterpart).

 

In that case we can calculate that the distance travelled = radius times angular velocity times time. (This is not like its linear counterpart since there are three terms in the equation for distance, not two).

 

If either the radius or the angular velocity varies we are into calculus, to obtain the distance travelled.

[Endercreeper01]

 

Is if the path of the object is a circle I can use the circle as some sort of vector diagram and calculate the magnitude of the velocity using angular velocity as some sort of ratio and the radius as a scaling instrument? s = r theta, v = r w, a = r alpha, that sort of thing? I understand intuitively why linear displacement can be found this way, but i only understand mathematically why the other quantities are found.

 

 

Yes, you can do that.

 

 

 

 

I suppose the further away a point is from the point of rotation the further it travels for a given change in theta, giving it a larger velocity than points which are closer, similarly a change in rate of ration would result in a bigger change in acceleration the magnitude of which would be determined by the points distance from the point of rotation.

 

That's true, because you have the arc length further away. You can imagine it as if the object was moving along the circumference of a circle. If it is further away from the circle (a bigger radius), it is traveling along a bigger length, since circumference is 2 pi * r, where r is radius. Because angular velocity is how much the angle change per unit time, this means that if you are further away and you have the same angular velocity as something else that is closer, you have a bigger velocity. This is because you are traveling across a bigger arc length in the same period of time. This means that velocity is the angular velocity times the radius v=wr

[decraig]

Angular displacement times r = |Linear displacement| ...

 

What if your angular displacement is 2[math]\pi[/math] ? The linear displacement is zero.

 

The equation should be about infinitesimals:

 

[math] \Delta \theta r = \Delta L[/math] where [math]\Delta L[/math] is small.

Edited December 1, 2013 by decraig

[Endercreeper01]

The angular displacement is (rcos(a), rsin(a)) where a is the angle displacement

[Subliminal]

What if your angular displacement is 2[math]\pi[/math] ? The linear displacement is zero.

 

The equation should be about infinitesimals:

 

[math] \Delta \theta r = \Delta L[/math] where [math]\Delta L[/math] is small.

It was incorrect of me to say that, what I meant was tangential velocity • time, that is the linear displacement it would have experience if it escaped the circular path.

 

I tried to use it to explain why tangential velocity equaled angular velocity ● radius before I knew it was cross multiplication occuring, I am still unsure if what kind of vector "r" represents, that was my original reason for thinking it was just a scaling device.

[studiot]

 

I tried to use it to explain why tangential velocity equaled angular velocity ● radius before I knew it was cross multiplication occuring, I am still unsure if what kind of vector "r" represents, that was my original reason for thinking it was just a scaling device

Please see my post#8 in your other thread http://www.scienceforums.net/topic/80386-is-there-angular-force-tangential-force-why-is-tangential-f-scaled-twice-for-torque/

Edited December 1, 2013 by studiot