Angular Velocity Question (1 Viewer)

[echelon4]

A point is moving in a circle. Prove that it's angular velocity about the centre of the circle is double it's angular velocity about any fixed point on the circumference.

THanks in advanced

 

[SeDaTeD]

Use the fact that the angle subtended by an arc to the centre is twice that of any angle subtended by the same arc to a point on the circumference, then use a limiting process.

 

[joesmith1975]

it would believe it but i was just looking at this now.

A+2B = Pi ( sum of triangle)

PS how do u make the gif as part of the post

 

[SoulSearcher]

Upload to website, then use [IMG ]URL[ /IMG] tags, without the spaces.

 

[joesmith1975]

im mean so u dont have to click a link and go to a new page but on the same post

 

[SoulSearcher]

That's how, load the image onto a website, then use the tags to show the image onto the post, as follows:


Or you could just quote my post and see how it works :uhhuh:

 

[SeDaTeD]

But we aren't looking at a traingle . Check your diagram properly.

 

[shsshs]

hi i have a question

so if you have a particle moving in circular motion

how do you find the accelerations of the particle

say u have x = cos@ , y = sin@

you differentiate wrt t twice,

but then how do u find acceleration?

sometimes i see pythagoras used, but sumtyms i see horizontal and vertical accelerations resolved tangentially and normally

i dont know what to use when

 

[SeDaTeD]

You would differentiate wrt to t twice keeping in mind that @ is a function of t, thus you would need the chain rule. There's different ways to get it though, you should end up with the component of acceleration perpendicular to the motion being r*omega^2 towards the centre of the circle and the component tangental to the motion being equal to r*d(omega)/dt .

 

[shsshs]

but after you get

d^2 x/dt^2 and d^2 y/dt^2

why dont you do

sqrt ( [d^2 x /d t^2]^2 + [d^2 y + d t^2]) to find acceleration i.e. pythagoras

 

[SeDaTeD]

By doing so you are only getting a scalar quantity, acceleration is a vector quantity. You must convert d^2 x/dt^2 and d^2 y/dt^2 into components tangental to the motion and perpendicular the motion. The direction of acceleration is not always towards the centre of the circle, it's only ("only" being used loosely) true if you are dealing with uniform circular motion.

What you get by applying pythagoras is the magnitude of the net acceleration, which is not very useful for solving problems.

 

[pesila]

A point is moving in a circle. Prove that it's angular velocity about the centre of the circle is double it's angular velocity about any fixed point on the circumference.

THanks in advanced

For the original question, call theta the angle subtended at the centre between the point and a fixed point on the circumference. Call phi the angle subtended at the circumference between the point and a fixed point on the circumference. theta = 2 phi
d (theta)/dt = 2 d(phi)/dt
This represents that the angular velocity at the centre is twice the angular velocity at the circumference.