establish a relation between linear velocity and angular velocity
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v = omega × r is the relation btw them
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Originally Answered: Intuitively, why is linear velocity equal to angular velocity times the radius?
I think it's probably easier to flip the question, and find why angular velocity is equal to the linear velocity divided by the radius:
ω=vrω=vr
From there, it's a simple algebraic manipulation. A much better term for linear velocity here would be perpendicular velocity, but I'll use your term hoping it matches up with your class content. :)
The first thing to recognise is that angular velocity is defined a little differently to linear velocity. With linear velocity, we have:
v=dxdtv=dxdt
This is a change in, say, meters per second. So we have some actual measure of distance present in the formula. With angular velocity, though, we have:
ω=dϕdtω=dϕdt
This is a change in angle per second which is a sort of relative measurement. If we want to relate linear and angular velocity, we need some way of 'converting' that angle into an actual distance. In fact, one way of intuitively looking at it would be that multiplying the angular velocity by the radius 'converts' the angles into a proper distance, making it a linear velocity.
Let's examine that a little more rigorously using this neat picture from the Angular velocity entry on Wikipedia:

We're interested in changes in the angle ϕϕ. Here, v∥v∥ and v⊥v⊥ are the components of the velocity of the particle whose angular motion we want to describe.
First, let's recognise that v∥v∥ won't be in this conversion factor, since it can't be responsible for a change in ϕϕ. No matter how tightly you pull a string toward you, it won't undergo circular motion; for that, you need a 'sideways' push given by v⊥v⊥ - the linear velocity you mention! This is therefore the only velocity that can appear in our formula later.
(see why perpendicular velocity would be a better name?)
Now let's set about deriving a specific formula by thinking about some particle completing a full circle (since our formula should hold no matter how far it travels). It will have completed an arc of 2π2π radians over some period of time, so we get:
dϕdt=2πdϕdt=2π
It's also traveled completely around the circumference of a circle, with length 2πr2πr, in that same period of time. Now, we know that its velocity around this circle must be v⊥v⊥ from our prior argument, so we get:
v⊥=2πrv⊥=2πr
And from there, it's easy to divide both sides by rr and form the equality:
ω=dϕdt=v⊥rω=dϕdt=v⊥r
Which rearranges nicely to answer your question! :) So now you've seen the math, let's run through our argument again to summarise the intuition:
We want to somehow express ωω (a change in angle per time) as a 'regular' velocity (a change in distance per time)
This must be your linear velocity (however bizarre that seems!), since anything else could not induce a change in angle
For a full circle, our total change in angle is 2π2π, and our total change in distance is 2πr2πr
If we took some time dtdt to traverse that circle, our angular and linear velocities would be ω=2πω=2π and v⊥=2πrv⊥=2πr
We therefore get ω=v⊥rω=v⊥r
I think the hardest step of all this is recognising that the perpendicular velocity is what carries you around the circle. It's very easy to fall into the trap of thinking it's "straight line" velocity from looking at that diagram.
Really, though, it's defined as the component of velocity perpendicular to the line segment formed between 'you' and the point you're rotating around. If you have angular velocity, this line segment is changing orientation, and therefore so is the perpendicular velocity, so it forms a curved path instead.
I think it's probably easier to flip the question, and find why angular velocity is equal to the linear velocity divided by the radius:
ω=vrω=vr
From there, it's a simple algebraic manipulation. A much better term for linear velocity here would be perpendicular velocity, but I'll use your term hoping it matches up with your class content. :)
The first thing to recognise is that angular velocity is defined a little differently to linear velocity. With linear velocity, we have:
v=dxdtv=dxdt
This is a change in, say, meters per second. So we have some actual measure of distance present in the formula. With angular velocity, though, we have:
ω=dϕdtω=dϕdt
This is a change in angle per second which is a sort of relative measurement. If we want to relate linear and angular velocity, we need some way of 'converting' that angle into an actual distance. In fact, one way of intuitively looking at it would be that multiplying the angular velocity by the radius 'converts' the angles into a proper distance, making it a linear velocity.
Let's examine that a little more rigorously using this neat picture from the Angular velocity entry on Wikipedia:

We're interested in changes in the angle ϕϕ. Here, v∥v∥ and v⊥v⊥ are the components of the velocity of the particle whose angular motion we want to describe.
First, let's recognise that v∥v∥ won't be in this conversion factor, since it can't be responsible for a change in ϕϕ. No matter how tightly you pull a string toward you, it won't undergo circular motion; for that, you need a 'sideways' push given by v⊥v⊥ - the linear velocity you mention! This is therefore the only velocity that can appear in our formula later.
(see why perpendicular velocity would be a better name?)
Now let's set about deriving a specific formula by thinking about some particle completing a full circle (since our formula should hold no matter how far it travels). It will have completed an arc of 2π2π radians over some period of time, so we get:
dϕdt=2πdϕdt=2π
It's also traveled completely around the circumference of a circle, with length 2πr2πr, in that same period of time. Now, we know that its velocity around this circle must be v⊥v⊥ from our prior argument, so we get:
v⊥=2πrv⊥=2πr
And from there, it's easy to divide both sides by rr and form the equality:
ω=dϕdt=v⊥rω=dϕdt=v⊥r
Which rearranges nicely to answer your question! :) So now you've seen the math, let's run through our argument again to summarise the intuition:
We want to somehow express ωω (a change in angle per time) as a 'regular' velocity (a change in distance per time)
This must be your linear velocity (however bizarre that seems!), since anything else could not induce a change in angle
For a full circle, our total change in angle is 2π2π, and our total change in distance is 2πr2πr
If we took some time dtdt to traverse that circle, our angular and linear velocities would be ω=2πω=2π and v⊥=2πrv⊥=2πr
We therefore get ω=v⊥rω=v⊥r
I think the hardest step of all this is recognising that the perpendicular velocity is what carries you around the circle. It's very easy to fall into the trap of thinking it's "straight line" velocity from looking at that diagram.
Really, though, it's defined as the component of velocity perpendicular to the line segment formed between 'you' and the point you're rotating around. If you have angular velocity, this line segment is changing orientation, and therefore so is the perpendicular velocity, so it forms a curved path instead.
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