Can't figure out how v=r*omega is derived

[slimnexus]

Homework Statement

Question comes from this book http://www.anselm.edu/internet/physics/cbphysics/index.html

In the section on rotational energy it gives the formula K$_{rolling}$ = K$_{translational}$+K$_{rotational}$ which it then expands to K$_{rolling}$ = $\frac{1}{2}$mv$^{2}$ + $\frac{1}{2}$Iω$^{2}$. Then it states that when an object that is rolling without slipping completes one rotation it travels a distance equal to its circumference so far so good. But then it says to divide both sides of the equation by 2πr, and on the left side we should have the speed of the object and on the right side its angular velocity. So I should end up with v = rω. But the author doesn't show any intermediate steps and I am not following this, I don't want to move on until I understand what's going on here.

So my question is how is v = rω derived from the earlier stated equation. Intermediate steps would be much appreciated. Thanks for any help you can provide.

 

[gneill]

Homework Statement

Question comes from this book http://www.anselm.edu/internet/physics/cbphysics/index.html

In the section on rotational energy it gives the formula K$_{rolling}$ = K$_{translational}$+K$_{rotational}$ which it then expands to K$_{rolling}$ = $\frac{1}{2}$mv$^{2}$ + $\frac{1}{2}$Iω$^{2}$. Then it states that when an object that is rolling without slipping completes one rotation it travels a distance equal to its circumference so far so good. But then it says to divide both sides of the equation by 2πr, and on the left side we should have the speed of the object and on the right side its angular velocity. So I should end up with v = rω. But the author doesn't show any intermediate steps and I am not following this, I don't want to move on until I understand what's going on here.

So my question is how is v = rω derived from the earlier stated equation. Intermediate steps would be much appreciated. Thanks for any help you can provide.

I'm not sure what the stated equation is supposed to be, but here's one way to arrive at where you want to go:

Suppose that the period for one revolution is T, so that in time T the distance traveled is one circumference. Thus:

$d = 2\pi\;r$ and dividing both sides by T gives $\frac{d}{T} = \frac{2\pi}{T}r$

On the left we see d/T which is the speed. On the right, we see $2\pi/T$ which is just ω in disguise...

 

[slimnexus]

I'm not sure what the stated equation is supposed to be, but here's one way to arrive at where you want to go:

Suppose that the period for one revolution is T, so that in time T the distance traveled is one circumference. Thus:

$d = 2\pi\;r$ and dividing both sides by T gives $\frac{d}{T} = \frac{2\pi}{T}r$

On the left we see d/T which is the speed. On the right, we see $2\pi/T$ which is just ω in disguise...

In the equations in my original post K is kinetic energy. So in words "Rolling kinetic energy is equal to translational kinetic energy plus rotational kinetic energy".

I can understand the method you just showed, that works out pretty simple. But I still don't see how the equations in the original post can be simplified to the same thing. Maybe someone could look at the book (it's free), it might make more sense than the way I worded it here.

Thanks for the help!

 

[rollcast]

From the textbook:

Distance traveled in one rotation = 2π r

Now if we divide both sides of this equation by the amount of time that it takes for the object to complete one rotation we obtain on the left, the speed of the object and, on the right, we can interpret the 2π as 2π radians and, since 2π radians is one rotation the 2π radians divided by the time it takes for the object to complete one rotation is just the magnitude of the angular velocity ω. Hence we arrive at

v =ωr

I took a look at the textbook and IMO he's now referring to another separate formula, c = πd or in his form, c = 2πr, where c is the circumference of the circle. The is means that c is the distance the wheel will travel in one rotation.

 

[slimnexus]

From the textbook:
I took a look at the textbook and IMO he's now referring to another separate formula, c = πd or in his form, c = 2πr, where c is the circumference of the circle. The is means that c is the distance the wheel will travel in one rotation.

I think I get it now. He's now referring to the formula distance = 2πr when he talks about dividing by time. Thanks alot! Now that I understand that I will be able to continue.