Ok! Let me do one more example and tack it on as a part 2 to my Linear and Angular Speed
video. And we have sort of a bicycle set up here. This could also be rotation of an engine
and the final rotation of the back axel. There are all kinds of mechanical engineering ideas
with this little/simple chain and gear set up. We have got... Let's say this is a bicycle
set up and you are up here peddling the bike and you have a front sprocket with a chain
going around it. And, you're peddling is 25 rotations, or 25 revolutions, in a 30 second
period. This is equivalent to 50 revolutions per minute. If you are peddling 25 times...you
know doing 25 revolutions in a 30 second period, how many times is that back gear spinning
to turn that back wheel. I am not going to do another... well I am going to do a two
part example but we are not going to talk about the outside diameter of the tire. We
are just going to discuss how many times that back sprocket is turning if you peddling 25
rotations every 30 seconds for the front sprocket. So I am not sure if this is the fastest way
to work out this problem. But this is how I am going to do it:) The back sprocket is
6 centimeters as a radius. Let's figure out what the angle speed is for this front sprocket.
Well angular speed is theta over time. Theta... radians...degrees, not rotations. So 25 rotations
times 2pi. Each full rotation is 2 pi radians or 360 degrees but I am working with radians.
This is over a 30 second period. Well that is going to be 50pi over 30 or 5pi over 3
radians per second. GREAT!!! This thing is turning at 5pi over 3 radians per second.
Well this turning of the sprocket is pulling a chain and whatever amount of chain you pull
around the front sprocket is the same amount of chain you are pulling around the back sprocket.
What I want to do now is find out how much chain I am moving in that 5pi over 3 radians
of rotation. I want to find arc length, how much chain is getting pulled around that front
sprocket. Again for the front sprocket, s is equal to the arc length, how much chain
I am pulling. It could also be...I could also just write v for linear velocity, how much
has that front sprocket (the teeth) moved in one second. But I think that since my kids
are thinking, "ok I am moving a chain around it and that same amount of chain is getting
moved around the back sprocket.", I have found that saying arc length instead of saying it
is linear velocity makes it easier to understand. So the front sprocket has a 15 cm radius and
in that hmm..that is the radius so it is r... and theta for a one second period... theta
is 5pi over 3. So in one second I am doing r times theta so 15 times 5pi over 3, the
fifteen and the three will cancel out to be five so we get 5 times 5 is 25 pi centimeters.
If I put per second, I need to call that linear speed so I again just say that it is r times
theta for arc length. But in one second you are moving 25 pi centimeters of chain. Now,
if you move 25pi centimeters of chain then you are moving 25pi centimeters of chain through
the back sprocket. We want the angular speed of the back sprocket. Well, v which is r times
w. And so maybe just for clarification for the video let's say that this is per second
and not call this arc length now but linear speed, the amount of chain you moved in one
second. Linear speed is equal to the radius times the angular speed. In that one second
period you have moved again 25pi centimeters of chain around a sprocket with a radius of
6 and.. you know... v equals r and w is the question. What is the angular speed per second?
How much is this rotating per second if the front sprocket is turning 5pi over 3 radians
per second? Divide both sides by six and that is 25pi over 6 radians per second for the
back sprocket. And this is faster, that is a larger radian per second value than we had
with the front sprocket because of course it is smaller. It has to turn faster to move
that same amount of chain that you are pulling around the front sprocket. So there you go!
A two part question going from Angular speed to Linear speed and back as you go from sprocket
to sprocket. BAM!!! I am Mr. Tarrou. GO DO YOUR HOMEWORK:D