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Rotational motion
A Nexus mindmap
0
1
This is analogous
to position
3
2
0
0
Just apply the chain rule and product rule as needed...
0
1
Analogous to velocity...
3
2
1
We can use the Taylor series to expand
any function around some point.
0
0
2
Expand each variable thats changing and
look at the overall change to first order
3
0
0
If the variation is being caused by a change in time then we arrive at the
differentiation that we got above for the velocities.
0
1
We can see how a small variation in x is generated from
small variations in the angle and radius by using partial
derivatives to determine the 'slope' of the angle and radius
with x. Same for y. Mathematicians call this the 'total
derivative'.
2
Faster way
1
We want to find the overall changes in
x
and
y
.
3
0
Incidentally, here is a useful way of looking at
changes - consider how small variations in one
parameter are generated by small variations in
another.
2
2
Here is a pictorial representation of what is
gong on with the variations.
1
0
1
2
1
0
Analogous to acceleration ...
2
Again, it's just an application of the chain rule and the product rule.. Note
that the last term in the velocity is a product of three functions of time - just
keep applying the product rule one term at a time...
0
1
2
0
0
i.e. the velocity is perpendicular
to the radial direction
0
0
0
This expression should be
familiar hopefully
1
1
1
2
0
0
1
0
0
0
0
Again, another familiar expression.
0
1
Note that the direction of the acceleration
is in towards the centre of the circle along
the the radius.
2
1
1
3
3
0
2
3
1
For linear motion
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2
0
3
1
4
Analogues
1
Constant angular
acceleration
4
0
0
0
We start off with (work) = (force)x(distance). If we call the expression in
brackets on the 3rd line the rotational analogue of force we will end up with
(work) = (analogue of force)x(analogue of distance)
1
0
1
5
For lots of particles the analogue still works
0
0
1
2
0
1
2
This now means that when we are looking at the static
situations there are two conditions that must be met - (i)
all the forces cancel, and (ii) all the torques cancel.
3
3
First start with (work) = (force)x(distance) then
equate that to the form of work we derived for
rotations. Looking at the appropriate components leads
to the formula.
0
0
2
The relationship between torque, force, and
distance (of the 'moment arm')
4
0
1
There's various ways of looking at the relationship. The torque is
(r)x(perpendicular component of F) or (perpendicular component of r)x(F) or
(r)x(F)x(sin of angle between them) or it's the vector crossproduct between r
and F ... (more on that later)
3
4
1
This is what we know for linear motion
0
0
Use
F
=
ma
to expand out the forces. It would be really neat if we could
rewrite this expression so that it looked like the torque came from the change
in time of the rotational analogue of momentum. That would really
strengthen the analogy...
0
1
It's easier to see this expression is possible by
working backwards
0
2
5
The result is that we CAN write the torque as the time derivative of the
angular momentum! How cool is that?
0
3
Similar to what we did for the torque, there are different ways of expressing
the angular momentum in terms of the linear momentum and the moment
arm. And yes, there is a vector cross product at work here too...
0
4
2
Consider an object which is rotating about some pivot
0
0
The total kinetic energy is the sum of the
kinetic energies of each particle
0
1
0
2
Hey! This looks like 1/2 m v^2 ... if call the term in the brackets the
rotational analogue of mass. Formally it's called the 'moment of inertia'.
0
3
0
4
0
When we consider continuous objects the sum becomes an
integral over all the little pieces of mass. A little practice
will teach you to unpack that integral into something that
you can compute. Just remember, an integral is basically
just a sum.
0
0
1
0
0
1
Newton's 3rd law meant that all the interal two-body forces cancelled
each other. The same trick works with torques too. And the total
external torque is the time derivative of the total angular momentum!
The analogues are still holding up.
3
3
3
0
1
Again we naturally arrive at a linear
analogue: (angular momentum) =
(angular mass)x(angular velocity)
2
2
2
4
0
1
0
Example calculation for moment of
inertia - a disk rotating about two
different axes.
2
0
1
3
3
Before you get too excited, the analogy between linear and rotational motion
breaks down in three dimensions... <sigh>
1
Actually, some of it survives.
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0
4
Rotational motion