Monthly Archives: March 2010

Velocity, Center of Mass and CIRCULAR MOTION

Velocity and Center of Mass

What happens if a rocket launches up into the air, then explodes? Where is the center of Mass? What about the velocities? Well, The velocity of the center of mass remains the same. But for instance, if you launch a rocket with two stages, and at the peak of the parabola, jettison the spent fuel canister? It’s velocity becomes zero, as the second stage kicks in and pushes off from the first stage. It picks up double the velocity (if they are equal weight, and is able to fly twice as far as the two masses would have gone together. The first stage falls straight down.


To keep track of Circular motion, we need to watch the same things we watch for Linear motion.

  • Position r (letter r)
  • Velocity ω (little omega)
  • Acceleration α (little alpha)

But now we are moving into Polar Coordinates, where it is much easier to keep track of a position on a circle in terms of radius and theta θ (Angle from the origin). If you use Cartesian coordinates you have to write x and y as r cos θ and r sin θ. So it’s easier to use Cartesian. Theta is also now always expressed in radians, or “length of arc over radius

What we do with the Polar Coordinates:

Theta in Radians equals length of arc over radius
1 ) Average Angular Velocity (little omega) equals change in Theta over change in Time.
2 ) Instantaneous Angular Velocity(little omega) equals the derivative of theta with respect to the derivative of time.
3 ) The derivative of Theta with respect to time is equivalent to the derivative of length with 4) respect to time multiplied by one over the radius.
5 ) The Tangential velocity at a time is equal to the derivative of the length with respect to time.
6 ) The Average Angular Velocity (little omega) equals the tangential velocity over the radius.
7 ) The Average Angular Acceleration (little alpha)  is equal to the change in Velocity (omega) over the change in time.
8 ) The angular instantaneous acceleration (little alpha) is equal to the derivative of angular velocity with respect to time.

You can use the same formula’s used for Linear constant acceleration, just replace the Linear velocity and acceleration with the Circular velocity and acceleration.

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Center of Mass

Finding the center of mass is essentially finding the weighted average of the component masses. Of course this can get more complicated depending on the number of masses and the amount of dimensions (1, 2 or 3).


Here cm means “Center of Mass”. You can expand this calculation out to x y and z, just consider the position (distance) of your mass from the origin in relation to the plane you are looking at.

For example an equilateral triangle, made from three masses connected by three straws. If you balance This contraption on a piece of cardboard, The center of mass along the x axis is Length over two and the center of pass along the y axis is 1/3(sqrt(3)/2) times L. The center of mass does not have to be where mass is.

The next part of the course combines lateral and circular motion and the center of mass is important for this.

Suppose you have a distributed mass of length L. And it’s a meter stick (just for fun). It has even linear density along the meter sticks, suppose you look at a very small quantity of mass, a little dm x distance from the origin (The start of the meter stick). If you add up all of those little bits of mass, and use the ratio dm/dx = M/L, you can find the linear density.

So, what this says, is that the center of mass, at some distance from the origin is the integral of the small change in mass times the distance, divided by the total mass. Which in the end figures out to the center of mass, is at one half the length.

Now think of a triangle. An Isosceles Right Triangle. Now to find the center of mass, divide it up into many little slices. (little dx’s and dy’s). Once again we can use the handy ratio. The little mass divided by the little area is equal to the big mass over the total area. dm/da = M/A.

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Review for Midterm #2

f is continuous if

The limit as x approaches a exists for f(x), the limit as x approaches a equals f(a) and it is continuous along the interval if true for every point a in the Interval.

f is differentiable if

The limit as h as f evaluated at x plus h minus f evaluated at x over h exists. Along an interval if limit exists for every point a in the interval.

A differentiable function is always continuous, but not the other way around. A continuous function can not have a derivative at a point (Break or sharp corner)


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Looking At Functions

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More on Recursive Functions

Ackermans Function – Interesting Recursive function example

Towers of Hanoi Problem

n-1 to tower B using C as temp storage.

n to tower C

n-1 to tower C using A as temp storage.

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Logarithmic Differentiation

When trying to solve a problem like y = (ln x)cos x use Logarithmic differentiation so you can use the product rule.

And a Trig identity I just recently grasped the usefulness of.

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Elastic Collisions

If two objects collide, one stationary and one moving, the resulting velocity of the two objects is given by:

What happens if one mass is much larger than the other? Say m2 >>> m1. We can use the same equation as before but we might as well ignore the very tiny m1 values. This gives us:

What if both objects are moving though? For example, a large mass m2 is struck by a little mass m1. →BANG!← We can go into a moving frame of reference. For instance, if we are standing on the big mass, our speed of reference becomes the velocity of big mass. Now the little mass is moving at a velocity relative to our frame of reference i.e. –(v+V).

Once we solve for the velocities, we can go back to the ground frame of references, by subtracting what we did in the first place to get into the moving frame of reference.

This principle, how a little mass colliding with a big mass manages to pick up so much extra velocity from the big mass is called “Sling shot” it is used commonly to launch space probes at a faster speed into deep space.

We should be able to use this principle to calculate the height of a tennis ball dropped at the same time as a basket ball and striking the basket ball as it hits the ground. (The tennis ball flies off to great heights.)

Now we switched to talking about the forces acting on a tennis ball as it is struck by a racket. If you plot this force, it has a large spike in force, for a short amount of time. So this is easier almost to calculate using change in momentum.

Now, what happens if we move into two dimensions? Consider the law of conservation of momentum as a queue ball strikes a pool ball and the two balls move off on different trajectories. The vector for conservation of momentum can be broken into it’s component parts.

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int rabbit(int n)
if (n==1) return 1;
else if (n==2) return 2;
else return rabbit(n-1)+rabbit(n-2)

All recursive solutions can be stated in iterative terms (loops)

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Potential Energy, Perfect Spring

I needed to keep my graphic from last lecture… We continued the problem, of a mass sliding down a slide, represented by an arc one quarter of a circle. So we were talking about how as the mass goes down the slide it makes these very small changes in distance, which are the same as making very small changes in the angle theta.

Total work done is equal to mass * gravity * radius.

So, looking at this graphic, we are able to say the total work done is equal to the change in potential energy, which is equal to one half mass * velocity squared at the bottom. This concept also applies for work which does not move in a smooth arch to the bottom.

How high should you start from in order to successfully make a loop-de-loop on a rollercoaster?

If you put a mass on a perfect spring, compress the spring and then release it, what is the potential energy in the spring at the top of the in the air fly time? k is the spring bounciness.

We briefly started to touch on momentum. Momentum is a vector, represented by the symbol “p”. It is mass times velocity. The professor noted how the conservation of energy does not actually happen to often. But conservation of momentum does tend to happen quite regularly.

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Implicit Differentiation, Tangent Lines, Inverse Trig Functions

Find the tangent line at point (3,2)

Now that we know dy/dx, solve for the slope using the given values for (x and y)

Now use the Tangent Line formula

Next Problem

Now to find the tangent line for this problem, with the coordinates (2,0)

A brief aside

Prove derivative of ln(x) using Implicit Differentiation

  1. The Problem
  2. Write in full form, including y
  3. Use log properties to arrange into x = form
  4. Take the derivative with respect to the change in x from both sides
  5. The derivative of “e to the y” remains “e to the y”, the derivative of x is one.
  6. Rearrange to solve for the derivative dy/dx.
  7. Substitute for y.
  8. “e to the ln of anything” is equal to the anything, in this case anything is x.

Find the derivative of arcsin x.

  1. The Problem
  2. Rearrange the problem using Trig
  3. Take the derivative from both sides
  4. x is equal to 1
  5. Solve for dy/dx
  6. Substitute for y in the solution.
  7. The professor kept going and somehow (using Trig?) managed to bring the answer to this.

This is what he did:

Not quite sure how cos y equaled x

Increasing and decreasing functions

For sin to be invertable, you need to restrict the domain to -pi/2 to pi/2
To see how a function looks, determine the invervals above and below zero. To do this solve for x.

You can look at the derivaties to decide in a function is concave up or concave down.

Second derivative is negative, concave down. Second derivative is positive, concave up

READ problem 2.7 #43 from the textbook.

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