Category Archives: PHY 125

April 5, 2010 · 9:15 pm

Collisions and Inertia

Collisions
Imagine two balls. One elastic and one inelastic. The balls move towards a mass and the elastic one pushes it over. The inelastic ball is unable to push over the mass of wood. This is because it is an inelastic collision and the change in mechanical energy is not conserved. With the elastic collision, energy is conserved and returned to the bouncy ball.

Torque
Torque is Force * Radius. The larger the radius the more force needed to move an object over a distance. m and R are constants. RF (Force times radius) is torque. It is a force that goes in a circle. Torque is a measure of rotational inertia. The Movement of inertia is I=mR2.


  1. Radial Force equals mass times radius times acceleration, using angular acceleration Radial Force equals mass times radius squared times angular acceleration (which is equal to acceleration tangential over Radius)
  2. Magnitude of Torque equals magnitude of radius times magnitude of Force
  3. Torque equals mass times radius squared
  4. Inertia equals torque ie Mass times radius squared.
  5. Angular acceleration equals Torque over mass times radius squared.

    Now if we look at this in terms of Kinetic Energy, and imagine we have a bunch of masses rotating around a center of mass. Changing the center point will change the inertia of the system.

    1. Kinetic energy equals one half inertia times angular velocity squared.
    2. Inertia total equals sum of masses times Radius squared

    Moment of Inertia of a thin stick of length L and a total mass M, rotated about one end.

    So if we take a little mass from the thin stick, and call this little mass dm, which is x distance from the origin and we need to express dm as the running variable, x. We use a ratio of total length to total mass to do this, and then substitute it back into the derivative of dI. So The total inertia when the thin stick is rotated about one end is the sum of all those little dI’s, which is the integral of dI.

    Leave a Comment

    Filed under PHY 125

    Tagged as ,

    March 26, 2010 · 4:28 pm

    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.

    CIRCULAR MOTION Chapter 10

    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.

    Leave a Comment

    Filed under PHY 125

    Tagged as , , ,

    March 24, 2010 · 2:15 pm

    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).

    Definition:

    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.

    Leave a Comment

    Filed under PHY 125

    Tagged as ,

    March 22, 2010 · 12:53 am

    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.

    1 Comment

    Filed under PHY 125

    Tagged as ,

    March 17, 2010 · 2:07 am

    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.

    Leave a Comment

    Filed under PHY 125

    Tagged as , ,

    March 15, 2010 · 7:31 pm

    Potential Energy, Conservative Force Field

    Poential Energy is represented by the symbol U.

    U = mgh
    Wby Gravity = -ΔU
    ΔU = UFinal – UInitial = 0 – mgh = -mgh
    ΔU = -mgh = Wby Gravity = ΔKE

    What is important is the change in potential energy.

    -ΔU = (UFinal – UInitial) = Wdone by Gravity = ΔKE = KEFinal – KEInitial

    Lost Potential Energy = gained Kinetic Energy

    KEFinal + UFinal = KEinitial + UInitial

    Total Mechanical Energy is Conserved

    Conservative Force Field: Total work done about a closed path sums to ZERO.

    Friction Force is non-conservative (You do not get energy back in a closed path)

    Imagine how force looks as a ball is dropped from the sky. think of dy as the many very small distances it travels over to fall to the ground. dW is the very small but of work

    ΔU = -W = -F
    dU = -dW  = -Fy
    dydU/dy = -Fy
    U(y)=mgy

    Read this “The potential Energy with respect to height is equal to mass times gravity times height.

    Now what about stretched springs? This story is a little different. The force exerted to do work grows as the spring stretch’s further.

    If you draw a graph, the potential energy of the spring as it is compressed and stretched would form a parabola.

    Now what about a child on a slide?

    The component of gravity parallel to the slide is mg cos θ. It is doing work. θ is changing all the time, as the mass of the child moves along the slide.

    Leave a Comment

    Filed under PHY 125

    Tagged as , , ,

    March 14, 2010 · 12:16 pm

    Chapter 6 – Centrifugal Acceleration Review

    How to convert acceleration in terms of g


    Centrifugal velocity equals two pi R over time.

    Time equals one over frequency (number of revolutions per second)

    Centrifugal force, on an object not experiencing forces along the y axis (Car driving in circle, etc…)

    Centrifugal force, on an object experiencing forces along the x and y axis (ball spun horizontally above your head)

    When you spin a ball vertically, at the top of the circle, the tension force can be at minimum, 0, or else the ball will fall out of the sky, thanks to gravity. This means, the velocity of the ball overhead is equal to:

    The max pull at the bottom of the circle is equal to:

    For banking turns,

    Friction Force

    μ could be μ static or μ kinetic. It is the coefficient of Friction

    Conic Pendulum

    Leave a Comment

    Filed under PHY 125

    Tagged as , , ,

    March 12, 2010 · 11:38 pm

    Work – Constant Force, and Changing Force

    Work increases or decreases over displacement.

    1. Kinetic Energy equals one half mass times velocity squared
    2. Work equals change in Kinetic Energy, which is equivalent to final kinetic energy minus initial kinetic energy.
    3. Work equals force over displacement (If the force and displacement are parallel).
    4. Work equals Force over displacement times cos theta (If the force and displacement are at an angle to each other. Theta is the angle formed from Force and displacement).

    Force is an example of a Dot Product or Scalar Product of two vectors, for example vector’s A and B.

    1. Vector A dot product Vector B is equal to the scalar value of magnitude A times magnitude B times cos theta.
    2. Work (scalar value) equals the dot product of Force times displacement, which is equal to magnitude of Force times magnitude of displacement times cos Theta.
    3. Dot Products distribute.

    So the point of learning about dot products is that work, can be broken down into it’s component parts, Fx, Fy, Fz and dx, dy, dz and you can use these component parts to find out the angle theta.

    If we think about a spring and the work done to pull a spring open, the Force needed to continue pulling the spring grows, the further you pull on the spring

    This makes the Force Applied equal to a constant “k” times displacement x.

    Leave a Comment

    Filed under PHY 125

    Tagged as , ,

    March 10, 2010 · 12:18 am

    Work – What is it?

    We started talking about work/energy today. To get started we decided Work = Force times displacement. We talked about how when force and displacement are parallel or anti-parallel things are simple to calculate. We also talked about the units of work. So if

    Word Done = Force times displacement the unit equals 1 Newton per Meter but not one actually says that. Instead they say 1 Joule, which is the unit of work and energy. So remember:

    1 Joule = 1 Newton per Meter

    Gravity can do work too.

    Fgrav = -mg

    displacement = change in height = -h (since the mass falls down thanks to gravity)

    Work = (-mg)(-h)=mgh

    But when you lift a mass, you have to apply upward force, equal to the amount gravity is pushing down with plus just a tiny bit more. So

    Work = (mg)(h) = mgh

    How much work did gravity do? If you think about lifting a mass at a constant speed, and how gravity pulls a mass out of the air, it is different. Gravity is getting faster and faster. So if you think about a very small bit of the work gravity is doing, and call it dW, you now have a very small bit of work, which moves the mass over a very small bit of distance.

    Read this: “The very small amount of work is equal to the mass times the current velocity times the very small change in velocity. PS In the above, dv/dt is acceleration aka the instantaneous rate of change of velocity.

    So, you can add up all of these little dW’s to get the amount of work done over a larger distance, that is not so very small. Adding up all of these little dW’s is called integrating. And that is what an integral is.

    Read this: The sum of all of the little works is equal to the total work done throughout the process, which is equal to the sum of all of the little changes of mass, times current velocity, times the very small change in velocity. Which is equal to 1/2 mass times velocity squared. (Mostly… the professor seemed to indicate there is actually more to it than this.)

    So we looked at the problem of the work done by gravity as it pulls an eraser out of the air. And we solved for velocity based on the work done by gravity.

    This solution for velocity happens to be the same as if we used one of the constant acceleration formulas. I think it only works here because in this case our Force was caused by constant acceleration.

    This says “ The total work done is the final amount of work, minus any work it initially had.”

    Kinetic Energy of a mass m moving with a speed v

    The unit of Kinetic Energy (KE) is joules.

    What happens when force displace is not parallel?

    Word equals Force times displacement times cosine x.

    Leave a Comment

    Filed under PHY 125

    Tagged as , , , ,

    March 8, 2010 · 2:35 am

    Friction forces on Banked Turns

    Today we talked about the friction force between your car and the tires, which prevents your car from leaving the road. As your wheel rotates it actually has the force of static friction, not kinetic friction, because as wheels rotate they are always in direct contact with the ground except at different points. Only if your wheel stops rotating and continues moving does kinetic friction take over.

    Friction force can vary, from 0 to μsFn. The Friction force can never exceed the value μsFn. μs depends on the surface.

    If the curve is flat:

    Maximum velocity is equal to the square root of the radius, times gravity, times the static friction component.

    Now, what happens if this road is banked?

    Q1: At what angle should a road be tilted so the friction force is not necessary to keep the car on the road?

    The angle which makes Friction force unnecessary is the arc tangent of maximum velocity squared divided by gravity times the radius.

    What happens if you go faster than the design speed?

    When driving around a banked curve, and wishing to not fly off it, and going fast enough that if friction force was to vanish you would leave the curve, Maximum velocity is equal to the square root of radius, times gravity times tangent theta plus the coefficient of static friction. Divided by one minus the coefficient of static friction times tangent theta.

    Leave a Comment

    Filed under PHY 125

    Tagged as ,