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Posts posted by willorn

  1. Quite well. I've enjoyed myself and my classes here. The course Modern Physics has just been removed from the curricula and sort of integrated into Quantum Physics I so I'll be taking that next semester. Yeehaw!

  2. [ATTACH=CONFIG]371[/ATTACH] We don't always work in the library.

    Maybe this thread could be about college and high school seniors could ask questions. I know I had a lot when I was taking AP C.

  3. I'm in it. Professors are awesome. I was reminded of Aplusphysics and a certain professor when I caught up to the Big Bang Theory episodes a few days ago. Honors Physics seems to be starting slow but I hear tales that it will kick my butt. More to follow.

  4. I used the same method as for part I but with Kinetic Energy having both rotational and linear components. I have begun the derivation using some geometric observations made in Part I of the problem. *note: this derivation will be hard to follow without first walking through the solution to Part I of the challenge problem.

    Finding the changed velocity: (This time around, I figured the kinetic energy at the time the ball leaves the basketball would have both rotational and linear kinetic energy because it is rolling)\

    Eventually I changed omega, rotational velocity, into a linear expression to solve for velocity.

    r^2)v^2 \Rightarrow 2mghr^2 = (I+mr^2)v^2 \Rightarrow v^2 =\frac{2mghr^2}{I+mr^2}

    Just like part I, substitute the velocity squared expression into the cosine expression.

    rg = \frac{2mhr}{I+mr^2}

    recalling the alternate expression from Part I, and recalling that the vertical height "h" is the vertical height "r"

    gif.latex?\frac{2mhr}{I+mr^2} = \frac{r-k}{r} \Rightarrow \frac{2mkr}{I+mr^2} = \frac{r-k}{r} \Rightarrow \frac{2mkr^2}{I+mr^2} = r-k

    gif.latex?2mkr^2 = (r-k)(I+mr^2) \Rightarrow 2mkr^2 = Ir -Ik +mr^3 -mkr^2 \Rightarrow 2mkr^2 +Ik +mkr^2= Ir +mr^3 \Rightarrow k(2mr^2+I+mr^2) = Ir+mr^3 \Rightarrow k= \frac{Ir+mr^3}{I +3mr^2}

    After isolating the "k" terms, I had found a solution.

  5. After the conclusion of my AP tests, I think I've finally found a nice solution to this problem! I will employ the wonderful Latex to derive my answer!

    I began with an FBD diagram, just like my momma (Cermak) taught me.


    I began by examining my FBD and drawing trigonometric conclusions. (After receiving word from Frach, I realized I neglected to mention the Normal; It is important to note that the Normal would necessarily be zero for the puck to fly off the basketball, and that the normal force is a function of the centripetal force and therefore:mg


    Then I used my given knowledge that the basketball is frictionless in order to determine that the total energy is conserved in the basketball puck system.


    Later of course, I realized I had unnecessarily taken the square root of velocity, oh well. I substitute the centripetal acceleration from the first eqution with its representative expression and substitute my new velocity! At this time I am very hopeful.


    And now I am stuck. But here comes the part that took some more observation, if we look at the FBD again, it can be seen that an equivalent theta exists accidentally in the very diagram I just drew! This comes as a result of the basic geometric concept of Alternate Interior Angles. The length that was previously ignored can be expressed as the difference between R, (the total radius) and "k" (the height at which the puck leaves the basketball).


    Using a new expression of the same angle theta, I can now set up and solve for what seems to be an answer that is either perfect or too good to be true. Note that I was able to substitute the "h" for "k" in the expression because k represents the vertical height the puck has fallen.


  6. I was attempting to recreate the situation on a real basketball but using a longer object on top (I didn't have a hockey puck), and it does indeed look as though the long object leaves the basketball before getting halfway down. I am beginning to think that this is because the fulcrum location on the puck changes as it moves down the basketball until the force of gravity on the puck causes it to rotate until it leaves the basketball (because it is orientated more vertically at this point).

  7. I have limited knowledge of the normal force, as I learned in physics B that it is simply a "reactionary force" always with equal magnitude to the initiating force unless deformation occurred. In physics C we spent perhaps 5 minutes learning that the normal force is a summation of resistance forces from everything that the initial force is "pushing on." With that in mind, the normal force is the only force in play that I imagine would be able to cause the puck to leave the basketball early. BUT I assume because there is no deformation to the basketball or the puck that the normal force can be no more than equal to the weight of the puck at any time, and so there is no acceleration outward. Please disprove my statement.

  8. Wait then how is it that the object is moving 2 m/s if the forces are the same? If the forces are the same, shouldn't the object be motionless?

    It would seem so. What you have to keep in mind Newton's Second Law of Physics F=ma. A force results only in an acceleration. AND we know that if an object has Fnet=0 (that means that there is either no forces acting on an object, or that the forces acting on the object cancel each other, as in this case) then the object is not accelerating in any direction. There are only two possibilities for an object that is experiencing no accelerations; this comes from Newton's First Law: an object will remain at rest or in constant motion unless acted upon by an outside force. Your object, however, also has a velocity. Ergo, your object is moving with constant velocity. Hope this helps!

  9. If you weren't a super duper physics student, it would seem so! However, Forces cause accelerations on an object. and if an object is not accelerating in any direction, two things can be happening to the object. It can either be motionless OR in constant motion; talk to yourself about the definition of constant velocity, thats how I wrapped my brain around that one. Velocity only changes with acceleration (forces), so a constant velocity means no forces or forces that cancel each other out. I hope this helps!

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  10. I'm hoping this is the right place to post for Physics C Problems.

    I thought I heard that people were confused by the billiards web assign so I thought I'd weigh in. To me it seems like the velocities of the cue ball and the 8-ball have been switched on the webassign, but maybe I'm crazy. Also, I only know from experience that perfectly spherical billiard balls that collide perfectly tangentially ricochet off each other at complimentary angles, but the mathematical explanation in the answer packet leaves a bit to be explained for me. Any clarification would be much appreciated!


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