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Rotational Stuff

  1. Mar 30, 2006 #1
  2. jcsd
  3. Mar 30, 2006 #2

    Doc Al

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    Mechanical energy (PE + KE) of the system is conserved. Figure out the change in gravitational PE.
     
  4. Mar 30, 2006 #3
    and how could question 9 be Postive i thought gravity always does negative work
     
    Last edited: Mar 30, 2006
  5. Mar 30, 2006 #4
    so MGH = 119.25 but that doesnt work out
     
  6. Mar 30, 2006 #5

    Doc Al

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    [tex]\Delta {PE} + \Delta {KE} = 0[/tex]
     
  7. Mar 30, 2006 #6

    Doc Al

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    Why would you think that? If the force and the displacement are in the same direction, the work done by the force will be positive; if they are in opposite directions, the work is negative.
     
  8. Mar 30, 2006 #7
    according to the formula Wgrav = -mgy

    Okay so if they ask if the ball swing from C to B is that still positve or would it be negative now
     
  9. Mar 30, 2006 #8

    Doc Al

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    But y can be positive or negative.

    You tell me.
     
  10. Mar 30, 2006 #9
    hmmm i would have to say postive since like u said up there force and displacment would be in the same direction. So can i safely assume for these type of problems on exam gravity will always be positive then?
     
  11. Mar 30, 2006 #10
    K problem 21. i Have (8)(9.8)(X) + (6)(9.8)(X) + .5 (8)(3^2) + .5 (6)(3^2) + .5 (2)*(3/.4)^2 = 0
    So where am i doin mistake or is this even right?
     
  12. Mar 30, 2006 #11

    Doc Al

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    Right. If the ball falls, the work done by gravity is positive.


    Don't know what you mean. Do you mean will the work done by gravity always be positive? Of course not. (Balls can rise as well as fall.)

    The way to understand that formula is as follows:
    [tex]W = Fd[/tex]
    [tex]W_{grav} = (-mg)(\Delta y)[/tex]

    This uses the convention that down is negative. So when something falls, [itex]\Delta y [/itex] is negative.
     
  13. Mar 30, 2006 #12

    Doc Al

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    The KE terms look good, but the PE needs some work. Realize that one mass goes down a distance h (what's its change in PE?), while the other goes up a distance h (what's its change in PE?).
     
  14. Mar 30, 2006 #13
    thanks alot, for your help. Question 24. This is a torque problem , so i make the fulcrum be zero and the equation would be -GL + -GL ( sin or cos). so (-1)(9.8)(1) + (X)(1) (sin/cos) I don't understand why Cos was used instead of sin. I had the answers narrowed down to D and E. Answer is E
     
  15. Mar 30, 2006 #14

    Doc Al

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    Depending on how you are taught to find the torque:

    (1) Torque = Force X moment arm; moment arm is the perpendicular distance between the line of the force and the pivot point. That perpendicular distance is L cos(35)

    (2) Torque = Force X distance sin(theta); the angle theta is the angle between the force (in this case downward) and the distance along the lever. Theta = 90 degrees + 35 degrees; mathematically, sin (90 + x) = cos (x).
     
  16. Mar 30, 2006 #15
    sorry for all these questions, i just wana know if i had the right logic for 15 on this exam. http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp04

    Okay so i figured the PE of the top and set that equal Kinetic and got a velocity of 5.77. so i assumed that it will start at a low V since the height has to be lower than 1.7 and has to work up to 5.77 inorder to equal the potential energy up there so speed has to increase. Thanks
     
  17. Mar 30, 2006 #16
  18. Mar 30, 2006 #17
  19. Mar 31, 2006 #18

    Doc Al

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    Don't really know what you're doing here. In problem 14, you calculated the work done by the 30N force. The work done will equal the change in total mechanical energy.

    Another way to solve this is to find the acceleration of the block using Newton's 2nd law.
     
  20. Mar 31, 2006 #19

    Doc Al

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    All they want is the position of the center of mass of the beam. The beam is uniform, so where is its center of mass?
     
  21. Mar 31, 2006 #20

    Doc Al

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    Total momentum is conserved. What is the momentum of the system before the collision? What's the x-component of that momentum? The y-component?

    You can figure the x and y components of the momentum of 2M after the collision, since you are given its velocity. Subtract that from the x and y components of the total momentum to find the momentum of M, and then its velocity.

    Remember: Momentum and velocity are vectors; direction and signs count.
     
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