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Work Energy Method for Rotational Motion

  1. Jul 29, 2012 #1
    1. The problem statement, all variables and given/known data
    I don't how how to work out for the Gravitational Energy "H".

    Info given is 1.5rev converting to 9.42rad/s
    S=rδ
    S=0.2x9.42
    =1.884

    *Should i multiply the axle radius or wheel?

    3. The attempt at a solution
    E1=K1+G1+S1
    =1/2(8.5)202+(20)(9.81)(H)+0


    Thanks
     

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  3. Jul 29, 2012 #2

    TSny

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    Check the units for the 9.42

    What are you trying to calculate above?

    Your expression for the kinetic energy is not correct. How do you calculate the kinetic energy of a rotating object? The hanging mass also has kinetic energy at the initial time.

    Since the mass is located at the "datum" at the initial time, you can consider the potential energy to be zero at that time.
     
  4. Jul 29, 2012 #3
    Sorry should be 9.42rad

    S=rδ
    S=0.2x9.42
    =1.884
    I am trying to calculate the distance for the height for gravitational energy

    How do you calculate the kinetic energy of a rotating object?
    K=Iω2

    Since v=rω
    2=0.1ω
    ω=20rad/s

    what do you mean by potential energy? do you mean kinetic?

    The equation should be:

    1/2(8.5)(20)2+1/2(20)(2)2+G1+0

    Rotational as "1/2(8.5)(20)2"
    Linear as "1/2(20)(2)2"

    G1-don't know how to form

    S1 as 0 because there is no spring
     
    Last edited: Jul 29, 2012
  5. Jul 29, 2012 #4

    TSny

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    OK, that looks good except for a factor of 1/2 in K.
    All right. I didn't know what the symbols S and G were standing for. As you say, there's definitely no spring potential energy. For the gravitational potential energy, note that the hanging mass is located initially at the "DATUM". That's the point where the gravitational potential energy is taken to be zero. So, G1 = 0.
     
  6. Jul 29, 2012 #5
    Did the box go down eventually? To my understanding yes because that question says in 1.5 rev" So that means the box travel 1.5 rev downwards.

    So i set my datum point at the 1.5 rev point, hence there is gravitational potential energy at the initial position which i am trying to calculate but i don't know whether to use the axle or the wheel radius.
     
    Last edited: Jul 29, 2012
  7. Jul 29, 2012 #6

    TSny

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    Yes. So, when you go to part (b) the box will be below the datum.

    [EDIT: Did you use the correct radius for finding this distance?]
     
  8. Jul 29, 2012 #7
    Did the box go down eventually? To my understanding yes because that question says in 1.5 rev" So that means the box travel 1.5 rev downwards.

    So i set my datum point at the 1.5 rev point, hence there is gravitational potential energy at the initial position which i am trying to calculate but i don't know whether to use the axle or the wheel radius.
     
  9. Jul 29, 2012 #8
    Hmm after some thinking, do you mean i have to follow the question datum and i cannot set the datum myself unless the question never state the datum in the 1st place?
     
  10. Jul 29, 2012 #9

    TSny

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    You don't want to change the datum point. The answers you gave are for the datum shown in the picture. So, if you want to get the same answers, you need to keep the datum as shown.

    The string tied to the box is wrapped around the pulley of radius 0.10 m. So, that's the radius you want to use to convert angle of rotation to linear distance traveled by the box.
     
  11. Jul 29, 2012 #10
    For Part B

    U1-2=T.δ+F.s
    =(P)(9.42)+(P)(0.942)
     
  12. Jul 29, 2012 #11

    TSny

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    For part (b) you just need to calculate the total energy at the final location. What is the kinetic energy there? What is the potential energy there?
    [Sorry, that's for part (c)]
     
  13. Jul 29, 2012 #12

    TSny

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    For part (b) you need to think about how to calculate the work done by a force. Do you know a formula for this?
     
  14. Jul 29, 2012 #13
    I don't know I am just trying to calculate the linear and rotatational energy. This is getting confusing :cry:
     

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  15. Jul 29, 2012 #14

    TSny

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    Confusion is normal. We'll get it. The questions in the problem are sort of guiding you through step by step. So, in part (a) you find the initial energy. In part (c) you will find the final energy. For part (b), you just need to find an expression for the work done by the force P.

    You have posted a couple of pages. One expresses the work in terms of force and distance, the other in terms of torque and angle. You can use either one of these expressions. They will yield the same answer.
     
  16. Jul 29, 2012 #15
    For Part B

    U1-2=T.δ+F.s
    =(P)(9.42)+(P)(0.942)

    Wrong answer :cry: I assume my linear work done is correct. But my δ is definitely wrong.



    *wait give me some time to work out
     
    Last edited: Jul 29, 2012
  17. Jul 29, 2012 #16

    TSny

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    Use either force times distance or torque times angle. If you decide to use force times distance, the distance must be linear distance in meters. If, instead, you decide to use torque times angle to get the work, then you will need to express the torque in terms of P.
     
  18. Jul 29, 2012 #17
    "Use either force times distance or torque times angle."

    I thought there is linear and rotational? The parcel is moving downwards (linear) and the wheel is rotating (rotational)

    ???
     
  19. Jul 29, 2012 #18

    TSny

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    The force P acts at one point and does some work. Your job is to express the work done by P. Since P acts in a direction opposite to the motion of the rim of the wheel, the work done is negative. The basic formula for work when the force acts opposite to the motion is

    work = -Force.distance = -P.s

    That will give you the total work done by P. (If you want to, you can express this work in terms of the torque exerted by P as work = -torque.angle).

    Let's just stick with work = -P.s What value of s should you use? (Hint: you already calculated it in your first post.) Does this give you the answer for (b)?
     
  20. Jul 29, 2012 #19
    I'm sorry i don't understand.

    The question ask "Write an expression for work done by force P in stopping the wheel and axle"

    Stopping the wheel - Wheel is rotating in the 1st place (Rotational Motion)
    Stopping the axle - Parcel is tied to the axle (Linear Motion)

    With reference to the screenshot below part C, the question ask for the work done by the frictional force on the drum,so it is only just the rotational motion and not linear motion.

    So why do you say i can use either force times distance or torque times angle? But to my understanding linear and rotational exist, shouldn't i use both?
     

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  21. Jul 29, 2012 #20

    TSny

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    The force P acts over a distance equal to the arc length, s, that a point of the rim travels as the wheel rotates through an angle δ. The relation between s and δ is s = r.δ

    The work done by P is just the force times the distance s: work = -P.s

    But note that we could express this as work = -P.(r.δ) = -(P.r).δ = -T.δ = -torque.angle.

    The force P does not do both "linear work" and "rotational work". The definition of work is just force times distance (when the force is parallel to the distance, as here). So, work = -P.s and that's it. It's just that sometimes you will see this work expressed as -T.δ. It represents the same work.
     
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