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Angular Velocity and Power Input Question

  1. Aug 27, 2012 #1
    Hi there,

    I've got these questions to do, and I've looked at the formulas and can't seem to plug in the numbers with the values I've been given. I think I'm missing something somewhere, if someone could just start me off, or be so kind as to go through the method so I know for the future?

    Thank you :)

    a) A wheel rolls up an incline of 15 degrees on its hubs without slipping and is pulled by the 100N force applied to the cord wrapped around its outer rim. The wheel starts from rest, so compute its angular velocity after its center has moved a distance of 3m up the incline. Radius to inner rim is 100mm, radius to outer rim 200mm.

    b) The wheel has a mass of 40 kg with center of mass at O and has a centroidal radius of gyration of 150mm. Determine the power input from the 100N force at the end of the 3m motion interval.
     
  2. jcsd
  3. Aug 27, 2012 #2

    Simon Bridge

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    Yep - plugging numbers into equations does not work ... you are missing the physics - get that right and the equations will take care of themselves.
    You should still show us how you are thinking about this so we can target the specific place you need help.

    Hint: It's rolling without slipping - so it's angular velocity and speed up the ramp are related. The force is applied at the rim - so it will make a torque.

    Power is energy over time... the energy comes from the work done by the force.
     
  4. Aug 27, 2012 #3
    Hi, thanks for your response.

    Torque. i knew there was something I was missing.

    So kind of working backwards, to work out torque I have the formula T = rFsinθ. Would the radius be the 200mm to the outside rim from the centre? and what is the centroidal radius of gyration because I've never heard the term before.

    So once I have the torque, the power would then be P = T x ω

    But to find the angular velocity ω, the formula I have is ω = v/r which is what I was given. Is v some sort of linear velocity?

    I'm very sorry, i know this is simple but I haven't done these sorts of problems since I was in school.

    Thanks again for your help

    EDIT: I'm using the formulas I was given, but if there are better ones, that'd be great.
     
    Last edited: Aug 27, 2012
  5. Aug 27, 2012 #4

    Simon Bridge

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    "Centroid of gyration" is related to "moment of inertia" ... look it up ;)

    Don't use formulas!
    You need to be able to understand the physics of what is going on - a torque is a twisting force. If it is created by a linear force [itex]\vec{F}[/itex] acting at some position [itex]\vec{r}[/itex] from the center of rotation, then the torque is the cross product of these: [itex]\vec{\tau}=\vec{r} \times \vec{F}[/itex] . So torque is a (pseudo-)vector.

    [itex]\omega[/itex] is the angular speed - under an unbalanced torque, this will change. So your equation for power isn't going to work right?

    Angle is the rotational version of position, rate of change of angle is [itex]\omega = d\theta /dt[/itex] is speed, [itex]\alpha = d\omega /dt[/itex] is acceleration and force is [itex]\tau = I \alpha[/itex] so I plays the role of inertia (mass) and it is called "moment of inertia" for historical reasons.

    I don't actually remember these equations - I just reconstructed them right now from the physics. All I did was describe the ideas using math like a language.

    I think you need to brush up on your physical understanding:
    http://www.sparknotes.com/physics/rotationalmotion/rotationaldynamics/section3.rhtml
    http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1025.0
    http://physics.info/rolling/summary.shtml
     
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