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Frictional Torque of a Flywheel

  1. Jul 10, 2013 #1
    1. A mass of 0.5 kg suspended from a flywheel, if the mass is released from rest and falls a distance of .5m in 1.5s calculate
    a) the linear acceleration of the mass
    b) the angular acceleration of the wheel
    c) the tension in the rope
    d) the frictional torque, resisting motion

    mass of wheel = 3kg
    outside radius of wheel = 300mm
    radius of gyration = 212mm

    2. Relevant equations
    I = Mk^2
    K.E. = 1/2 Iw^2
    θ = 1/2at^2
    T = work done/angular desplacement
    T= m(a+g)



    3. The attempt at a solution
    a)
    v = .5/1.5 = .33m/s
    a = v-u/t
    .33/1.5 = .22m/s^2

    b) angular acceleration = a/r
    0.22/0.3 = .73m/s^2

    c) T = m(a+g)
    0.5(.22+9.8) = 5.01 N


    d)Frictional Torque
    moment of interia I = mk^2
    0.3 x (.212)^2 = .135 kg.m^2

    displacement θ = 1/2 at^2
    1/2 (0.73)(1.5)^2 = .821 rad

    K.E = 1/2 Iw^2
    1/2 (.135) (1.1)^2 = .0816J

    T = work done / angular displacement
    .0816/0821 = .0993 Nm


    Plz let me know if I am doing it right, specially Frictional Torque...
     
  2. jcsd
  3. Jul 10, 2013 #2
    Hi PoorTech

    I uaually like to check my answers makes sense before proceeding to a next step, by plugging in the values I get into another equation to see if the values all turn out correct.

    Three equations of lnear motion with constant acceleration that you must memorize and know how to manipulate are:
    Velocity and time:
    v = v[itex]_{0}[/itex] + at

    Displacement and time
    x[itex]_{1}[/itex]= x[itex]_{0}[/itex] + v[itex]_{0}[/itex]t + 1/2 a t^2

    Velocity and displacement
    v^2 = v[itex]_{0}[/itex]^2 + 2 a d

    Your equation of v = d/t = 0.5m/1.5s to get the velocity of 0.33m/s does appear to calculate the average velocity of the mass from start to finish, that is as if the mass had a constant velocity with zero acceleration.
    The second equation appears to have a unit problem and error.

    see if you can calculate the linear acceleration by choosing one of the above equations.

    Cherios,
     
    Last edited: Jul 10, 2013
  4. Jul 10, 2013 #3
    i did it that way, bcaz given velocity is in /1.5s, so i do need to convert it in m/s
    we cant use V = Vo + at, as a is still not calculated...

    am i right?
     
  5. Jul 10, 2013 #4
    so, I guess you cannot use that formula, your right.

    The problem gives you the DISTANCE the mass fell and the TIME starting from rest.
    For a) they ask to find the linear acceleration, a.
    Which of the three formulas can you use to find the acceleration, knowing time and distance?
     
  6. Jul 10, 2013 #5
    2nd formula i guess.

    S= ut + 1/2 at^2

    a = 2S/t^2 = 0.44m/s^2

    is it ok now?

    if yes, what you think about rest of the calculation?
     
    Last edited: Jul 10, 2013
  7. Jul 10, 2013 #6
    a) I guess? It looks OK.

    b) seems to be OK - just use the new acceleration

    c) also seems to be in error due to a sign, which for falling masses on a rope makes it tricky.
    What you have, by adding g + a together is an acceleration of the falling mass greater than freefall if there was not a rope - is that possible?

    The forces on the mass are its weight W=mg downwards, and the tension in the rope, T, upwards.
    The net force causes the mass to accelerate.
    Fnet = m(-a) <== if positive direction is chosen as up, then downwards weight and acceleration are negative.

    So, from a FBD of the mass,
    T-W = -ma


    d) Frictional torque is caused by friction. If the flywheel was frictionless, then the falling mass could accelerate it with all the torque the mass could muster. Friction wlll cause the flywheel to accelerate not as fast.
    Perhaps take a look at this site, to see an example for study.

    http://www.physics247.com/physics-tutorial/frictional-torque.shtml
     
  8. Jul 10, 2013 #7

    haruspex

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    In addition to 256bits' comments:
    What are the units of angular acceleration? What units does dividing linear acceleration in m/s2 by radius in m give you?
    It's 3, not 0.3, but the answer is right.
    Not if this is the same a as before. Maybe you mean α.
    But as 256bits points out, calculating the displacement is not the most logical way to solve this part. Think about the difference the friction makes to the acceleration.
     
  9. Jul 11, 2013 #8
    its rad/s^2, that was a typing mistake

    that was a typing mistake as well.

    thnx a lot 4 ur help, yaa i am trying on that.
     
  10. Jul 11, 2013 #9
    i re-calculated everything and now stuck on the last one, i tried the link u provided, actually over there they have both the accelerations without and with friction, but in this example we have only 1..
     
  11. Jul 11, 2013 #10
    OK.
    Well, you can calculate the frictional torque using your methed with energy, which I have not checked for correct application.

    Or you can try the the method from the site.

    To turn the flywheel a net torque has to act on it, and set it spinning,
    T = I alpha
    There are 2 torques on the flywheel.
    One is the torque from the falling mass, which you have already calculated.
    The other, in the opposite direction is the frictional torque.

    Torque from the mass - torque from friction = angular momentum of the flywheel.

    You also have calculated the angular momentum of the flywheel,

    Are you now able to calculate the friction torque?

    ( Note how similar this is to F=ma for an object being pulled by a rope on the ground with friction.
    . where Frope - Ffriction = ma
    the difference for rotation is that the forces turn into torque, the mass turns into I, and a becomes alpha )

    Let us know how you did.
     
    Last edited: Jul 11, 2013
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