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Frictional talk - mass suspended from a flywheel

  1. Sep 6, 2013 #1
    frictional talk -- mass suspended from a flywheel

    1. The problem statement, all variables and given/known data
    a mass of 0.5kg is suspended from a flywheel. if the mass is released from rest and falls a distance of 0.5m in 1.5sec calculate the frictional torque.
    mass of wheel = 3kg
    Radius= 0.3m
    Radius of gyration = 0.212m


    2. Relevant equations
    Torque = mk^2a


    3. The attempt at a solution
    From a prevoius question i calculated the angular acceleration to be 1.48rads.
    So torque= mk^2a = 3x0.212^2x1.48

    So Torque = 0.2Nm if friction is ignored.

    So i have a torque 0.2Nm and angular acceleration of 1.48rads^2

    Am i right in thinking Torque-frictional torque= angular acceleration?
    so 0.20- frictional torque = 1.48?. This doesnt seem to work out for me. any help would be appreciated
     
    Last edited by a moderator: Sep 8, 2013
  2. jcsd
  3. Sep 6, 2013 #2
    so torque net-frictional torque = angular acceleration
    so 0.2-frictional force= 1.48rads^2
    rearrange to Tf = 1.48+0.2
    Frictional torque = 1.68Nm??
     
  4. Sep 7, 2013 #3

    haruspex

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    You have the right angular acceleration, but the system being accelerated is not just the flywheel. The descending mass is also being accelerated. There are two ways to deal with this. You could introduce a variable for the tension and consider the forces and accelerations of mass and flywheel separately, but easier is to consider the effective moment of inertia of the mass about the flywheel's axis and add that in. What do you think its moment of inertia is?
    When you have that right, consider how gravity comes into the equation. You appear to have taken the friction as causing the acceleration, but friction opposes relative motion.
     
  5. Sep 7, 2013 #4
    I worked out the moment of inertia as follows:

    I = mk^2

    I = 3x0.212^2

    I = 0.135 Is this correct?
     
  6. Sep 8, 2013 #5

    haruspex

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    That's the moment of inertia of the wheel, but the wheel is not the only object being accelerated by the weight of the suspended mass. The mass is also accelerating.
    As I said, you have two ways you can take this into account. pick one.
    1. Introduce an unknown for the tension in the cable. The torque from that tension (less the frictional torque from the axle) accelerates the wheel. The acceleration of the mass comes from the force of gravity on the mass minus the tension.
    2. Figure out the moment of inertia of the descending mass about the wheel's axle and add that to the moment of inertia of the wheel.
     
  7. Sep 8, 2013 #6
    So I need to add the moment of inertia of the mass = F=ma I think.

    So f = 0.5 x 0.222 = 0.111.

    so 0.111 + 0.135 = 0.246.

    so total torque = I@ (ANGULAR ACCELERATION) ?
    The above formula gives me the total torque (forces) involved in the question. Now i realize gravity must come into play as the object descends. Not sure how inertia and gravity ties in?
    Not understanding the physics behind it!
     
  8. Sep 10, 2013 #7

    haruspex

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    F=ma? where's the moment of inertia in that? And I've no idea where you are getting the .222 from. You need a mass multiplied by the square of a distance. The mass is obviously .5kg, but what distance is relevant? The mass is accelerating vertically, so in what direction should the distance of interest be?
    If you can't figure that out, switch to the other option I gave you. It's important that you use a method in which you understand the physics.
     
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