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Math Help - Linear and angular acceleration of flywheel

  1. #1
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    Linear and angular acceleration of flywheel

    Hi, I have the following question and was wondering if anyone could help:

    A mass of 0.5kg is suspended from a flywheel, of mass 3kg, outside radius 0.3m and radius of gyration 0.212m. If the mass is release from rest and falls a distance of 0.5m in 1.5 seconds, 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 the motion

    Equations (I think!); a = g[(2m/(M+2m)] v=s/d w=v/r T=mg[M/(M+2m)] I=mk2

    I think (although I could be wrong) that (a) can be calculated using a = g[(2m/(M+2m)] where m = 0.5kg, M = 3kg and g = 9.81m/sec2. This would give an answer of 2.45m/sec2, unless I'm being stupid and its as straight forward as acceleration due to gravity = 9.81m/sec2?

    For (b) Angular velocity = linear velocity/radius = (0.5/1.5)/0.3 = 1 rad/sec

    Therefore, angular acceleration = change in angular velocity/time = (1-0)/1.5 = 0.6 rad/sec2

    (c) Does T, tension=mg[M/(M+2m)] = 3.78 N ?

    (d) I'm not sure how to approach this one, I know frictional torque = total torque - accelerating torque and moment of inertia and radius of gyration must be included at some point but haven't solved it yet - help!?
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  2. #2
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    Re: Linear and angular acceleration of flywheel

    Quote Originally Posted by chunkylumber111 View Post
    Hi, I have the following question and was wondering if anyone could help:

    A mass of 0.5kg is suspended from a flywheel, of mass 3kg, outside radius 0.3m and radius of gyration 0.212m. If the mass is release from rest and falls a distance of 0.5m in 1.5 seconds, 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 the motion

    Equations (I think!); a = g[(2m/(M+2m)] v=s/d w=v/r T=mg[M/(M+2m)] I=mk2

    I think (although I could be wrong) that (a) can be calculated using a = g[(2m/(M+2m)] where m = 0.5kg, M = 3kg and g = 9.81m/sec2. This would give an answer of 2.45m/sec2, unless I'm being stupid and its as straight forward as acceleration due to gravity = 9.81m/sec2?

    For (b) Angular velocity = linear velocity/radius = (0.5/1.5)/0.3 = 1 rad/sec

    Therefore, angular acceleration = change in angular velocity/time = (1-0)/1.5 = 0.6 rad/sec2

    (c) Does T, tension=mg[M/(M+2m)] = 3.78 N ?

    (d) I'm not sure how to approach this one, I know frictional torque = total torque - accelerating torque and moment of inertia and radius of gyration must be included at some point but haven't solved it yet - help!?
    m = hanging mass

    d = linear displacement of the hanging mass

    M = wheel mass

    R = wheel radius

    r_g = radius of gyration

    (a) linear kinematics ...

    d = \frac{1}{2}at^2

    a = \frac{2d}{t^2}

    (b) assuming no slippage ...

    \alpha = \frac{a}{R}

    (c) Newton's 2nd law ...

    mg - T = ma

    T = m(g-a)

    (d) \tau_{net} = I \alpha

    TR - \tau_f = M(r_g)^2 \alpha

    solve for the frictional torque, \tau_f
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    Re: Linear and angular acceleration of flywheel

    That's great, thanks a lot for your help. I think I've got some answers now but I'm still not 100% sure on how (d) is worked out, could you give me a little more help please?

    So (a) = 0.44m/s^2

    (b) = 0.44/0.3 = 1.47 m/s^2

    (c) 0.5(9.81-0.44) = 4.69N

    (d) Is the T in (TR - Tf) the tension in the rope worked out from part (c)?

    or is it Tnet = Ia and since =M(k^2) where a = alpha and k = radius of gyration

    Therefore Tnet = M(k^2)a = 3x(0.212^2)x(1.47) = 0.198 Nm

    in which case Tf = (0.198x0.3) - 3(0.212^2)x1.47 = -0.139Nm ?

    Is the above correct?
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    Re: Linear and angular acceleration of flywheel

    (d) torque caused by the hanging mass , \tau = T \cdot R , where T is the tension in the cord attached to the hanging mass

    torque due to friction in the axle (what you want to find) ... \tau_f

    net torque is the difference between the torque generated by the hanging mass and the frictional torque ...

    T \cdot R - \tau_f = I \alpha

    solve for \tau_f ...

    \tau_f = T \cdot R - I \alpha

    finish it.
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  5. #5
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    Re: Linear and angular acceleration of flywheel

    Ok now I get it! So Tf = 1.407 - 0.198 = 1.209Nm? Thanks
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