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Angular acceleration of a wheel

  1. Nov 30, 2005 #1
    A wheel of radius 0.2m is mounted on a frictionless horizontal axis. The rotational inertia of the wheel about the axis is 0.05kg(m^2). A massless cord wrapped around the wheel is attached to a 2kg block that slides horizontally. If a horizontal force of magnitude P=3N is applied to the block, what is the angular acceleration of the wheel?

    r=0.2m
    I=0.05kg(m^2)
    m=2kg
    P=3.0N

    [tex]T = rFsin \theta[/tex]
    [tex]T = 0.6[/tex]
    [tex]T=I \alpha[/tex]
    [tex]T = 12RPM^2[/tex] Units?

    im guessing this is wrong because I never had to use the mass anywhere

    Can anyone help?
     
  2. jcsd
  3. Nov 30, 2005 #2
    Any help would be appreciated
     
  4. Nov 30, 2005 #3

    Fermat

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    I'm a bit unclear what a diagram of this looks like.

    The cord that is wrapped around the wheel. It comes off the top or bottom of the wheel ?
    Is the cord at an angle when it comes off the wheel, or is is horizontal ?
    Is this cord then horizontally attached to the 2kbg block ?
    Is the 2 kg block resting on a frictionless surface ?
     
  5. Nov 30, 2005 #4
    its horizontal
    yes its horizontally attached
    yes its on a frictionless surface
     
  6. Nov 30, 2005 #5

    Fermat

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    OK.
    You have a force P applied to the 2 kg block.
    This force accelerates both the 2kg block and the Wheel.
    Divide up this force into two components, F1 and F2. Obviously F1 + F2 = P

    Let F1 be the accelerating force on the 2kg block.
    Let F2 provide the accelerating torque on the wheel.
    Since they are both attached, then the linear acceleration of the circumference of the wheel matches the linear acceleration of the 2 kg block.

    Can you work things out from this ?
     
  7. Nov 30, 2005 #6
    hmm..im still not surbe I understand
    Im trying ot find F2?
    P is known
    how would i calculate F1?
     
  8. Nov 30, 2005 #7

    Fermat

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    You don't calculate either F1 or F2.
    You create expressions involving F1 and F2 and then add the expressions together to equal P. F1 +F2 = P.

    F1 is the accelerating force on the 2kg block, ergo

    F1 = ma (m = 2kg)

    F2 provides the torque on the wheel, ergo

    F2*r = I*alpha

    ergo²,

    F1 + F2 = P
    ma + I*alpha/r = P

    The only unknowns now are a and alpha, the linear and angular accelerations.

    You have to find a relatoinshipo between them. Remember my comment in my earlier post.
     
  9. Nov 30, 2005 #8
    so a=alpha(r)
     
  10. Nov 30, 2005 #9

    Fermat

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    Yep, that's it :smile:

    The rest should fall out now.
     
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