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Wheel and spoke

  1. May 15, 2008 #1
    1. The problem statement, all variables and given/known data

    a wheel (hoop) of 10lbs has one 10lb spoke which is horizontal and rests on the ground (held). it is connected to an axle. the radius of the wheel (length of spoke) is 1ft. when the wheel is let go, what is the force of the axle on the wheel?

    (the problem also asked for the rotational inertia at the axle).

    the answer given is 18lbs.

    2. Relevant equations

    Rotational Inertia at Axle = 1/3(10/32 slugs)(1ft)^2 + (10/32 slugs)(1ft)^2

    Sum of horizontal forces = m(a-radial) = Tension of spoke

    torque at axle = RI * alpha = radius * spoke-weight

    3. The attempt at a solution

    RI = 0.417slug*ft^2

    i set up an fbd of the wheel.

    torque at axle = (0.417slug*ft^2)(alpha) = (1ft)(5lbs) (i assume that the weight of the spoke is equally distributed between the attachment on the wheel and the axle)

    i then get rotational accelerate (alpha) = 12 rad/s^2

    for the horizontal force, i have (20/32 slugs)(a-radial) = Tension

    vertical forces, I have (20/32slugs)(a-tangential) = vertical-Axle-force - 5lbs

    a-tangential = radius*alpha = 12ft/s^2

    vertical-axle-force = 12.5 lbs.

    thats as far as i got...

    dunno how to get the 18lbs (upward) which is the answer

  2. jcsd
  3. May 16, 2008 #2
    actually, i think i realized that solving the problem involves just the free body diagram of the spoke.

    the axle applies a force with unknown vertical component V and horizontal H

    the net force on the vertical for the spoke becomes something like:

    (10/32 slugs)(a-cm) = 10lb - V

    where a-cm is the acceleration at the center of mass


    by already knowing alpha, i can solve for a-cm:

    a-cm = (1/2ft)alpha

    plugging in, i get a-cm = 6 ft/s^2

    6(10/32) lb = 10lb - V

    V = 8lb which is the wrong answer :(
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