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14 October, 15:58

If the 250-lb block is released from rest when the spring is unstretched, determine the velocity of the block after it has descended 5 ft. The drum has a weight of 50 lb and a radius of gyration of k0 =.5 ft about its center of mass 0

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  1. 14 October, 19:42
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    1080 rpm

    Explanation:

    The drum will follow a constant acceleration for a given time and then maintain its speed.

    The constant acceleration will follow:

    w (t) = w0 + γ * t

    Where

    w0 = initial angular speed (zero in this case)

    γ = angular acceleration

    The angular acceleration will depend on the torque applied and the inertia moment of the drum

    T = γ * J

    The block will apply a torque of:

    T = P * r

    T = 250 * 0.5 = 125 lb*ft

    The drum weights 50lb, so it has a mass of

    m = P/g = 50/32.2 = 1.55 lb*s^2/ft

    The drum has a moment of inertia of

    J = 1/2 * m * r^2

    J = 1/2 * 1.55 * 0.5^2 = 0.194 lb*s^2*ft

    (remember that lb is a unit of force not mass, but lb*s^2/ft is an indirect unit of mass)

    So, the angular acceleration is

    γ = T/J

    γ = 125/0.194 = 644 rad/s^2

    Now we need to know how long will it take the block to fall 5 ft.

    Since wee know the drum has radius r = 0.5 f, we need to find how much it needs to spin to release 5 ft of rope.

    The circumference is 2π*r = 3.14 feet, it will take it 5/3.14 = 1.59 turns to release 5 feet of rope. 1.59 turns is 2π*1.59 = 10 radians.

    The equation for the rotation is:

    a (t) = a0 + w0 * t + 1/2 * γ * t^2

    a0 and w0 are zero

    10 = 1/2 * 644 * t^2

    t^2 = 0.031

    t = 0.176 s

    Then, the angular speed is:

    w (0.176) = 644 * 0.176 = 113.3 rad/s

    Which is 113.3/2π = 18 turns per second or

    18 * 60 = 1080 rpm
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