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Homework Help: Bucket of water of is suspended by a rope wrapped around a pulley

  1. Jan 3, 2010 #1
    how do you do the last part.

    A 13.0 kg bucket of water of is suspended by a rope wrapped around a pulley, which is just a solid cylinder 0.300 m in diameter with mass 17.0 kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 12.0 m to the water. Neglect the weight of the rope.

    (a) What is the tension in the rope while the bucket is falling?
    50.367 N
    (b) With what speed does the bucket strike the water?
    11.925 m/s
    (c) What is the time of fall?
    2.013 s
    (d) While the bucket is falling, what is the force exerted on the cylinder by the axle?

    Attempt
    F=Mg-Ma
    then implied:
    a=(F-Mg)/M
    which seems unsound, it should have said:
    a = (Mg - F)/M
    ----------------------
    Now, let's begin and use the previous gentleman's notation:
    Mass of bucket=M
    mass of cylinder=m
    Tension force=F
    So, regarding the bucket:
    F + Ma = Mg
    F = Mg - Ma
    a = (Mg - F)/M
    Also, regarding the cylinder:
    T = Iα

    (By the dot product of torque, radial vector crossed with force, r cross F, which is Frsinθ for θ = 90 degrees)
    T = Iα = Fr = (1/2)(mr2)(a/r)
    where α = a/r
    And T = Fr because that's what torque is, just the radial distance times the force.
    So
    a = 2F/m
    and we also have:
    a = (Mg-F)/M
    So 2F/m = (Mg-F)/M
    2F = mg - mF/M
    2F+(mF)/M = mg
    F(2+m/M) = mg
    F = mg/(2+m/M)
    F = (167 kg*m/s2)/(2+1.31kg/kg)
    F = 50.367 N
    That is to say, the tension that the rope induces on the bucket is 50.5 Newtons in the upward direction (positive).
    --------------
    So we have solved Part (A)
    F = 50.367 N
    ---------------
    We know that:
    F = Mg-Ma
    So
    a = (Mg - F)/M
    where F = 50.367 N
    So
    a = [(13.0kg)(9.8m/s2) - 50.367N]/13.0kg
    a = (127N - 50.367N)/13.0kg
    a = 5.925 m/s2
    --------------------
    Now that we have a, let's find t,
    a = a
    v = at + vyo
    y = yo + vyot + (1/2)ayt2
    Counting positive y to be downward, (our acceleration will be positive then)
    12.0 m = 0 + 0t + (1/2)(5.925m/s2)t2
    t2 = (12.0m)/(2.963 m/s2)
    t2 = 4.051 s2
    t = 2.013 s
    That is to say, it takes the bucket 2.10 seconds to reach the water surface below.
    --------------
    So we have solved Part (C)
    t = 2.013 s
    ---------------
    a = a
    v = at + C
    But C, the constant of integration, is the initial velocity, is zero.
    so
    v = at
    where t =2.013 s
    a = 5.925 m/s2
    so
    v = (5.925m/s2)(2.013s)
    v = 11.925 m/s
    That is to say, the bucket is travelling at 11.925 m/s when it reaches the surface of the water.
    --------------
    So we have solved Part (B)
    v = 11.925 m/s
    ---------------
    The force exerted on the cylinder by the axle is just the torque (which is in the z direction, like a pole coming out of the page).
    We know that the torque is:
    T = Iα
    where moment of Inertia for a cylinder is
    I = (1/2)mr2
    and the angular acceleration is
    α = a/r
    with a = 5.925m/s2
    r = 0.150 m
    So
    T = Iα
    T = (1/2)(17.0kg)(0.150m)2[5.925m/s2]/(0.150m)
    T = (8.5kg)(0.150m)(5.925m/s2)
    T = 7.555 N*m
    or
    T = 7.555 J

    T= Fr
    F=T/r
    F= 7.555/0.150
    F= 50.367 N
     
  2. jcsd
  3. Jan 3, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Since the axle is frictionless, it exerts no torque. (If it did exert a torque, your previous work would be incorrect since it ignored it.)

    Hint: Consider the forces acting on the cylinder. What must be the net force?
     
  4. Jan 3, 2010 #3
    so is it just the force of tension. that means the answer to a and d are the same.
     
  5. Jan 3, 2010 #4

    Doc Al

    User Avatar

    Staff: Mentor

    No. You're overlooking one of the forces acting on the cylinder.
     
  6. Jan 3, 2010 #5
    oh would i do
    force of gravity + tension = force on cylinder
    mg + 50.367 =
    17(9.8) + 50.367=
    216.967 N
     
  7. Jan 3, 2010 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Good.
     
  8. Jan 3, 2010 #7
    thank you for your help
     
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