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Homework Help. ^_^

  1. Oct 28, 2006 #1
    I've tried these problems but I need some help.

    44. a small mass m slides without fritction around the looped aparatus show in the picture below. If the object is to remain on the track, even at the top of the circle (whose radius is r), from what minimum height H must it be released?


    71. A paratrooper fell 370m after jumping from an aircraft without his parachute opening. He landed in a snowback, creating a crater 1.1m deep, but survived with only minor injuries. Assuming the paratroopers mass was 80kg and his terminal velocity was 30m/s, estimate the work done on him by air resistance as he fell.

    74. In a film of Jesse Owens's famous long jump in the 1936 Olympics, it is observed that his center of mass rose 1.1m from launch point to the top of the arc. What minimum speed did he need at launch if he was also noted to be traveling at 6.5m/s at the top of the arc?

    76. A ball is attached to a horizontal cord of length L, whose other end is fixed, (picture below). (a) If the ball is released, what will be its speed at the lowest point of its path?(b)A peg is located a distance H directly below the point of attachment of the cord. If H=.8L, what will be the speed of the ball when it reaches the top of its circular path about the peg?


    81. Water flows over a dam at the rate of 550 Kg/s and falls vertically 80m before striking the turbine blades. Calculate: the rate at which mechanical energy is transfered to the turbine blades, assuming 60% efficiency.

    85. A 75Kg student runs at 5.0 m/s, grabs a rope, and swings out over a lake. He releases the rope when is velocity is zero. (a) What is the angle (-)(theta) when he releases the rope? (b) What is the tension in the rope just before he releases it? (c) What is the maximum tention in the rope?

    Thanks to all who respond. ^_^
  2. jcsd
  3. Oct 28, 2006 #2


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    Homework Helper

    Well, if you're stuck somewhere, you could perhaps point out where the problem is.
  4. Oct 28, 2006 #3
    N + mg = mv^2 /r -at the top of the loop
    N goes to zero and
    mg = (mv^2)/r

    so v minimum = √rg

    now use energy conservation
    mgh + 0 = 0.5m(√rg)^2
    h = 0.5r

    and your H = 2r+ h = 2.5*r
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