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I really need help

  1. Oct 27, 2004 #1

    I tired this question but i don't get it :frown:

    A 100 kg pilot sits on a scale, as the plane reaches the bottom of the loop, what does the scale read? The plane is travelling at 138.88 m/s. The radius is 200m

    So i did this -->

    Fc= Fscale - Fg

    (100)(138.88)^2 / 200 = Fscale - 980
    Fscale = 10623.87 N

    I have no clue if this right..since the teacher never gave us the answer..just the question.. Can someone plz help..he said this question mite be on a quiz :frown:

    PLz! ...Thank u !

  2. jcsd
  3. Oct 27, 2004 #2


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    Homework Helper

    Hi, Tanya.

    First off, you need to keep straight on the dimensions you're talking about. Weight is a force, for instance, while the 9.80 or 980 is an acceleration.

    Second, you need to think about what's going on in the loop. Imagine swinging a rock on the end of a string in a vertical circle. (If you've never done something like that, go do it. It'll help a lot.) How hard does the string pull on your hand as it goes around? If you do it, you'll notice that it pulls harder when the rock is at the bottom of the loop than when it is at the top. You can, in fact, swing it just right so that it doesn't pull at all when it's at the top.

    Think of it in terms of centripetal force - the force that acts towards the center of the circle, that's necessary for any object to undergo circular motion. When the rock is at the top, gravity is acting directly towards the center of the circle. If you're swinging it at the right speed, gravity will provide all of the centripetal force necessary, and the string won't have to pull at all - no tension in the string. In that case, at the bottom of the loop, gravity is pulling directly away from the center of the circle. In that case, the string must provide both the centripetal force as well as counteracting gravity - the tension in the string will be twice the weight of the rock. It will vary from zero at the top to 2W at the bottom as it moves through the loop, going up as the rock goes down and vice versa.

    For your problem: First, figure out the centripetal acceleration necessary for a loop of that radius at that speed. You should have a formula for that. Once you have it, compare it to gravitational acceleration (9.80 m/s^2). At the bottom of the loop, the acceleration will be g + a(centripetal). Multiply that by the mass of the pilot and you'll have the scale reading you need.

    Hope this helps.
  4. Oct 28, 2004 #3
    Thank u Diane

    Ohhhh now i see!! Thank u Diane :biggrin:

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