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Two Extra Credit Rotation Questions

  1. Dec 8, 2004 #1
    1) The pulley on an Atwood machine has a moment of inertia of 5.0 KG and a radius of .5 meters. The cord supporting the masses does not slip and the axle is frictionless. (A) Find the acceleration of each mass if M1 = 2Kg and M2 = 5Kg. (B) Find the tension in the cable supporting M1 and the tension in the cable supporting M2.

    My Analysis:

    Obviously M2 is going to accelerate it down. M1 and the Pulley are what is resisting the pull. I know this but am having a hard time wiht how to set up the formula to solve it.

    2) A model airplane whose mass is .75 KG is tethered by a wire so that it flies in a circle 30 meters in radius. The airplane engine provides a net thrust of .8 newtons perpendicular to the tethering wire. (A) Find the torque the net thrust produces about the center of the circle. (B) Find the acceleration oif the airplane when it is in level flight. (C) Find the linerar acceleration of the airplane tanget to its flight path.

    My Work:

    I got part A and C correctly-
    A) 24 N/m
    C) 64 Rad/s

    Again I am confused on what exactly they are asking for in part B, and am confused with what formula to use. I originally thought it would be the tangential acceleration added to the centripetal acceleration, but it came off way wrtong. The answer comes out to be .0356 Rad/s, but again, I cannot figure it out.
  2. jcsd
  3. Dec 8, 2004 #2


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    Homework Helper

    1) There are three "masses" in the system - M1, M2 and the pulley. Do a free-body diagram for each of them. Remember that, for the masses, F = ma and for the pulley, T = IA (T = torque, A = alpha, the angular acceleration. I have got to learn to do Greek letters on this thing.)

    2) I didn't check your answers on a and c, so no guarantees there. On b, you know that the airplane is not moving at constant velocity, so there must be an acceleration. In steady-state, the speed will be constant - so what kind of acceleration must be acting on it, given that it is moving in a circle (hint hint). That should be relatively easy to calculate.
  4. Dec 8, 2004 #3
    1) Sum of the forces will equal (weight of M2) - (weight of M1) - (torque of the pulley?)

    2) Centripetal acceleration...V^2/R.

    R=30 M but I get confused with what to put in for V
  5. Dec 8, 2004 #4


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    No. The sum of the forces on M1 will equal M1(a), where a is the acceleration of M1. The sum of the forces on M2 will equal M2(a), where a is the acceleration of M2. You should be able to see why the acceleration is the same in both cases.

    The sum of the torques on the pulley will be IA (again, A is the angular acceleration). There is a relation between a and A - it should be in your text, and is easy to figure out if not.

    You'll end up with three equations and (I think - I haven't checked) two unknowns. That is eminently solvable.

    Is that enough?

    Edit: Oh - and be sure to keep track of your directions. These are vector equations, after all. The easiest way to do it is to define a direction along the rope - I would suggest making towards M2 positive and towards M1 negative.
  6. Dec 8, 2004 #5
    I have been sick with the flu for about a week which may be why I am not getting these basic things.

    Wont the pully only accelerate at the same rate as the blocks? Yes two unknowns would be solvable.

    If down is positive then

    Sum of forces on M2 = M2a
    Sum of forces on M1 = M1a
    Confused on the torque equation, sorry for seeming sort of slow, really appreciate the help.

    And on #2
    2) Centripetal acceleration...V^2/R.

    R=30 M but I get confused with what to put in for V
    Last edited: Dec 8, 2004
  7. Dec 8, 2004 #6


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    Honey, you don't want to use the vertical for your direction. The problem is that one block will accelerate up and the other down, but the system is really only accelerating in one direction. That's why I suggested using the direction of the rope for your directions. Does that make sense?

    As to the pulley accelerating at the same rate as the blocks - well, yes and no. The pulley will not actually accelerate - it's stationary. It will start to turn, though. The angular acceleration of the pulley will be caused by the acceleration of the blocks (through the medium of the rope), so in that sense you're right. But you're going to have to use angular quantities with the pulley.

    As far as the v^2/r - well, that's a valid equation for centripetal acceleration, but that's not what we're dealing with here. There will be a centripetal acceleration, but it doesn't act on the pulley. The pulley is rotating, not moving in a circle. Do you see the difference? What you're looking for is a relation between the acceleration of a point on the edge of the pulley and the angular acceleration of the pulley as a whole. It's very similar to the relation between the tangential velocity of a point on the rim of a wheel and the angular velocity of the wheel.

    Let me know if this is still confusing, and I'll get a little more specific. It is better if you can work through it on your own, though. :)
  8. Dec 8, 2004 #7
    looks like your right, triedo ut a few things, got it to work out. Thanks for all the help
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