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Rotational motion of a door on a moving car

  1. Dec 8, 2015 #1
    1.
    A car door is left open at an angle theta with the side of the car. The door is uniform of mass M and length L. The car accelerates with linear acceleration a. How long does it take the door to close.

    2. t
    heta = .5(angular acceleration)t^2
    angular acceleration = tangential acceleration/radius

    3. tangential acceleration = a sin (theta)

    plug in and solve for t? not sure if this makes sense.
     
    Last edited by a moderator: Dec 8, 2015
  2. jcsd
  3. Dec 8, 2015 #2

    berkeman

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    The variable a is the linear acceleration, not the angular acceleration. The Moment of Inertia (MOI) probably needs to be considered in this problem. Show us the Relevant Equations for MOI and angular acceleration, and also please draw the Free Body Diagram (FBD) of the door for this problem. :smile:
     
  4. Dec 8, 2015 #3

    haruspex

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    i don't think there's a confusion between linear and angular in the proposed solution. Rather, it's a confusion between the linear acceleration of the car and that of the door. It treats a as the linear acceleration of the mid point of the door in the tangential direction.
    @physicstime , are you comfortable using non-inertial frames? If so, treat the car's acceleration as a force on the door and consider torque.
     
  5. Dec 9, 2015 #4
    i don't know where to apply the torque. because the car is applying a force at the hinge i tried getting the tangential component there and used the the center of mass as the pivot point. but i can't find a force that makes the center of mass want to go to the right towards the car.
     
  6. Dec 9, 2015 #5

    TSny

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    Don't assume that the direction of the force on the door from the hinge is parallel to the acceleration of the car.

    As haruspex pointed out, the problem is easier when analyzed in the frame of reference of the accelerating car. A uniformly accelerated frame acts like a non-accelerated frame with an added gravitational field in the direction opposite the acceleration of the frame.

    But if you are not familiar with this idea, then you can stick with the frame of reference of the earth.
     
  7. Dec 9, 2015 #6
  8. Dec 9, 2015 #7
    because the force of the car is on the hinge i'm breaking the force into a component normal to the door and one parallel to the door.
     
  9. Dec 9, 2015 #8
    Maybe it would easier to consider a linear force, of constant direction, applied to the end of
    a stick on a frictionless surface.
     
  10. Dec 9, 2015 #9

    TSny

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    OK, good. Can you see what causes the CM of the door to move to the right?
     
  11. Dec 9, 2015 #10

    haruspex

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    i assume that's using the inertial frame, so your linear force is the door hinge, yes? But that force will not be constant.
     
  12. Dec 9, 2015 #11
    no it won't be constant and it goes to 0 as the door closes.
     
  13. Dec 9, 2015 #12
    My initial comment about a stick on ice appears to be rather useless.
    Although I am unfamiliar with non-inertial frames, it seems that the solution to
    this would be the same as that of finding the period of a physical pendulum
    in a frame where g is replaced by the acceleration of the car and that is
    a rather formidable problem since you can't use the small-angle approximation
    when the initial angle is pi / 2.
     
  14. Dec 9, 2015 #13

    haruspex

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    Agreed.
     
  15. Dec 10, 2015 #14
    ok, so here's what i tried. using W = KEf - KEi = 1/2 I (omega - final)^2 and W = ∫τdθ I can solve for the final angular velocity when the door has closed. I'm just not sure how to get the time it took for it to close from there.
     
  16. Dec 10, 2015 #15

    haruspex

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    That formula is wrong, but since omega here will be zero you will get away with it.
    You cannot find the time taken without considering the force/torque/aceleration equations. Even then, as J Hann noted, the differential equation you get cannot be solved unless you assume the initial angle of the door is small.
     
  17. Dec 10, 2015 #16
    the initial angular velocity is 0, i knowingly excluded it. ok, i tutor mathematics. while tutoring AP Calc a student posed this problem to me. he is in AP Physics. is there NO way to solve this in a manner a high school student would follow?
     
  18. Dec 10, 2015 #17

    haruspex

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    As J Hann mentioned, the problem is equivalent to a simple pendulum. If the course covers that, it covers this too. But finding the period of a simple pendulum (at this level) requires the small angle approximation, resulting in the standard differential equation for simple harmonic motion. The question as stated did not specify that theta is small.
     
  19. Dec 10, 2015 #18

    TSny

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  20. Dec 10, 2015 #19
    I said a "physical" pendulum, but I think the derivation is similar but uses the moment of inertia instead of the mass.
     
  21. Dec 10, 2015 #20

    TSny

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    Right. The percent error caused by using the small angle approximation for the period is the same for simple and physical pendulums.
     
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