force perpendicular to the motion-really need help

Discussion in 'Undergraduate Math' started by something_about_math, Sep 23, 2005.

  1. hello

    this is really confusing me

    Object in uniform circular motion is moving around the perimeter of the circle with a constant speed and while the speed of the object is constant , its velocity is changing . The direction is always directed tangent to the circle due to acceleration towards the center


    Why is it that if the direction of the force remains perpendicular to the motion will the direction change without the speed changing ?

    Is it that since centripetal force only pulls on object for such a brief period of time ,that it only has time to change velocity's direction and then centripetal force already changes its position and the story again repeats itself ?

    So if object A is moving with constant speed along the x axis with velocity vector V ( V for now only has x component ) and something starts pulling it along the y axis with force F1 , then for a brief instant when force F1 is perpendicular to velocity vector V force F1 should change direction of V so that now V also has y component , but still has same magnitude .
    And after that ( after it is no longer perpendicular to velocity vector )F1 should also start changing V's magnitude ( by increasing y component )

    My point being that if it is true that if direction of force remains perpendicular to the motion will the direction change without the speed changing , then magnitude of x component of an object A shouldn't be equal to the magnitude of velocity object A had prior to pulling on A with force F1 , since F1 already changed the direction of velocity V

    I hope someone can clear this up for me

    thank you
     
    something_about_math, Sep 23, 2005
    #1
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  2. The centripetal force acts all the time, not for "brief periods". You
    should think of the process as a continuous one, not one that moves in
    discrete jumps. (However, discretising it and taking limits is one way
    to do the maths.)
    Pretty much, provided you realise that "brief" instant means "an
    instant", i.e. no time whatsoever. This situation arises, for example,
    when a ball is thrown in a parabolic trajectory. At the top of the
    trajectory the ball travels horizontally for an "instant". The duration
    of the period when it's travelling horizontally is zero.
     
    matt271829-news, Sep 23, 2005
    #2
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  3. And, as a consuquence, there is NO time when V has a non-zero y
    component but still has the same magnitude.
     
    matt271829-news, Sep 23, 2005
    #3
  4. something_about_math

    Jim Spriggs Guest

    Will it be deemed irrelevant to introduce Cartesian coordinates? Let's
    do so anyway, because it makes things so easy to understand.

    Newton's second law says that if a force (F) acts on a body then the
    body accelerates (a) in the direction of the force and the acceleration
    is proportional to the force. The constant of proportionality is called
    (inertial) mass (m). Hence

    F = ma

    Remember: F and a are vectors, m is a scalar. Let's swing a body (B)
    around on the end of a piece of string. The only significant force
    acting on the body is the tension in the string which we'll assume is
    constant. Now consider the Cartesian coordinates:

    y
    ^
    |
    | B
    | /
    | /T
    | /
    |/theta
    O-------------->x

    where vector T from B to O represents the tension. Resolving along the
    x and y axes:

    F = (-T sin theta, -T cos theta) = ma.

    So

    a = (1/m)(-T sin theta, -T cos theta).

    Acceleration of the derivative of velocity (v) w.r.t. time, so

    a = (1/m)(d/dt)(T cos theta, -T sin theta),

    v = (1/m)(T cos theta, -T sin theta).

    Speed is the magnitude of velocity. The magnitude of a vector w is

    the positive square root of w dot w

    so

    speed = sqrt((1/m)(T cos theta, -T sin theta)
    dot (1/m)(T cos theta, -T sin theta))

    = sqrt(T^2/m^2(cos^2 theta + sin^2 theta))

    = T/m.

    So we see that the speed is constant if the tension and mass are.

    Furthermore, two vectors are perpendicular if their dot product is
    zero. Consider

    a dot v = (1/m)[(-T sin theta, -T cos theta)
    dot (T cos theta, -T sin theta)]

    = (T/m)(-sin theta cos theta + cos theta sin theta)

    = 0.

    So acceleration and velocity are perpendicular.

    Note the assumptions: Newton's second law holds, mass is constant, the
    force is constant and central.
     
    Jim Spriggs, Sep 24, 2005
    #4
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