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Ls 發問於 科學及數學物理學 · 3 年前

Question about circular motion on a horizontal road?

Why can't we take the moment on the left wheel?

In the case,

2d(N2) - mgd

= mgd + m(v^2)h/r - mgd

=m(v^2)h/r ≠ 0

Shouldn't there be an anticlockwise moment about the left wheel?

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1 個解答

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  • 天同
    Lv 7
    3 年前
    最愛解答

    Be aware that the Principle of Moment, which states that the clockwise moment about a point on an object equals to the counter-clockwise moment, is only applicable to an object in STATIC EQUILIBRIUM, i.e. an object at rest. But in this question, the car is NOT in static equilibrium. It is in motion under an acceleration (the centripetal acceleration). Hence, the application of the Principle of Moment in this case is very limited.

    In physics, an object, which is under an acceleration, is said to be in a "non-inertia reference system"(非慣性系統). In such situation, one needs to include the "inertia force (慣性力)", or sometimes called "pseudo force", into consideration. In this example, the "inertia force" is the centrifugal force (離心力) that acts radially outward on the centre of gravity of the car. This is the physical situation that is experienced by the driver or passenger inside the car. By including the inertia force, the laws of physics that is originally applicable to "inertia reference system (慣性系統)" (i.e. objects at rest or in motion at constant velocity) will equally apply.

    Now, if we take moment about the left wheels as what you want, there are three moments arising from three forces, namely: the normal reaction from the right wheels N2 (counterclockwise moment), the weight of the car (mg) (clockwise moment) and the centrifugal force at the centre of gravity of the car Fc (clockwise moment).

    To an observer (driver or passenger) inside the car, the car is in equilibrium to him (i.e. the car has no relative motion to him), and hence the Principle of Moment does apply. We have,

    N2.(2d) = mg.(d) + (Fc).(h)

    But Fc = mv^2/r

    thus, 2d.(N2) = mgd + m(v^2)h/r

    i.e. N2 = mg/2 + m(v^2)h/2dr = (m/2)[g + (v^2)h/rd]

    This is exactly the equation for N2 obtained in the example.

    As said above, the result of your derivation is expected, as you have tried to apply the Principle of Moment to an accelerated car without considering the effect of centrifugal force. The "net moment" will thus turn out not to be zero.

    Should you include the centrifugal force into your calculation, the equation would become,

    Net moment = 2d(N2) -[mgd + m(v^2)h/r] = [mgd + m(v^2)h/r] - [mgd - m(v^2)h/r]

    = 0

    This is the expected result.

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