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Equation of motion - choosing directions for the coordinate axis

Physics Asked by Jyotishraj Thoudam on July 13, 2021

I’m not sure this equation is the right one if we invert the direction of the positive $y$-axis.

Am I right if we assume the positive direction of the $y$-axis downwards, that the correct equation of motion would be

$$mddot{y}=mg-{frac{rho A C}{2}dot{y}^2}$$

as the spacecraft tries to land downwards?

Where would be the direction of acceleration due to the net force?

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3 Answers

It is up to you to decide which coordinate system you'd like to use.

In your case, you have chosen the downward direction of the y-axis to be the positive direction. Therefore, the sign of the $mg$ will be positive.

Since the $mg$ force is acting downwards, the term will be positive. The force which is acting upwards will be negative.

$mddot{y}=mg-{frac{rho A C}{2}dot{y}^2}$

Your diagram however, hints that you should use the upward direction as the positive direction. In this case, the signs of the quantities will flip.

$mddot{y}={frac{rho A C}{2}dot{y}^2} - mg$

Answered by Yashas on July 13, 2021

I believe that the original equation as presented in the OP is ill-formed.

A correct vector equation should not be affected by an arbitrary choice as to the positive direction for the vectors.

In the equation given, the LHS will change sign under such a change in the positive direction. A landing space ship that is slowing to a stop will have a numerically positive upward acceleration, or a numerically negative downward acceleration.

Likewise the acceleration of gravity (second term, RHS) will numerically change in sign. If up is positive, $g=-9.8$: if down is positive, $g=9.8$

However, the drag term, with the square of the velocity, is always non-negative. It does not switch signs, and the equation fails.

The inclusion of a unit vector in the drag term, with a direction opposed to the velocity, would make the equation correct...

Answered by DJohnM on July 13, 2021

The correct form of the equation is

$$ mddot{y}=mg-{frac{rho A C}{2}|dot{y}|,dot{y}} $$

or

$$ begin{cases} mddot{y}=mg-{frac{rho A C}{2}dot{y}^2} & dot{y}>0 mddot{y}=mg+{frac{rho A C}{2}dot{y}^2} & dot{y}<0 end{cases} $$

This is because air resistance must oppose the motion $dot{y}$.

Answered by John Alexiou on July 13, 2021

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