Projectile Motion with Drag: Understanding F=ma and dv/dx in Physics

[Helena54321]

Hi First ever post on Physics Forum.

N2: F=ma can be written as dv/dx=-gy-(bv/m)

g=grav. constant
b=constant where D(total drag force opposing dir. of
trav.)=-bv
y is simply y/vertical component
m=mass

I know dv/dt=a but what is dv/dx-working with the units I just got m^2s^-1. Sooo please can anyone enlighten me on this? Just started uni nd this is in lecture notes.

Ahh merci x x x

 

[Andrew Mason]

Hi First ever post on Physics Forum.

N2: F=ma can be written as dv/dx=-gy-(bv/m)

g=grav. constant
b=constant where D(total drag force opposing dir. of
trav.)=-bv
y is simply y/vertical component
m=mass

I know dv/dt=a but what is dv/dx-working with the units I just got m^2s^-1. Sooo please can anyone enlighten me on this? Just started uni nd this is in lecture notes.

Ahh merci x x x

That should be dV/dx. The V stands for potential energy. If one does work against gravity, there is an increase in potential energy dV = -dW where W is work. F = -dV/dx follows from dV = -dW = -Fdx .

And, by the way, welcome to Physics Forums!

AM

 

[Helena54321]

Hi Andrew, thank you very much for replying to my post :). In the lecture the lecturer said that it was an error in the notes and it should be dv/dt.

I just did a test in another lecture and walked out crying. There was one question (which probably seems very easy to you :) ) were you had to convert polar equations to cartesian ones. We also had to draw the cartesian graphs (just 2D).

a) rcos(th)
b)r=2asin(th)
c)r^2sin2(th)=2k
d)rsin(th+(pi/4))=a2^(1/2)

th=theta.

In our lecture notes for this course we have derivations for conversion from cylindrical-cartesian (3D) and spherical-cylindrical. (-=either way)

But when it comes to 2D cartesian and polar I'm like ?. I have no clue what to do. I know a) is a straight line where x=a and b) is a circle but only because my friend told me.

I feel quite lost. How do I approach polar and cartesian egtn conversions?? Thank you.

 

[Andrew Mason]

Hi Andrew, thank you very much for replying to my post :). In the lecture the lecturer said that it was an error in the notes and it should be dv/dt.

I just did a test in another lecture and walked out crying. There was one question (which probably seems very easy to you :) ) were you had to convert polar equations to cartesian ones. We also had to draw the cartesian graphs (just 2D).

a) rcos(th)
b)r=2asin(th)
c)r^2sin2(th)=2k
d)rsin(th+(pi/4))=a2^(1/2)

th=theta.

In our lecture notes for this course we have derivations for conversion from cylindrical-cartesian (3D) and spherical-cylindrical. (-=either way)

But when it comes to 2D cartesian and polar I'm like ?. I have no clue what to do. I know a) is a straight line where x=a and b) is a circle but only because my friend told me.

I feel quite lost. How do I approach polar and cartesian egtn conversions?? Thank you.

Polar co-ordinates are just another way of identifying a point on a plane. Rather than identifying the point by its position relative to the x and y axes, the point is identified by its displacement from the origin (distance and angle). You could give a position as "86.0 km east and 43.0 km north of the airport or you could say "97.5 km 30 degrees north of east of the airport".

To convert from polar to cartesian co-ordinates, $x = r\cos\theta \text{ and }y = r\sin\theta$. To go the other way:
$$r = \sqrt{x^2 + y^2)} \text{ and }\theta = \arctan (y/x)$$.

Polar co-ordinates are useful for graphing relations that are symmetrical about the origin.

In cartesian co-ordinates, the equation for a circle is
$$y^2 = R^2 - x^2$$
but in polar co-ordinates it is just much simpler r = R (ie. r is constant).

To describe a straight line in the cartesian plane y = mx+b. In polar co-ordinates, it is not so easy unless it passes through the origin (ie $\theta = constant$).

As to your particular questions, a) is not an equation. But if rcos(th) = a, (for $-\pi/2 < \theta < \pi/2$) the graph would be a straight line x = a. Just do the conversion above. For the rest of them, plot a few points for each and see if you can visualize them and try to work them out using the above.

AM

 

[PJC01]

Hi there,

Welcome to the Physics Forum! Projectile motion with drag can be a complex topic, but let's break it down step by step.

First, let's start with the equation F=ma. This is known as Newton's Second Law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In projectile motion with drag, we are interested in understanding the forces acting on an object as it moves through the air.

The first term in the equation you provided, -gy, represents the force of gravity acting on the object. This is the same as the acceleration due to gravity, g, multiplied by the object's mass, m. This term accounts for the downward acceleration of the object as it moves through the air.

The second term, -bv/m, represents the drag force acting on the object. This is a bit more complicated, as it involves the constant b, which takes into account the shape and size of the object, the density of the fluid (in this case, air), and the speed of the object. The negative sign in front indicates that the drag force acts in the opposite direction of the object's motion.

Now, let's look at the units of dv/dx. As you correctly stated, dv/dt represents acceleration, with units of m/s^2. When we take the derivative with respect to position, x, we end up with units of m/s. This represents the change in velocity (dv) over a small change in position (dx).

I hope this helps to clarify the equation for you. Keep in mind that this is just one part of understanding projectile motion with drag, and there are many other factors that can affect the motion of an object. Keep learning and exploring, and don't hesitate to ask for help when needed. Good luck with your studies!