# Homework Help: Just Need Help With Some Wording

1. Sep 19, 2010

### Hotwir3

1. Problem

My professor has given four problems, all of which I understand how to do, but he is asking us to "Write down the equation of motion for the system."

The force is given or must be contrived in each problem.

2. Relevant equations

We're using Newton's Second Law
F=ma

3. The attempt at a solution

So should I set the Force equation on the object equal to "m(dv/dt)" and integrate so that it's basically the integral of:

Fgiven=ma
Fgiven=m dv/dt

Fgiven/m (dt) = dv

which gives an equation for the velocity.

So basically, is an equation for the velocity what the professor means by "equation of motion for the system"?

2. Sep 19, 2010

### RoyalCat

The equation of motion for the system, in this context, is simply the formulation of Newton's Second Law, or of the conservation of mechanical energy in the system, where applicable.

The equation of motion is a differential equation which relates the object's position to its time derivatives. Given the proper initial conditions you can describe the time evolution of the system from this equation.

To give two prominent examples, let's look at two simple systems.

The first is a point mass under the influence of a constant force (Constant in direction and in magnitude), starting at rest.
Its equation of motion would read:

$$ma = F$$

Introducing the notation:
$$\ddot x = \frac{d ^2 x}{dt^2}$$

$$m\ddot x = F$$

$$\ddot x = \frac{F}{m}$$

Integrating once: $$v-v_0 = \frac{F}{m}t$$
Integrating once more, remembering that $$v=\frac{dx}{dt}$$
$$x=\tfrac{1}{2}\frac{F}{m}t^2 + v_0 t + x_0$$
This should be familiar to you as the good ol' kinematic equation for 1d motion at constant acceleration.

Another example, simple harmonic motion.
You get simple harmonic motion when you have an acceleration directly proportional to the displacement. For instance, one such system would be a point mass attached to a spring.
Newton's Second Law reads: $$F=-kx$$
And the equation of motion would be: $$\ddot x = -\omega ^2 x$$ where we have defined $$\omega ^2 \equiv \frac{k}{m}$$

Here the solution isn't quite as trivial, and until you learn how to solve differential equations properly, you just need to memorize that the solution to $$\ddot x = -\omega ^2 x$$ is in general $$x=A\cos{(\omega t + \phi)}$$ where $$A, \phi$$ are constants to be determined by the initial conditions.

3. Sep 19, 2010

### Hotwir3

Thanks for the response!

It just seems like "Equation of motion" would refer to the velocity equation, since that describes the motion, but either way, it's just another integral.

Can anyone answer this follow-up question?

I have one question that has two forces. It's a projectile that has the force of gravity acting on it and linear drag force.

What is F?

I have F = -mg - bv

The linear drag makes sense because it is the negative drag coefficient (b) times the velocity vector, so the force direction opposes the velocity direction.

HOWEVER, how do I represent that -mg is in the negative y-direction? With a j? Unit vector in the y-direction?

The way this equation is written out right now would only work for a projectile shot straight up.

4. Sep 19, 2010

### RoyalCat

The equation of motion is any differential equation which relates the position of the mass to the different derivatives of the position (Usually with respect to time).
Motion isn't just the velocity.

Your force equation isn't making sense because you haven't made a clear free-body diagram.
You need to break the forces down into two components. In the x direction, $$\Sigma F_x = -bv_x=-b\frac{dx}{dt}$$
Which leads to: $$m\ddot x = -b\dot x$$

In the y direction, $$\Sigma F_y = -mg -bv_y$$
Which leads to: $$m\ddot y = -mg -b\dot y$$

5. Sep 19, 2010

### Hotwir3

I'll just give the whole problem.

A projectile subject to linear air resistance is launched from the origin with velocity vo at angle (theta)

a. Apply Newton's second law and write down the equation(s) of motion for the projectile.
b. Solve the equation for the x-component of the motion
c. Solve the equation for the y-component of the motion

Last edited: Sep 19, 2010
6. Sep 19, 2010

### Hotwir3

So the final equations that you gave above for x and y directions, would those be the equations of motion (part a) and I integrate in order to get down to the position equation?

7. Sep 20, 2010

### RoyalCat

If we're on the subject of wording, when you solve a differential equation, it isn't always by integrating (Although for many simple systems, that is the most elegant solution). In many cases it takes a bit of guesswork and skulduggery to get to the solution, and an integral wasn't always part of that solution.

The term we use is to solve the differential equation, and the solution we are usually after is the position as a function of time, from which we can easily derive the velocity and acceleration as functions of time as well.

Before you start trying to solve the equations I posted, draw the free body diagram and convince yourself why these are the equations of motion for the body in question, and how they come from Newton's Second Law.