mechanics.org

Classical Mechanics: Notes

Newton’s Laws

Mass

resistance to being accelerated

Weight

gravitation force of an object

Momentum

mass * velocity

1 Newton

force to accelerate 1 kg by 1 m/s^2

First Law

In the absense of forces - a point mass moves with constant velocity v (= zero acceleration)

Second Law

The net force on a particle is F=ma
Where a = dv/dt = d²r/dt² \ In other words, force is a change in momentum p. ie. F= dp/dt

Solve for velocity and position

Ex: Accelerting a particle with force F₀ we can write out the equation for the second law

F₀ = m ⋅ a(t) = m ⋅ d²x(t)/dt²

Then by integration we can solve

∫F₀ = ∫m⋅d²x(t)/dt²
∫F₀/m = ∫d²x(t)/dt² \ [F₀/m] t = v₀ + dx(t)/dt \

With this, given a starting velocity v₀ and a time t we can solve for the velocity dx/dt

∫[F₀/m] t = ∫v₀ + dx(t)/dt
[F₀/2m] t² = v₀t + x₀ + x(t) \

Integrating once more we get an equation for the position. We just need a starting position x₀ to solve

Third Law

If Object1 exerts a force F₂₁ on Object2
Then Object2 always exerts an opposite force F₁₂ \ Such that F₁₂ = -F₂₁

Conservation of momentum

Givens:

• If you have a system with many particles α = [1,2 … N]
• They extert an influence on each other particle F_α{ β}
• They are under some external force F^ext_α{}

The total force on each particle will be:

F_α{} = ∑_β{≠α} F_α{β} + F^ext_α{}

This will equal the change in momentum for that particle - dp_α{}/dt (b/c Force = change in momentum)

dp_α{}/dt = ∑_β{≠α} F_α{β} + F^ext_α{}

The total momentum of the system will be

P = ∑p_α{}

The corresponding change in momentum of the system will be

dP/dt = ∑dp_α{}/dt

(the derivative of a sum is the sum of derivatives)
And then we just plug in..

dP/dt = ∑_α{}∑_β{≠α} F_α{β} + ∑_α{}F^ext_α{}

But for every term F_ab = -F_ba, so all the terms cancel out in the first sum

dP/dt = 0 + ∑_α{}F^ext_α{}
dP/dt = F^ext

And the total change in momentum is just the sum all the external forces. The forces internal to the system can’t change the momentum of the system as a whole!

leveraging cartesian coordinates

We can also break up forces along coordinates to make our lives easier by using unit vectors ex: x^{^}, y^{^}

F = F_xx^{^} + F_yy^{^}

And these components can be treated separately b/c they differentiate independently

r = x ⋅ x^{^} + y ⋅ y^{^}
dr/dt = dx/dt ⋅ x^{^} + dy/dt ⋅ y^{^} \ d²r/dt^2 = d²x/dt^2 ⋅ x^{^} + d²y/dt^2 ⋅ y^{^}

So our F=ma can be rewritten

F = F_xx^{^} + F_yy^{^} = m ⋅ d²x/dt^2 ⋅ x^{^} + m ⋅ d²y/dt^2 ⋅ y^{^}

And each force can be looked at explicitely

F_x = m ⋅ d²x/dt²
F_y = m ⋅ d²y/dt²

So if you have a sliding block on a plane.. you set your coordinate system so that x^{^} is along the plane and y^{^} is orthogonal to the plane and then you can look at the forces separately.

In the y direction there is no movement so the normal force cancels out with the force of gravity

F_y = N - mg cos(θ) = 0
so we can solve for the normal force N \ N = mg cos(θ)

In the x direction there is gravity and friction - where friction is a function of the normal force N

f = μN
f = μmg cos(θ)

And gravity we already know how to calculate

gravitation force in the direction of x^{^} = mg sin(θ)

So the total force in x^{^}

F_x = m ⋅ d²x/dt² = mg sin(θ) + f
m ⋅ d²x/dt² = mg sin(θ) + μmg cos(θ)

And we can solve for d²x/dt², dx/dt, x

d²x/dt² = g sin(θ) + μg cos(θ)
dx/dt = [g sin(θ) + μg cos(θ)]t \ x(t) = (1/2)[g sin(θ) + μg cos(θ)]t^2

(these assume the initial velocity and position are 0)

Projectiles

A class of problems involve projectiles traveling through air. These undergo two forms of drag - quadratic and linear grad - with the former dominating for large object and the latter dominating for small things (like small oil drops and mist)

Linear Drag

This kind of drag decreases the speed of an object proportional to the current speed

dv/dt = -kv

When you have a variable that changes as a function of itself then you should think of the exponential e^-kt. It has the property that its derivative is equal to itself

v = Ae^-kt

Dropping a weight

F_falling = F_gravity + F_drag
m⋅dv(t)/dt = mg - bv(t)

Solving this is a bit more difficult. You proceed by subsitution..

u=(v(t)-mg/b)
du/dt = dv(t)/dt \ \therefore{} \ m⋅dv(t)/dt = mg - bv(t) \ m (dv(t)/dt) = -b ((v(t)-mg/b) \ m du/dt = -b u \

We have gotten it into the same form as the simpler problem and therefore

u = Ae^-tb/m = (v(t)-mg/b)
\therefore{} \ v(t) = Ae^-tb/m+mg/b \ and \ dv(t)/dt = -A(b/m)e^-tb/m

A is still a free variable, which we constrain by adding a starting velocity so that at t=0 -> v(0) = v₀

v₀ = Ae⁰+mg/b
A = v₀ - mg/b

So the final velocity is

v(t) = (v₀ - mg/b)e^-tb/m+mg/b

..
Pluggin back into the original equation to double check

m⋅dv(t)/dt = mg - bv(t)
-mAe^-tb/m = mg - b(Ae^-tb/m+mg/b) \ ✓

Note: When the net force is zero and acceleration stops (the terminal velocity)

m⋅dv(t)/dt = mg - bv(t)
0 = mg - bv_terminal \ v_terminal = mg/b

Alternately you can put in t=∞ into our equation for v(t)

v(t) = (v₀ - mg/b)e^-tb/m+mg/b
v(∞) = (v₀ - mg/b)e^-∞{b/m}+mg/b \ v(∞) = mg/b