Notes
Center of mass
- For particles: $\displaystyle x_{cm} = \frac{\sum m_ix_i}{\sum m_i}$.
- For a continuous object: $\displaystyle x_{cm} = \frac{\int x\,dm}{\int dm}$, with $dm = \lambda\,dx$ for a rod of linear density $\lambda$.
Newton's laws with calculus
- $\displaystyle \vec F_{net} = m\vec a = \frac{d\vec p}{dt}$.
- Ramps: $F_N = mg\cos\theta$ and the component of gravity down the ramp is $mg\sin\theta$.
- Atwood machine (two masses over a pulley): $\displaystyle a = \frac{(m_2 - m_1)g}{m_1 + m_2}$ for an ideal pulley.
- Systems: treat connected objects as one system to find $a$, then isolate one object to find tension.
Resistive (drag) forces
A force like $F = -bv$ leads to a differential equation:
- Solution starting from rest: $\displaystyle v(t) = \frac{mg}{b}\left(1 - e^{-bt/m}\right)$.
- Terminal velocity happens when drag balances weight: $\displaystyle v_T = \frac{mg}{b}$.
Circular motion and gravitation
- Net radial force $\displaystyle = \frac{mv^2}{r}$. On a banked curve or loop, use the real forces (normal, gravity, friction, tension).
- At the top of a vertical loop, the minimum speed to stay on track satisfies $\displaystyle mg = \frac{mv^2}{r}$, so $\displaystyle v = \sqrt{gr}$.
- Universal gravitation $\displaystyle F = \frac{Gm_1m_2}{r^2}$. Circular orbit speed $\displaystyle v = \sqrt{\frac{GM}{r}}$.
Practice
$a = g\sin 30°$.
$\displaystyle a = \frac{(5 - 3)(10)}{8}$.
Set $mg = bv_T$.
As $v$ grows, drag grows and the net force shrinks.
$\displaystyle x_{cm} = \frac{\int_0^L kx^2\,dx}{\int_0^L kx\,dx} = \frac{L^3/3}{L^2/2} = \frac{2L}{3}$.
$\displaystyle v = \sqrt{gr}$.
$\displaystyle v \propto \frac{1}{\sqrt r}$.
$a = 3t$, so $\displaystyle v = \frac{3}{2}t^2 = 6$ m/s.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A ball of mass $m$ is dropped from rest and experiences a drag force $F = -bv$.
(a) Write the differential equation for $v(t)$.
Show the answer and scoring guide
$\displaystyle m\frac{dv}{dt} = mg - bv$.
Check off the points you earned:
(b) Solve it for $v(t)$.
Show the answer and scoring guide
$\displaystyle \int_0^v \frac{dv}{mg - bv} = \int_0^t \frac{dt}{m} \Rightarrow -\frac{1}{b}\ln\frac{mg - bv}{mg} = \frac{t}{m} \Rightarrow v(t) = \frac{mg}{b}\left(1 - e^{-bt/m}\right)$.
Check off the points you earned:
(c) Sketch (describe) the graph of $v(t)$.
Show the answer and scoring guide
It starts at 0 with slope $g$, then curves and levels off approaching the horizontal asymptote $\displaystyle v_T = \frac{mg}{b}$.
Check off the points you earned: