Notes
Circular motion
- Period $T$ and frequency $\displaystyle f = \frac{1}{T}$. Speed $\displaystyle v = \frac{2\pi r}{T}$; velocity is tangent to the circle.
- Centripetal acceleration $\displaystyle a_c = \frac{v^2}{r}$, pointing toward the center.
- Centripetal force $\displaystyle F_c = \frac{mv^2}{r}$ is the net inward force, supplied by tension, friction, gravity or a normal force. There is no real outward force.
Universal gravitation
- $\displaystyle F = \frac{Gm_1m_2}{r^2}$ with $G = 6.67 \times 10^{-11} \text{ N·m}^2/\text{kg}^2$. Inverse-square: double $r$, force $\div 4$.
- Field strength $\displaystyle g = \frac{GM}{r^2}$; $9.8 \text{ m/s}^2$ at Earth's surface.
Orbits
- Gravity provides the centripetal force: $\displaystyle v = \sqrt{\frac{GM}{r}}$, independent of the satellite's mass.
- Higher orbits are slower with longer periods. Orbiting astronauts are weightless because they are in free fall.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle a_c = \frac{2^2}{4} = 1 \text{ m/s}^2$.
$\displaystyle F_c = \frac{0.2 \times 25}{0.5} = 10 \text{ N}$.
The sensation comes from inertia; the real force (from the seat or door) points left, toward the center.
Doubling $r$ divides the force by $2^2 = 4$: $\displaystyle \frac{36}{4} = 9 \text{ N}$.
$\displaystyle g \propto \frac{1}{r^2}$; halving $r$ multiplies $g$ by 4.
$\displaystyle v = \sqrt{\frac{GM}{r}}$ grows as $r$ shrinks; the satellite's mass does not matter.
Free response
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Free-response question
A 1,200 kg car drives around a flat, circular curve of radius 40 m at a constant speed of $14 \text{ m/s}$.
(a) Find the car's centripetal acceleration and its direction.
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$\displaystyle a_c = \frac{v^2}{r} = \frac{14^2}{40} = \frac{196}{40} = 4.9 \text{ m/s}^2$, directed toward the center of the curve.
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(b) What force provides the centripetal force, and how large must it be?
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Static friction between the tires and road provides it. $F_c = ma_c = 1200 \times 4.9 = 5880 \text{ N}$.
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(c) Find the smallest coefficient of static friction that lets the car make the turn. If the road's coefficient is exactly this value and the car goes twice as fast, can it still make the turn? Explain.
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$\displaystyle \mu_s mg = \frac{mv^2}{r} \Rightarrow \mu_s = \frac{a_c}{g} = \frac{4.9}{9.8} = 0.50$. At twice the speed the needed force is 4 times larger ($23520 \text{ N}$), but friction can supply at most $0.50 \times 1200 \times 9.8 = 5880 \text{ N}$, so the car would skid outward along a nearly straight path.
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