Circular Motion, Orbits, and Gravity Notes

A particle moving at a constant speed around a circle of radius $r$ is in uniform circular motion.

In uniform circular motion, the speed is constant, but the velocity is not because the direction is always changing.
Centripetal acceleration is present in uniform circular motion.

QuickCheck 6.1: A ball swung in a horizontal circle accelerates because its direction is changing.
QuickCheck 6.2: The direction of the ball's acceleration is toward the center of the circle.

Period (T): The time interval for an object to complete one revolution around a circle.
Frequency (f): The number of revolutions per second. The SI unit is inverse seconds or Hertz (Hz).
Relationship between period and frequency: $f = \frac{1}{T}$
Speed (v) of an object in circular motion: $v = \frac{2\pi r}{T}$
Also written as: $v = 2\pi r f$
Centripetal acceleration: $a = \frac{v^2}{r}$

Riders on a circular carnival ride are accelerating and thus experience a net force.
Net force equation: $F_{net} = ma = m\frac{v^2}{r}$
A particle of mass $m$ moving at speed $v$ around a circle of radius $r$ must have a net force of magnitude $m\frac{v^2}{r}$ pointing toward the center.
This net force is due to familiar forces like tension, friction, or the normal force.

QuickCheck 6.3: The centripetal acceleration of a ball swung in a horizontal circle is produced by tension in the string.
QuickCheck 6.4: The direction of the net force on the ball is toward the center of the circle.
QuickCheck 6.5: For an ice hockey puck swung in a circle, tension in the string produces the centripetal acceleration.

A 1500 kg car turns a curve of radius 20 m on a level road without sliding. Find the maximum speed.

A curve on a racetrack of radius 70 m is banked at an angle. Find the speed at which a car can take this curve without friction.
With no friction, the horizontal component of the normal force causes the centripetal acceleration.

Centrifugal force does not appear on free-body diagrams and is not included in Newton's laws.
When a car turns, the force of the car door pushing inward causes you to turn. What you feel is your body trying to move in a straight line.

Your sensation of weight changes on roller coasters.
The force felt is the contact force that supports you, usually the normal force.
At the bottom of a loop: The net force points upward, so $n > w$ . Her apparent weight is greater than her true weight.
Newton's second law at the bottom: $n - w = m\frac{v^2}{r}$
Apparent weight: $n = w + m\frac{v^2}{r}$
Newton's second law at the top: $n + w = m\frac{v^2}{r}$
Solving for apparent weight: $n = m\frac{v^2}{r} - w$
Critical speed is the speed for which $n = 0$ .
Critical speed equation: $v_{critical} = \sqrt{gr}$

QuickCheck 6.10: (Pendulum at the bottom) - Correct free-body diagram.
QuickCheck 6.11: (Car coasting over a hill) - Correct free-body diagram.
QuickCheck 6.12: (Roller coaster at the top of a loop) - Correct free-body diagram.

Centrifuges separate components of a liquid with different densities.
They produce centripetal accelerations many times greater than free-fall acceleration (g).
Separate cells/components in minutes/hours instead of days.

An 18-cm-diameter ultracentrifuge produces a centripetal acceleration of 250,000g.

QuickCheck 6.13: (Coin on a turntable) - Friction acts in the plane of the turntable.
QuickCheck 6.14: (Coin on a turntable - free-body diagram) - Correct diagram.
QuickCheck 6.15: (Car on a banked road) - Correct free-body diagram.

A 1500 kg car goes over a hill at 20 m/s. The hill is approximately circular with a radius of 60 m.
- What is the force of gravity on the car?
- What is the normal force of the road on the car at the highest point?

The force of gravity on a projectile is directed toward the center of the earth.

If the launch speed of a projectile is sufficiently large, the curve of the trajectory and the curve of the earth are parallel, resulting in a closed trajectory called an orbit.
An orbiting projectile is in free fall.
The force of gravity is the force that causes the centripetal acceleration of an orbiting object.
For an object moving in a circle of radius $r$ at speed $v$ : $v = \sqrt{\frac{GM}{r}}$
Orbital speed of a projectile skimming the surface: $v = \sqrt{\frac{GM}{R}}$
Period of the satellite's orbit: $T = \frac{2\pi r}{v}$
Actual Low Earth Orbit (LEO) is approximately 7800 m/s, with a period of about 90 minutes.

QuickCheck 6.16: Astronauts on the International Space Station are weightless because they are in free fall.