Circular Motion, Orbits, and Gravity Flashcards

In uniform circular motion, a particle's speed is constant, but its velocity is not, as the direction is always changing.
This change in velocity indicates the presence of centripetal acceleration.

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 of frequency is inverse seconds, or Hertz (Hz).
f = \frac{1}{T}
The speed (v) of an object in circular motion is related to the circumference of the circle and the period:
v = \frac{2\pi r}{T}
v = 2\pi r f
Centripetal acceleration can be expressed as:
a = \frac{v^2}{r}

A table saw blade with a diameter of 25 cm spins at 3600 rpm.
Time for one revolution:
- T = \frac{1}{f} = \frac{1}{3600 \text{ rpm}} = \frac{1}{60 \text{ rps}} \approx 0.0167 \text{ s}
Speed of a tooth at the edge of the blade:
- v = 2 \pi r f = 2 \pi (0.125 \text{ m})(60 \text{ Hz}) \approx 47.1 \text{ m/s}
Acceleration of the tooth:
- a = \frac{v^2}{r} = \frac{(47.1 \text{ m/s})^2}{0.125 \text{ m}} \approx 17750 \text{ m/s}^2

Riders on a circular carnival ride are accelerating, implying a net force acting on them.
Newton’s second law applies: \vec{F}_{\text{net}} = m \vec{a}
The net force producing centripetal acceleration is:
F_{\text{net}} = ma = m\frac{v^2}{r}
A particle of mass m moving at constant speed v around a circle of radius r must have a net force of magnitude
F_{\text{net}} = m\frac{v^2}{r} pointing toward the center.
The net force is due to familiar forces like tension, friction, or normal force.

Engineers design curves and dips in roads as segments of circles.
When a car goes through a dip at constant speed, the normal force at the bottom of the dip is greater than the car’s weight.
When a car turns a corner, the force providing the centripetal acceleration is typically friction between the tires and the road.

In the hammer throw, an athlete spins a mass at the end of a chain and then releases it.
For a 7.3 kg mass at the end of a 1.2 m chain, swung at 29 m/s, the tension in the chain is:
- T = m\frac{v^2}{r} = (7.3 \text{ kg}) \frac{(29 \text{ m/s})^2}{1.2 \text{ m}} \approx 5100 \text{ N}

The maximum speed for a car to turn a corner without sliding depends on the radius of the curve and the coefficient of static friction between the tires and the road.

A car can take a banked curve without assistance from friction at a specific speed determined by the banking angle and the radius of the curve.

When a car turns a corner, the force of the door pushing inward causes a passenger to turn.
The sensation of being pushed outward is due to the body's inertia, not a centrifugal force.
Centrifugal force should not be included on a free-body diagram or in Newton’s laws.

Sensation of weight changes on roller coasters due to changes in contact forces.

The minimum speed at the top of the loop is determined by setting the normal force to zero and equating weight to centripetal force. When N = 0, skater's weight is the centripetal force:

Used to separate components of a liquid with different densities by producing large centripetal accelerations.

An ultracentrifuge with a diameter of 18 cm produces a centripetal acceleration of 250,000g.
Frequency in rpm:
- a = \frac{v^2}{r} = (2 \pi r f)^2/r = 4 \pi^2 r f^2
- f = \sqrt{\frac{a}{4 \pi^2 r}} = \sqrt{\frac{250000 \cdot 9.8 \text{ m/s}^2}{4 \pi^2 (0.09 \text{ m})}} \approx 2620 \text{ Hz}
- f \approx 2620 \text{ Hz} \cdot 60 \text{ s/min} \approx 157,000 \text{ rpm}
Apparent weight of a 0.0030 kg sample:
- w_{\text{app}} = ma = (0.0030 \text{ kg})(250000 \cdot 9.8 \text{ m/s}^2) \approx 7350 \text{ N}

The force of gravity on a projectile is directed toward the Earth's center.
As initial speed increases, the projectile travels farther before hitting the ground.
At a sufficiently large launch speed, the trajectory and the Earth's curvature are parallel, resulting in a closed trajectory called an orbit.
An orbiting projectile is in free fall.

The force of gravity provides the centripetal acceleration for an orbiting object:
G \frac{Mm}{r^2} = m \frac{v^2}{r}
An object moving parallel to the surface with orbital speed will have centripetal acceleration if:
v = \sqrt{\frac{GM}{r}}
The orbital speed of a projectile skimming the surface of a smooth, airless Earth is: