Circular Motion, Orbits, and Gravity Flashcards

Dynamics of Uniform Circular Motion

• 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, Frequency, and Speed

• 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}

Example: Spinning Table Saw Blade

• 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

Dynamics of Uniform Circular Motion

• 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.

Forces on a Car

• 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.

Example: Hammer Throw

• 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}

Finding the Maximum Speed for a Car to Turn a Corner

• 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.

Finding a Car’s Speed on a Banked Turn

• 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.

Apparent Forces in Circular Motion

Centrifugal Force

• 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.

Apparent Weight in Circular Motion

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

Example: Skateboard in a Full Pipe

• 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:

Centrifuges

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

Analyzing the Ultracentrifuge

• 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}

Car Going Over a Hill

• A car of mass 1500 kg goes over a hill with a radius of 60 m at a speed of 20 m/s.

  • Force of gravity on the car:

  • F_g = mg = (1500 \text{ kg})(9.8 \text{ m/s}^2) = 14700 \text{ N}

  • Normal force of the road on the car:

  • F{\text{net}} = Fg - N = m\frac{v^2}{r}

  • N = F_g - m\frac{v^2}{r} = 14700 \text{ N} - (1500 \text{ kg})\frac{(20 \text{ m/s})^2}{60 \text{ m}} = 14700 \text{ N} - 10000 \text{ N} = 4700 \text{ N}

Circular Orbits and Weightlessness

Orbital Motion

• 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.

Orbital Speed

• 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:

Weightlessness in Orbit

• Astronauts and their spacecraft are in free fall, experiencing weightlessness.

Example: Orbit of the Moon

• The Moon is