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Uniform Circular Motion AP Physics 1

Centripetal Acceleration In order for an object to follow a circular path, a force needs to be applied in order to accelerate the object Although the magnitude of the velocity may remain constant, the direction of the velocity will be constantly changing As a result, this force will provide a centripetal acceleration towards the centre of the circular path

How can we calculate centripetal acceleration?

Centripetal Force Like the centripetal acceleration, the centripetal force is always directed towards the centre of the circle The centripetal force can be calculated using Newton’s Second Law of Motion

Problem – horizontal circle A student attempts to spin a rubber stopper (m = 0.050kg) in a horizontal circle with a radius of 0.75m. If the stopper completes 2.5 revolutions every second, determine the following: –The centripetal acceleration –The centripetal force

The stopper will cover a distance that is 2.5 times the circumference of the circle every second Determine the circumference Multiply by 2.5 Use the distance and time (one second) to calculate the speed of the stopper

Use the speed and radius to determine the centripetal acceleration Then use the centripetal acceleration and mass to determine the centripetal force

Problem – vertical circle A student is on a carnival ride that spins in a vertical circle. –Determine the minimum speed that the ride must travel in order to keep the student safe if the radius of the ride is 3.5m. –Determine the maximum force the student experiences during the ride (in terms of number of times the gravitational force)

Problem – vertical circle

Vertical Circle While travelling in a vertical circle, gravity must be considered in the solution While at the top of the circle, gravity acts towards the centre of the circle and provides some of the centripetal force While at the bottom of the circle, gravity acts away from the centre of the circle and the force applied to the object must overcome both gravity and provide the centripetal force

Vertical Circle To determine the minimum velocity required, use the centripetal force equal to the gravitational force (as any slower than this and the student would fall to the ground) To determine the maximum force the student experiences, consider the bottom of the ride when gravity must be overcome

At the top of the circle, set the gravitational force (weight) equal to the centripetal force Solve for velocity

At the bottom of the circle, the net force is equal to the sum of the gravitational force and the centripetal force Solve for number of times the acceleration due to gravity

Road Design You are responsible to determine the speed limit for a turn on the highway. The radius of the turn is 55m and the coefficient of static friction between the tires and the road is 0.90. –Find the maximum speed at which a vehicle can safely navigate the turn –If the road is wet and the coefficient drops to 0.50, how does this change the maximum speed

Diagrams

The maximum speed at which a vehicle can safely navigate the turn

Coefficient drops to 0.50, how does this change the maximum speed

Planetary Motion – Orbits & Gravity

Johannes Kepler (1571-1630) Worked for Tycho Brahe Took data after his death Spent years figuring out the motions of the planets Came up with… Three Laws of Planetary Motion

1 st Law: Planets move in elliptical orbits with the Sun at one foci Sun Foci (sing. Focus) PerihelionAphelion Average distance from the Sun = 1 Astronomical Unit (1 A.U.) = approx. 150 000 000 km

2 nd Law: Planets move faster at perihelion than at aphelion OR a planet sweeps out equal areas in equal time periods. 1 Month

3 rd Law: Period is related to average distance T = period of the orbit r = average distance kT 2 = r 3 Longer orbits - greater average distance Need the value of k to use the formula k depends upon the situation Can be used for anything orbiting anything else

Special version of Kepler’s third Law – If the object is orbiting the Sun T – measured in years, r – measured in A. U., then…. T 2 = r 3

For planets A and B, Kepler’s 3 rd Law can look like this…

Isaac Newton (1642-1727) Able to explain Kepler’s laws Had to start with the basics - The Three Laws of Motion

1. Law of Inertia - Objects do whatever they are currently doing unless something messes around with them.

2. Force defined F = ma F=force m=mass a=acceleration (change in motion)

3. For every action there is an equal and opposite reaction. The three laws of motion form the basis for the most important law of all (astronomically speaking) Newton’s Universal Law of Gravitation

F g = force of gravity G = constant (6.67 x 10 -11 Nm 2 /kg 2 ) M 1, M 2 = masses R = distance from “centers” Gravity is the most important force in the large- scale Universe

An Inverse Square Law…

Newton’s Revisions to Kepler’s Laws Newton agreed with 1 st law of motion Defined bound orbits (i.e. circular, elliptical) and unbound orbits (i.e. hyperbolic, parabolic) with Sun at one focus Used conic sections to describe orbits

Newton’s Revisions to Kepler’s Laws Newton agreed with 2 nd law of motion Believed planetary motion to be non- constant acceleration due to varying distance between planet and Sun Force causing acceleration was gravity

4π2 and G are just constant #s (they don’t change) M 1 and M 2 are any two celestial bodies (could be a planet and Sun) Importance: if you know period and average distance of a planet, you can find mass of Sun (2 x 10 30 kg) or any planet! Mass of Sun is 2 000 000 000 000 000 000 000 000 000 000 kg Mass of Earth is 6 000 000 000 000 000 000 000 000 kg Mass of Mr. J is 100 kg! WOW! Newton’s Revisions to Kepler’s Laws Newton extended 3 rd law to…

Newton’s Mountain Horizontal projectile launched at 8km/s How far does the projectile fall in one second? How far does the Earth “fall” away from the projectile? –Assume that arc length and chord length are equal over the 8km distance and the Earth’s radius is 6400km

Newton’s Mountain Shortly after developing the Universal Law of Gravitation, Newton began a series of thought experiment involving artificial satellites Newton’s thought was that if you had a tall enough mountain and launched a cannonball fast enough horizontally, it would fall towards the Earth at the same rate the Earth would “fall” away This would result in the cannonball orbiting the Earth

Geostationary Satellite A geostationary (or geosynchronous) satellite, will always be above the same spot on Earth What is the orbital radius, altitude and speed of a geostationary satellite? –Use Newton’s Version of Kepler’s Law to solve for orbital radius –Subtract Earth’s radius from orbital radius to determine altitude –Set gravitational force equal to centripetal force and solve for orbital speed

Weightlessness

The International Space Station orbits at an altitude of 226km; determine the force of gravity on an astronaut (65.kg) at this altitude and compare this to their weight on the surface of the Earth 594N at the ISS 638N on Earth Is the astronaut “weightless?”

Weightlessness Weightlessness occurs because objects are all falling towards the surface of the Earth at the same rate NASA simulates this on the “Vomit Comet” a high altitude aircraft that plunges toward Earth

Another way to look at “g”…

Another way to look at gravitational potential energy of an object… (h is height but since it is arbitrary, it can be chosen as the distance from the center of the Earth to the position of the object…or r)

Some important orbital applications… Geosynchronous means having an orbit around the Earth with a period of 24 hours

Gravitational Field The strength of a gravitational field can be determined using a test mass (m t ) The mass should be very small compared to the mass creating the field A gravitational field will be measured in the units of N/kg

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