Circular Motion

AIM: How is this even possible?????

But First…Facts about Circular Motion• The distance around one lap of a circle is the

circumference ()• What is ½ a lap?• What is ¼ of a lap?

• The displacement for one lap around a circle is zero.• What is ½ a lap?• What is ¼ of a lap?

• Time once around a circle is the Period (T)

More Facts about Circular Motion• Speed while moving in a circle

• Acceleration while moving in a circle

• Net force while moving in a circle

Even More Facts about Circular MotionVelocity

(tangent to the circle at all times)

Centripetal Acceleration and

Centripetal Force(always towards the center of the circle)

V

Fc ac

What Provides Centripetal Force?

• Centripetal force is a NET force which means it is calculated – either by Newton’s 2nd Law (Fc=mac) – Or by drawing a free body diagram and

determining the net force from the orientation of the forces on the diagram.

• Because it is a net force, you should use the same steps you were using (hopefully) last topic.

Friction as a Centripetal Force(running, driving)

• On a level surface, friction IS the force that keeps the object (car, motorcycle, person) moving in a circle.– This is a STATIC friction force because the tires/shoes are

not skidding across the surface.– Does the mass of the car affect its maximum turning

speed? JUSTIFY!

0=

Vertical

=

Horizontal

Ff

FN

Fg

Examples

1. A 200Kg motorcycle with rubber tires wants to complete a turn of radius 12m on a level dry asphalt road as fast as possible.

a. What is the max static friction force between the road and the tires?

b. What is the max speed the motorcycle can take the turn?c. Compared to the speed of the motorcycle, what would

be the maximum speed a 400kg car could take the turn?

2. A 65Kg runner takes a turn of radius 15m at 7.5m/s. a. What is the friction force required to make this turn?b. What is the coefficient of friction between the shoe and

the track?c. Is this a static or kinetic coefficient? WHY?

Friction on an Inclined Surface

Ff

FN

Fg

𝑭⊥=𝒎𝒈cos𝜽

𝑭∥=𝒎𝒈 sin 𝜽

0=

Vertical

===

Horizontal

Tension as the Centripetal Force• An object moves in a horizontal circle on a

frictionless surface while attached to a string of length l

Table top

T

Tension as the Centripetal Force• An object moves in a vertical circle on a at a

constant speed while attached to a string of length l top

TFg

Tension as the Centripetal Force• An object moves in a vertical circle on a at a

constant speed while attached to a string of length l bottom

T

Fg

Tension as the Centripetal Force• When an object is moving in a vertical circle, where in the

motion is the string most likely going to break? JUSTIFY!

As seen in the equations above, the tension in the rope at the top of the circle is the centripetal force minus the weight of the object because at that point, gravity is providing some of the centripetal force. At the bottom, the tension in the rope is the centripetal force plus the weight because at that point, the tension not only has to hold up the weight of the object, it must also provide the centripetal force.

𝑇 𝑡𝑜𝑝=𝑚𝑣2

𝑙−𝑚𝑔 𝑇 𝑏𝑜𝑡𝑡𝑜𝑚=𝑚𝑣2

𝑙+𝑚𝑔

Tension as the Centripetal Force• An object moves in a conical pendulum while

attached to a string of length l making an angle with the vertical

T

Fg

Ty

Tx

Normal Force as a Centripetal Force(Gravitron Ride)

• When you are riding the gravitron, the centripetal force is provided by the wall of the ride!

• Does the mass of the rider matter? – PROVE IT MATHEMATICALLY!

Ff

FN

Fg

=

Vertical Horizontal

𝜇𝑠𝑚𝑣2

𝑟=𝑚𝑔

Examples

1. A 60Kg student rides a gravitron with a radius of 7m. The ride completes 10 revolutions in 1 minute

a. What is the average speed of the rider?b. What is the normal force provided by the wall?c. What is the coefficient of friction between the rider and

the wall

2. An average person can tolerate an acceleration of 4g before passing out. If you wanted to build a gravitron ride with a radius of 12m

a. What is the maximum average speed you can set the ride to?

b. What would have to be the coefficient of friction between the wall and the person to make the ride successful?

Circular motion in a Vertical Loop• At the TOP of the loop

FN Fg

If you ‘just make it’ FN=0N

Circular motion in a Vertical Loop• At the BOTTOM of the loop

FN

Fg

Circular motion in a Vertical Loop• Where in the loop do you

feel the heaviest? JUSTIFY!TopBottom

You feel heaviest at the bottom of the loop because your ‘apparent weight’ is the normal force. At the bottom the normal force has to counteract the weight and provide the centripetal force. At the top, some of the centripetal force is provided by gravity.

Examples

1. A 60Kg student rides a rollercoaster with a vertical loop with a radius of 15m.

a. What is the minimum speed needed to complete the vertical loop?

b. If the actual speed of the ride at the top is 20m/s, how heavy (FN) does the rider feel at the top of the loop?

c. If the ride’s speed is 31m/s at the bottom, how heavy do they feel?

2. A 3kg set of keys attached to a 0.75m long string in being swung in a vertical loop.

a. Assuming the maximum tension the string can provide is 140N, how fast can the keys be swung without breaking the string?

Circular motion on a Hill• At the TOP of the hill

FN

Fg

If you ‘lose contact with the seat’ FN=0N

Examples

1. A 50Kg student rides a rollercoaster with a circular hill that has a radius of 20m

a. What is the maximum speed the coaster can take the hill before the student loses contact with the seat?

2. If you want a 300kg motorcycle to be able to stay in contact while driving over a circular hill at 40m/s. What is the radius of the hill?

Gravity as a Centripetal Force

Newton’s Law of Gravity

• Fg Force of gravitational attraction between any two objects.

• G Universal Gravitational Constant=6.67x10-11 Nm2/kg2

• m the mass of the objects (kg)• r the distance between the CENTERS of the

objects.

𝐹 𝑔=𝐺𝑚1𝑚2

𝑟 2

Acceleration Due to Gravity

General g

Orbital Speed

=

General Orbital Speed-> Kepler’s 3rd Laws

𝑣𝑜𝑟𝑏𝑖𝑡=√𝐺𝑀𝑜𝑟𝑏𝑖𝑡𝑒𝑑

𝑟 𝑜𝑟𝑏𝑖𝑡

2𝜋 𝑟𝑜𝑟𝑏𝑖𝑡𝑇

=√𝐺𝑀𝑜𝑟𝑏𝑖𝑡𝑒𝑑

𝑟𝑜𝑟𝑏𝑖𝑡

( 2𝜋𝑟𝑜𝑟𝑏𝑖𝑡𝑇 )2

=𝐺𝑀𝑜𝑟𝑏𝑖𝑡𝑒𝑑

𝑟𝑜𝑟𝑏𝑖𝑡4𝜋 2𝑟𝑜𝑟𝑏𝑖𝑡

2

𝑇 2 =𝐺𝑀𝑜𝑟𝑏𝑖𝑡𝑒𝑑

𝑟𝑜𝑟𝑏𝑖𝑡𝑟𝑜𝑟𝑏𝑖𝑡

3

𝑇 2 =𝐺𝑀𝑜𝑟𝑏𝑖𝑡𝑒𝑑

4𝜋 2

Kepler’s Laws of Planetary Motion1. The law of ellipses: all planets orbit the sun in

elliptical paths with the sun at one focus.2. The law of equal areas: As a planet orbits the sun,

it sweeps out equal areas in equal amounts of time. This means the closer the planet is to the sun, the faster it is moving.

3. The law of harmonies: the orbits of all the planets’ orbits are related

𝑟𝑜𝑟𝑏𝑖𝑡3

𝑇 2 =𝐺𝑀𝑜𝑟𝑏𝑖𝑡𝑒𝑑

4𝜋 2

𝒓 𝑨𝟑

𝑻 𝑨𝟐=

𝒓 𝑩𝟑

𝑻𝑩𝟐

Questions• How could have scientists figure out the mass of the

Earth?

• How could have scientists figure out the mass of the sun?

• How could have scientists figure out the distances to all the other planets in our solar system?

𝒈=𝑮𝑴𝒑𝒍𝒂𝒏𝒆𝒕

𝑹𝒑𝒍𝒂𝒏𝒆𝒕𝟐

𝒗𝒐𝒓𝒃𝒊𝒕=√𝑮𝑴𝒐𝒓𝒃𝒊𝒕𝒆𝒅

𝒓 𝒐𝒓𝒃𝒊𝒕

𝒓 𝑨𝟑

𝑻 𝑨𝟐=

𝒓 𝑩𝟑

𝑻𝑩𝟐