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**Centripetal Force and Acceleration**

Circular Motion Centripetal Force and Acceleration

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**Tangential Velocity Tangential Velocity (Vt)- object’s speed along**

an imaginary line that is drawn tangent to the circular path Depends on distance from the object to the center of the circular path When the tangential speed is constant then the motion is described as uniform circular motion

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**Tangential Velocity r=radius of the circle**

Vt=2πr/T r=radius of the circle T=Period (amount of time to complete one circle)

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Example A plane makes a complete circle with a radius of 3622 m in 2.10 min. What is the speed of the plane? 180 m/s

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**Centripetal Acceleration**

When an object is moving in a circular path, the direction changes If the object has a constant SPEED, Acceleration is due to the direction changing This is called centripetal acceleration

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**Centripetal Acceleration**

Centripetal acceleration is always directed towards the center of the circle Centripetal=center seeking

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**Centripetal Acceleration**

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**Tangential Accleration**

Centripetal acceleration results from change in direction When the speed changes in a circle it is called tangential acceleration Consider a car moving in a circle

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**Centripetal Force Remember!**

When an object is accelerating there is a net force If there is centripetal acceleration, there is a net force- Centripetal Force This is not a new force Net force that accelerates an object towards the center of a circle

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Examples If a mass is twirled in a circle, at the end of a string, the centripetal force is provided by the tension When a car rounds a corner on a highway, the centripetal force is provided by friction When the moon orbits the Earth, the centripetal force is provided by gravity

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**Centripetal Force Centripetal Force =mass x centripetal acceleration**

Fc=mac Substitute ac Fc=m V2/r Substitute Vt Fc=m 4π2r/T2

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Example- Standard A 0.50 kg mass is whirled in a circle of radius m at 2.3 m/s. Calculate the centripetal force acting on the mass 13.2 N

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Example-Honors A 0.50 kg mass sits on a frictionless table and is attached to hanging weight. The 0.50 kg mass is whirled in a circle of radius 0.20 m at 2.3 m/s. Calculate the centripetal force acting on the mass . Calculate the mass of the hanging weight 13.2 N Fg=T=Fc=mg =m9.8 1.35 kg

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Example A car traveling at 14 m/s goes around an unbanked curve in the road that has a radius of 96 m. What is its centripetal acceleration? What is the minimum coefficient of friction between the road and the car’s tires? 2.04 m/s2 Ff=Fc=mac μmg=mac μg=ac μ9.8= coefficient=.21

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Clicker Question Calculate the centripetal force acting on a 925 kg car as it rounds an unbanked curve with a radius of 75 m at a speed of 22 m/s.

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Clicker Question An amusement park ride has a radius of 2.8 m. If the time of one revolution of a rider is s, what is the speed of the rider?

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Clicker Question A 2.7x103 kg satellite orbits the Earth at a distance of 1.8x107 m from the Earth’s centre at a speed of 4.7x103 m/s. What force does the Earth exert on the satellite?

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Clicker Question . An object moves in a circle at a constant speed. Which of the following is not true of the object? A. Its centripetal acceleration is constant. B. Its tangential speed is constant. C. Its velocity is constant. D. A centripetal force acts on the object.

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Clicker Question A car traveling at 15 m/s on a flat surface turns in a circle with a radius of 25 m. 2. What is the centripetal acceleration of the car? A. 2.4 10-2 m/s2 B m/s2 C. 9.0 m/s2 D. zero

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Centrifugal Force When a driver takes a sharp left turn, the passenger slides to the right of the car and into the door, Why? Centrifugal=center-fleeing Apparent force that causes a revolving or rotating object to move in a straight line However-Newton’s first law states- an object in motion will stay in motion until a force acts on it Centrifugal force does not really exist!

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Describing Motion What happens when centripetal force disappears? Where does the object go? A ball that is on the end of a string is whirled in a vertical circular path. If the string breaks at the position shown in (a), the ball will move vertically upward in free fall. If the string breaks at the top of the ball’s path, as in (b), the ball will move along a parabolic path.

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**Vertical Versus Horizontal**

Draw a free body diagram of a mass moving in a vertical circle, when the object is at the top of the path and the bottom of the path As with any object moving in a circle there is a net force acting on it towards the center of the circle The net force is the centripetal force At the top Fnet (Fc)=Ft +Fg At the bottom Fnet (Fc)=Ft-Fg

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Example A 1.7 kg object is swung from the end of a 0.60 m string in a vertical circle. If the time of one revolution is 1.1 s, what is the tension in the string A) at the top? B) at the bottom? Top- 17N Bottom- 50 N

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Constant Velocity When you are trying to find the minimum speed of an object at the top of its circular arc we can use the equation Fc=Fg so The centripetal acceleration equals the acceleration due to gravity (ac=9.8 m/s/s) So g=Vt2/r V=√(gr)

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Example An object is swung in a vertical circle with a radius of 0.75 m. What is the minimum speed of the object at the top of the motion for the object to remain in circular motion? 2.7 m/s

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Circular Motion Gravitation

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**What is Gravity? The force that pulls us to the earth**

It is much more than that Gravity is the force that attracts two bodies that have mass and energy to each other Newton discovered that gravity attracts any two objects depending on their masses and their distance apart

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Gravity The gravitational forces that two masses exert on each other are always equal in magnitude and are opposite in directions. Newton’s Third Law

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**Gravity Is Proportional to the two masses**

Is inversely proportional to the square of the distance between their center of mass So if two objects are twice as far away from each what happens to the gravitational force by? Fg=Gm1m2 r2 G=6.67 x N m2/kg2 M1= mass of the first object M2= mass of the second object r= distance between the two centers of mass It gets reduced by 1/4

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Example Calculate the force of gravity between two 75 kg students if their centers of mass are 0.95 m apart 4.2 x N

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Example 2 A satellite weighs 9000 N on Earth’s surface. How much does it weigh if its mass is tripled and its orbital radius is doubled? Fg is the same thing as weight Fg= 9000 N Mass is on the top of the equation and the mass is proportional to the Fg, so the Fg goes up by a factor of 3 The radius is inversely proportional to the square of the distance, so if the radius is doubled then the Fg is reduced by a factor of ¼ So Fg X ¾ = 6750 N

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Clicker Question . Earth (m = 5.97 1024 kg) orbits the sun (m = 1030 kg) at a mean distance of 1.50 1011 m. What is the gravitational force of the sun on Earth? (G = N•m2/kg2) A 1032 N B 1022 N C 10–2 N D 10–8 N

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Clicker Question Gravitational force F exists between point objects A and B separated by distance R. If the mass of A is doubled and distance R is tripled, what is the new gravitational force between A and B? A) 9/2 F B) 2/3 F C) 2/9 F D) 3/2 F

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**Common Misconception: Mass Versus Weight**

Amount of matter Constant everywhere Gravitational attraction (Fg) Changes depending on location

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**Clicker Question Which of the following statements is correct?**

A. Mass and weight both vary with location. B. Mass varies with location, but weight does not. C. Weight varies with location, but mass does not. D. Neither mass nor weight varies with location.

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Satellites Are constantly “falling” when in orbit They are in freefall

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Gravitational Field Gravity is an example of a field force (ie not a contact force) A force that is exerted which has no direct contact A field- an area of influence Think about a campfire As you approach it As you increase the size

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Gravitational Fields Fields can be described as vectors or scalars, depends on what kind Gravitational fields are field forces so.. They are vectors Gravitational fields are represented by arrows, magnitude is represented by how many arrows are present

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**Gravitational Field Strength**

Gravitational field strength=acceleration due to gravity Recall Fg=mg So… g=Fg/m g=acceleration due to gravity, gravitational field strength, 9.80m/s2 near the Earth’s surface g varies with distance g=Gm1 r2 Measured in N/kg, same as m/s2

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Example What is the gravitational field strength on the Earth’s surface of the moon? The mass of the moon is 7.35 x 1022 kg and the radius of the moon is 1.74 x 106 m 1.62 m/s2 or N/kg

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Clicker Question . Which of the following is a correct interpretation of the expression? A. Gravitational field strength changes with an object’s distance from Earth. B. Free-fall acceleration changes with an object’s distance from Earth. C. Free-fall acceleration is independent of the falling object’s mass. D. All of the above are correct interpretation

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**The Period of a Satellite**

T=√(4π2r3) Gm m refers to the mass of the center of the orbit

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Example A satellite orbits the Earth at a radius of 2.2 x 107 m. What is its orbital period, when the mass of the earth is 5.98 x 1024 kg. What is the speed of the satellite? 32500 s 4251 m/s

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