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Vocabulary

Vocabulary

Vocabulary

By the End of Today’s Class You Should Be Able To… Calculate Centripetal acceleration and Centripetal force of an object in uniform circular motion.

Centripetal force keeps an object in circular motion.

Uniform circular motion is the motion of an object traveling at a constant (uniform) speed on a circular path. Period T is the time required to travel once around the circle, that is, to make one complete revolution. r 9

Example -Speed of a revolving ball A 150 -g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0. 600 m. The ball makes 2. 00 revolutions per second. Calculate its speed of the ball. r

Relationship between angular and linear velocity (Tangential Velocity )

Example 1: A Tire-Balancing Machine The wheel of a car has a radius of r = 0. 29 m and is being rotated at 830 revolutions per minute (rpm) on a tire-balancing machine. Determine the period and speed (in m/s) at which the outer edge of the wheel is moving. . 14

ØThe speed is constant, ‘v’ ØThe velocity is not! ØThere is an acceleration ØThere is a net force

Constant Speed not Velocity: The direction of v changes continually! The velocity is always tangent to the path

T = Period (time per revolution) f = frequency (revolutions per unit time) v = speed r = radius

Particle moving with uniform speed v in a circular path with radius r has an acceleration a. C : - The acceleration points to the center of the circle! - Centripetal acceleration

For an object to be in uniform circular motion, there must be a net force acting on it. We already know the acceleration, so can immediately write the force: (5 -1)

We can see that the force must be inward by thinking about a ball on a string:

Centripetal acceleration Centripetal Force

• The velocity of the particle is always _____ • The centripetal acceleration is towards the _____ • The centripetal force acting on the particle is towards the ______ • Centripetal force causes a change in the _______ but no change in ______ • The magnitude of the centripetal acceleration is: a = _____ • Newton’s law: The force on the particle is (centripetal force) F = m·a = _______

A particle is moving in a circular path. If the force on the particle would suddenly vanish (string cut) in which direction would the ball fly off?

Net forces that can cause centripetal acceleration Friction of tires on road is a centripetal force Tension of cord is a centripetal force Gravity keeping object in orbit Source: physicsclassroom. com

Mike flies his model plane, the radius of the orbit is 50 m and the 2 kg plane is flying at 20 m/s. 1. What was the centripetal acceleration of the plane? 2. What is the tension he feels in his hand? 3. At the position shown draw in vectors showing the direction of the force, acceleration, and velocity.

Horizontal Circle Lauren rotates a stone, m =. 50 kg that is attached to the end of a 1. 5 m cord above her head in a horizontal circle. a. If the cord can hold 50 N of tension, at what maximum speed will it rupture. b. Which force “provides” the centripetal force?

A 900 -kg car moving at 10 m/s takes a turn around a circle with a radius of 25. 0 m. Determine the acceleration and the net force acting upon the car. 1. a = (v 2)/R a = (10. 0 m/s)2/(25. 0 m) a = (100 m 2/s 2)/(25. 0 m) a = 4 m/s 2 2. F = ma = (900 kg) x (4 m/s 2) F = 3600 N A car is traveling on a highway at 20. 0 m/s and encounters a curve in the road during which the direction of the car’s velocity changes by 90. 0° in 30. 0 s. (a) Find the car’s centripetal acceleration.

A 95 -kg halfback makes a turn on the football field. The halfback sweeps out a path which is a portion of a circle with a radius of 12 m. The halfback makes a quarter of a turn around the circle in 2 s. Determine the speed, acceleration and net force acting upon the halfback.

Motion On A Flat Curve • On a flat, level curve, thethe friction between thethe tires and thethe road supplies thethe centripetal force. • If • the If the tires areare worn smooth or or thethe road is icy or or oily, this friction force will notnot bebe available. • The carcar will notnot bebe able to to move in in a circle, it will keep going in in a a straight line and therefore gogo offoff thethe road.

Jeff Gordon leads his race and must drive into a curve at top speed to win it all. The radius of the curve is 1000 m and the coefficient of static friction between his tires and the dry pavement is 0. 50. a. Which force “provides” the centripetal force? b. Find the maximum speed he can have and still make the turn.

Motion On A Banked Curve • FN sin can supply the required centripetal force. • Vehicles can make a sharp turn more safely if the road is banked. If the vehicle maintains the speed for which the curve is designed, no frictional force is needed to keep the vehicle on the road.

Motion On A Banked Curve FN cosƟ = FW = mg a car to travel in a

The rotor is an amusement park ride where people stand against the inside of a cylinder. Once the cylinder is spinning fast enough, the floor drops out. What force keeps the people from falling out the bottom of the cylinder? a. Centripetal force b. Friction c. Normal force

At amusement parks there is a popular ride where the floor of a rotating cylinder room falls away, leaving the backs of the riders “plastered” against the wall. Suppose the radius of the room is 3. 3 m and the speed of the wall is 10 m/s when the floor falls away. 1. 2. How much centripetal force acts on a 55 kg rider? What is the minimum coefficient of static friction that must exist between the rider’s back and wall, if the rider is to remain in place when the wall falls away?

Conical Pendulum No acceleration in the vertical direction The centripetal force is provided by A component of the tension provides the centripetal Force

Vertical Circular Motion Usually, the speed varies in this stunt. “non-uniform” 41

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ØThe normal forces between the roller coaster and tracks ØThe normal forces between you and the roller coaster

Sign Convention: ØTo the center is positive ØAway from the center is negative

Fc = w - n = mv 2/r mg - n = m v 2 / r n = mg - m v 2 / r Calculate the normal force in each case if the mass of the kids plus car is 2, 000 kg, v = 3 m/s, and r = 5 meters Fc = n – mg = m v 2 / r n = mg + m v 2 / r

Vertical Circle

Geosynchronous Satellites • To serve as stationary relay station the satellite must be placed at a certain height above the earth surface l they have an orbital period equal to the rotational period of the earth, 24 hours.

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