Aim: How do we explain centripetal motion?

Uniform Circular Motion An object that moves in a circle at constant speed is said to undergo uniform circular motion (centripetal motion).

Uniform Circular Motion How do we know that the object (blue particle) is accelerating? What is the direction of the object’s acceleration? What is the direction of the object’s instantaneous velocity?

Uniform Circular Motion The centripetal acceleration of an object is always directed toward the center. The instantaneous velocity of an object in centripetal motion is always directed tangent to the circular path.

Equation for Centripetal Acceleration and centripetal force Centripetal Acceleration = ac Speed = v Radius of circular path = r ac = v2/r

Centripetal Force When an object is in uniform circular motion, the net force acting on the object is called the centripetal force. Fnet=Fc=mac=mv2/r The centripetal force can be equal to a single force or the sum of a combination of forces. The centripetal force always has the same direction as the centripetal acceleration.

Centripetal vs centrifugal Centripetal force is the overall force acting on an object to sustain its circular motion Centrifugal force is the apparent force pushing outwards on an object undergoing centripetal motion. It is called a fictitious force.

Thought question 1 You are riding on a Ferris wheel that is rotating with constant speed. The car in which you ride always maintains its correct upward orientation-it does not invert. What is the direction of your centripetal acceleration when you are (a) at the top of the wheel? (b) At the bottom of the wheel? What is the direction of the normal force on you from the sear when you are (c) at the top of the wheel? (d) At the bottom of the wheel? (e) At which point, top or bottom, is the normal force largest in magnitude? a) Downward b) upward c) upward d) upward e) bottom

Thought Question 1

ThoughT question 2 A car travels on a circular roadway of radius r. The roadway is flat. The car travels at a high speed v, such that the friction force causing the centripetal acceleration is the maximum possible value. If this same car is now driven on another flat circular roadway of radius 2r, and the coefficient of friction between the tires and the roadway is the same on the first roadway, what is the maximum speed of the car such that it does not slide off the roadway? √2v

Thought question 3

Thought Question 3 High-speed curved roadways are banked-that is, the roadway is tilted toward the inside of tight curves. Why is this? Instead of just friction supplying the centripetal force, when the road is banked, there is a component of the normal force which also supplies the centripetal force. Thus the maximum speed that a car can travel at around the roadway increases

Thought question 3

Problem 1-How Fast can it spin? An object of mass 0.500 kg is attached to the end of a cord whose length is 1.50 m. The object is whirled in a horizontal circle. a) If the cord can withstand a maximum tension of 50 N, what is the maximum speed of the object can have before the cord breaks? b) Calculate the tension in the cord if the speed of the object is 5.00 m/s a) 12.2 m/s b) 8.33 N

Problem 1

Problem 2-The conical pendulum A small object of mass m is suspended from a string of length L. The object revolves in a horizontal circle of radius r with constant speed v. (Because the string sweeps out the surface of a cone, the system is known as a conical pendulum.) Find (a) the speed of the object, and (b) the period of revolution, defined as the time needed to complete one revolution. a) v =√(Lgsinθtanθ) b) T = 2π√(Lcosθ/g)

Problem 2

Problem 2

Problem 3-What is the maximum speed of the car? A 1500 kg car moving on a flat, horizontal road negotiates a curve whose radius is 35.0 m. If the coefficient of static friction between the tires and the dry pavement is 0.500, find the maximum speed the car can have in order to make the turn successfully. On a wet day, the car described in this example begins to skid on the curve when its speed reaches 8.00 m/s. What is the coefficient of static friction in this case? a) 13.1 m/s b) 0.187

Problem 3

Problem 3

Problem 4 A pilot of mass m in a jet aircraft executes a “loop-the-loop” maneuver. The aircraft moves in a vertical circle of radius 2.70 km at a constant speed of 225 m/s. Determine the force exerted by the seat on the pilot at (a) the bottom of the loop and (b) the top of the loop. Express the answers in terms of the weight mg of the pilot. a) 2.91 mg b) 0.911 mg

Problem 4