Slide 1

Circular Motion and Gravitation Chapter 5 Slide 2 Kinematics of Uniform Circular motion An object that moves in a circle at constant speed, v, is said to be under uniform circular motion. The magnitude of velocity remains constant. The direction of velocity changes continuously. Rate of change in direction is acceleration. Objects revolving in a circle are continuously accelerating. What causes acceleration? A net force. What direction? Toward the center. Slide 3 Centripetal Acceleration Acceleration toward the center of a circular path is called centripetal or radial acceleration, a R. Since acceleration depends on velocity and distance, Slide 4 Acceleration and Velocity for circular motion The acceleration vector points toward the center of the circle. The velocity vector points in the direction of the motion. Acceleration and velocity are always perpendicular at each point in the path of uniform circular motion. Velocity is tangential to the Path of the circular motion. Slide 5 Centripetal Acceleration Lets take a look at some examples we see in everyday life. http://www.youtube.com/watch?v=-G7tjiMNVlc Slide 6 Period and Frequency The frequency, f, of a revolving object is the number of revolutions it makes each second. The period, T, of an object is the time for one complete revolution. If an object revolves at 3 rev/s, then each revolution takes 1/3 s. For an object revolving in a circle at constant speed, v, we say oSince an object travels one circumference in one revolution. Recall: So: Slide 7 Example 1 A 150 g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600m. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration? Solution: The centripetal acceleration is a R = v 2 /r. First we determine the speed of the ball, v. If the ball makes 2 complete revolutions each second, then the ball travels in a complete circle in 0.500s, which is its period, T. Since the distance equals 2 r, where r is the radius of the circle, the speed is 2 r/T. Slide 8 Solution Example 1 Therefore the ball has a speed The centripetal acceleration is Slide 9 Example 2 The moons nearly circular orbit about the Earth has a radius of about 384,000km and a period T of 27.3 days. Determine the acceleration of the Moon toward the Earth. Slide 10 Solution example 2 In orbit around the Earth, the Moon travels a distance of 2 r, where r = 3.84 x 10 8 m is the radius of its circular path. The speed of the Moon in its orbit about the Earth is v = 2 r/T. The period T in seconds is T = (27.3d)(24.0h/d)(3600s/h) = 2.36x10 6 s. Therefore, In terms of g = 9.80m/s 2, a R = 2.78x10 -4 g. Slide 11 Dynamics of Uniform Circular Motion Circular motion still follows Newtons laws, especially Newtons Second. An object moving in a circle must have a force applied to it to keep it moving in that circle. A net force gives a circularly moving object, centripetal acceleration. Since a R is toward the center, the net force must be directed toward the center of the circle. Slide 12 Newtons First law revisited If NO net force acted on the circling object, then it would continue in a STRAIGHT LINE path, NOT in a circle! Since the direction of the straight line path continually changes, the direction of the force must continually change so that it is always directed toward the center of the circle. This is the centripetal force, the net force. Slide 13 What applies the force? The centripetal force on an object must be applied by a different object. In a rock circling at the end of a string over a persons head, the person pulls on the string and the string exerts the centripetal force on the rock. Slide 14 How does the object stay out there? Is there a force keeping the revolving object out there? A common misconception is the sense of a center fleeing force, or centrifugal force. This is incorrect! The inertia of the circling object causes it to continue in a straight line. You keep pulling inward, changing the path, but what if the string breaks? The rock NOT flying outward from the center disproves the centrifugal force idea Slide 15 Example 3 Estimate the force a person must exert on a string attached to a 0.150kg ball to make the ball revolve in a horizontal circle of radius 0.600m. The ball makes 2.00 revolutions per second. Slide 16 Example 3 Solution First draw a free body diagram for the ball showing the 2 forces acting on the ball, F g = mg and the tension force, F T from the string. (the balls weight makes it impossible to twirl the string truly horizontal, but if it was small enough we could ignore it and F T can act horizontally and provide the force to give the centripetal accel. F x = ma x or Where we round off because we ignore balls mass. Slide 17 Example 4: Tetherball anyone? The game of tetherball is played with a ball tied to a pole with a string. When the ball is struck, it whirls around the pole. In what direction is the acceleration of the ball, and what causes the acceleration? Slide 18 Example 4 Solution The acceleration points horizontally toward the center of the balls circular path. The force responsible for the acceleration may not be obvious at first, since there seems to be no force pointing directly horizontal. But it is the net force (sum of mg and F T ) that must point in the direction of the acceleration. The vertical component of the string tension balances the balls weight, mg. The horizontal component of the string tension, F Tx, is the force that produces the centripetal acceleration. Slide 19 Example 4 Illustrated F Ty FTFT F Tx mg Slide 20 5-3: A car rounding a curve One example of centripetal acceleration occurs when an automobile rounds a curve. The car must have an inward force exerted on it if it is to move in a curve. On a flat road, this force is supplied by friction between the tires and the pavement. As long as the tires are not slipping, the friction is static as one part is stationary for an instant. If friction is not great enough, the car will skid out of the curve in a nearly straight path. Slide 21 Skidding on a curve problem A 1000 kg car rounds a curve on a flat road of radius 50m at a speed of 50km/h (14m/s). Will the car make the turn, or will it skid if: (a) the pavement is dry and the coefficient of static friction, s = 0.60; (b) the pavement is icy and s = 0.25? Slide 22 Racecar solution The normal force on the car = its weight since the road is flat and no vertical acceleration: F N = mg = (1000 kg) (9.8 m/s 2 ) = 9800 N In the horizontal direction, the only force is friction and we must compare it to the force needed to produce the centripetal acceleration. (a) (F FR ) max = s F N = (0.60)(9800N) = 5900 N, so the car can make the turn. Slide 23 Racecar solution part b (b): (F FR ) max = s F N =(0.25)(9800N) = 2500 N The car will skid because the ground cannot exert sufficient force (3900 N is needed) to keep it moving in a curve of radius 50 m. If the wheels lock (stop rotating) when the brakes are applied too hard, the situation gets worse! Tires slide and the friction force is now kinetic which is less than static. Slide 24 The banking of curves helps reduce chance of skidding because there is a component of the normal force toward the center of the circle. The banking angle of a road, , is chosen so the horizontal component of the normal force, F n sin, is just equal to the force required to give centripetal acceleration, mv 2 /r. Slide 25 Your turn to Practice Please do Chapter 5 Review pg 138 #s 1,2,4,5,6 Please do Ch 5 Rev p 139 #s 1-4,6,7,9,14