Ch5. Uniform Circular MotionExample 1: A Tire-Balancing MachineSlide 3Slide 4Centripetal AccelerationSlide 6Slide 7Conceptual Example 2: Which way will the object go?Slide 9Slide 10Example 3: The Effect of Radius on Centripetal AccelerationSlide 12Conceptual Example 4: Uniform Circular Motion and EquilibriumSlide 14Check your understanding 1Slide 16Centripetal ForceSlide 18Slide 19Example 5: The Effect of Speed on Centripetal ForceSlide 21Conceptual Example 6: A Trapeze ActSlide 23Example 7: Centripetal Force and Safe DrivingSlide 25Slide 26Slide 27Check your understanding 2Slide 29Banked CurvesSlide 31Slide 32Example 8:The Daytona 500Slide 34Satellites in Circular OrbitsSlide 36Example 9: Orbital Speed of the Hubble Space TelescopeSlide 38Global Positioning System(GPS)Example 10: A Super-massive Black HoleSlide 41Slide 42Slide 43Slide 44Example 11: The Orbital Radius for Synchronous SatellitesSlide 46Check your understanding 3Apparent Weightlessness and Artificial GravityConceptual Example 12: Apparent Weightlessness and Free-FallExample 13: Artificial GravitySlide 51Example 14: A Rotating Space LaboratorySlide 53Slide 54Check your understanding 4Vertical Circular MotionSlide 57Slide 58Concepts & Calculation Examples 15: AccelerationSlide 60Slide 61Concepts & Calculation Example 16: Centripetal ForceSlide 63Slide 64Problem 4Slide 66Slide 67Problem 43Slide 69Problem 46Slide 711Ch5. Uniform Circular MotionUniform 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.Trv2r2Example 1: A Tire-Balancing MachineThe wheel of a car has a radius of r = 0.29m and is being rotated at 830 revolutions per minute (rpm) on a tire-balancing machine. Determine the speed (in m/s) at which the outer edge of the wheel is moving.The speed v can be obtained directly from , but first the period T is needed. It must be expressed in seconds. Trv23830 revolutions in one minuterevolutionsrevolutionmin/102.1min/83013T=1.2*10-3 min, which corresponds to 0.072ssmsmTrv /25072.0)29.0(224Uniform circular motion emphasizes that 1. The speed, or the magnitude of the velocity vector, is constant. 2. Direction of the vector is not constant. 3. Change in direction, means acceleration 4. “Centripetal acceleration” , it points toward the center of the circle.5Centripetal AccelerationMagnitude ac of the centripetal acceleration depends on the speed v of the object and the radius r of the circular path. ac=v2/r6v in velocity divided by the elapsed time or a= /tvtSector of the circle COP. is very small, the arc length OP is approximately a straight line whose length is the distance v traveled by the object.tt7COP is an isosceles triangle. Both triangles have equal apex angles .rtvvv tv  /ac=v2/rThe direction is toward the center of the circle.8Conceptual Example 2: Which way will the object go?An object on a guideline is in uniform circular motion. The object is symbolized by a dot, and at point O it is release suddenly from its circular path.If the guideline is cut suddenly, will the object move along OA or OP ?9Newton’s first law of motion guides our reasoning. An object continues in a state of rest or in a state of motion at a constant speed along a straight line unless compelled to changes that state by a net force. When the object is suddenly released from its circular path, there is no longer a net force being applied to the object. In the case of a model airplane, the guideline cannot apply a force, since it is cut. Gravity certainly acts on the plane, but the wings provide a lift force that balances the weight of the plane.10As a result, the object would move along the straight line between points O and A, not on the circular arc between points O and P.In the absence of a net force, then, the plane or any object would continue to move at a constant speed along a straight line in the direction it had at the time of release. This speed and direction are given in Figure 5.4 by the velocity vector v.11Example 3: The Effect of Radius on Centripetal AccelerationThe bobsled track at the 1994 Olympics in Lillehammer, Norway, contained turns with radii of 33 m and 24 m, as the figure illustrates. Find the centripetal acceleration at each turn for a speed of 34 m/s, a speed that was achieved in the two-man event. Express the answers as multiples of g=9.8m/s2.12From ac=v2/r it follows thatRadius=33mRadius=24mgsmmsmac6.3/3533)/34(22gsmmsmac9.4/4824)/34(2213Conceptual Example 4: Uniform Circular Motion and EquilibriumA car moves at a constant speed, and there are three parts to the motion. It moves along a straight line toward a circular turn, goes around the turn, and then moves away along a straight line. In each of three parts, is the car in equilibrium?14An object in equilibrium has no acceleration, according to the definition given in Section 4.11. As the car approaches the turn, both the speed and direction of the motion are constant. Thus, the velocity vector does not change, and there is no acceleration. The same is true as the car moves away from the turn. For these parts of the motion, then, the car is in equilibrium. As the car goes around the turn, however, the direction of travel changes, so the car has a centripetal acceleration that is characteristic of uniform circular motion. Because of this acceleration, the car is not in equilibrium during the turn. In general, an object that is in uniform circular motion can never be in equilibrium.15Check your understanding 1The car in the drawing is moving clockwise around a circular section of road at a constant speed. What are the directions of its velocity and acceleration at (a) position 1 and (b) position 2?16(a) The velocity is due south, and the acceleration is due west.(b) The velocity is due west, and the acceleration is due north.17Centripetal Force18Concepts at a glance: Newton’s second law indicates that whenever an object accelerates, there must be a net force to create the acceleration. Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration. As the Concept-at-a-glance chart, the second law gives this net force as the product of the object’s mass m and its acceleration v2/r. This chart is an expanded version of the