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Transcript
LECTURE 14 CIRCULAR MOTION Instructor: Kazumi Tolich Lecture 14 2 ¨ Reading chapter 6-5 ¤ ¤ Velocity Uniform circular motion n ¤ ¤ Acceleration and speed Centripetal and tangential acceleration of non-uniform circular motion Dynamics of circular motion n n n Centripetal force Unbanked and banked curves Loop-the-Loop Circular motion 3 Motion along a circular path or a segment of a circular path is called circular motion. ¨ An object in a circular motion is always accelerating. ¨ Velocity of circular motion 4 ¨ ¨ Consider the motion of a pendulum. The direction of the velocity is always tangential to the path, perpendicular to the circle’s radius. Uniform circular motion 5 ¨ ¨ ¨ ¨ Motion in a circle at constant speed (not constant velocity!). Constantly accelerating since direction of the velocity is changing. The direction of acceleration in uniform circular motion is always toward the center of the circle, “centripetal” (center-seeking) acceleration. The magnitude of the centripetal acceleration is given by v2 acp = r Centripetal acceleration v ac Why don’t satellites fall into Earth? 6 Q. A. Why don’t satellites fall into Earth because of Earth’s gravity? They do. The tangential velocity of the satellite is fast enough so that the distance the satellite falls towards Earth, and the distance it travels in the tangential direction in a given time follow the satellite’s orbit. Example: 1 7 ¨ A spinner runs at v = 9.2 m/s around a circular track with a centripetal acceleration of acp = 3.8 m/s2. What is the track radius, r? Acceleration of non-uniform circular motion 8 ¨ The total acceleration of a particle moving in a circular path with tangential acceleration is given by ! ! ! atotal = acp + at http://www.youtube.com/watch?v=vT1mLpD6H_c Clicker question: 1 Centripetal force 10 ¨ ¨ ¨ To keep an object in a circular motion, force must be applied toward the center of the circle. Any form of such force (normal force, tension, gravity, etc) is called “centripetal force.” Centripetal force is just a name, so you should NEVER label a “centripetal force” in a FBD. v2 ∑ Fcp = m r Demo 1 11 ¨ Circle with Gap Direction of travel 12 ¨ ¨ ¨ ¨ Imagine a rock is whirled around in a uniform circular motion by a string. The tension in the string supplies the centripetal force necessary for the rock to follow the circular path. The velocity of the rock is always tangent to the path. If the string breaks, the rock will move along the straight line, tangent to the circular path. Recall Newton’s first law. no Clicker question: 2 Example: 2 14 ¨ Igor is a cosmonaut-engineer in a spacecraft orbiting Earth at an altitude h = 520 km with a speed v = 7.6 km/s. Igor’s mass is m = 79 kg. a) b) What is his acceleration? What (centripetal) gravitational force does Earth exert on Igor? Unbanked curves 15 On a unbanked road, static friction by the road on the car provides the centripetal force necessary for the car to follow a curve. ¨ It is static friction because no slipping occurs at the point of contact between road and tires. ¨ ! N Unbanked curves: 2 16 ¨ The maximum speed the car can go is limited by the maximum static friction the road can exert on the tire. x direction : − fs,max = − µs N = −macp,max 2 vmax = −m r y direction : N − mg = 0 → N = mg 2 vmax µs N = m r 2 vmax µs ( mg ) = m r vmax = µs gr ! N ! N +x Example: 3 17 ¨ A bicyclist travels at a constant speed of v = 9.00 m/s in a circle of radius r = 25.0 m on a flat ground. The combined mass of the bicycle and rider is m = 85.0 kg. Calculate the magnitude of the force of friction exerted by the road on the bicycle. Banked curves 18 On a banked road, the normal force of the road will have a component in the centripetal direction. ¨ Banking is useful for an icy road. ¨ The banking angle is usually chosen so that no friction is needed for a car to complete the curve at the specified speed. ¨ N Banked curves: 2 19 ¨ Banking angle: v2 x direction : N sin θ = macp = m r y direction : N cosθ − mg = 0 → N cosθ = mg v2 N sin θ m r = N cosθ mg v2 tan θ = rg Banking angle N ! N Loop-the-Loop 20 ¨ ¨ If you are going fast enough, the normal force on the bike is nonzero, i.e, the bike is touching the track. If you are barely making it, the normal force at the top is zero. At the top : N = 0 2 vmin ∑ F = N + mg = 0 + mg = m r vmin = gr At the top mg N Loop-the-loop failure 21 ¨ Loop-the-loop fails when you do not have enough speed at the top. ¤ http://www.youtube.com/watch?v=tzQJNeqiGG4 Example: 4 22 ¨ The radius of curvature of the track at the top of a loop-the-loop on a rollercoaster ride is r = 12.0 m. At the top of the loop, the force that the seat exerts on a passenger of mass m is 0.40mg. How fast is the roller-coaster car moving as it moves through the highest point of the loop?