Transcript Document
Chapter 5 Circular Motion; Gravitation Units of Chapter 5 •Kinematics of Uniform Circular Motion •Dynamics of Uniform Circular Motion •Highway Curves, Banked and Unbanked •Nonuniform Circular Motion •Centrifugation •Newton’s Law of Universal Gravitation •Gravity Near the Earth’s Surface; Geophysical Applications •Satellites and “Weightlessness” •Kepler’s Laws and Newton’s Synthesis •Types of Forces in Nature 5-1 Kinematics of Uniform Circular Motion Uniform circular motion: motion in a circle of constant radius at constant speed Instantaneous velocity is always tangent to circle. 5-1 Kinematics of Uniform Circular Motion Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that (5-1) This acceleration is called the centripetal, or radial, acceleration, and it points towards the center of the circle. Example 5-1 A 150 kg 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 in a second. What is its centripetal acceleration? v2 aR = r 2r 2(3.14)(0.600 m) v= = = 7.54 m/s T (0.500 s) v 2 (7.54 m/s) 2 aR = = = 94.7 m/s 2 r (0.600 m) Example 5-2 The moon’s nearly circular orbit about the earth has a radius of about 384,000 km and a period T of 27.3 days. Determine the acceleration of the moon toward the earth. r = 3.84x10 8 m, T = (27.3 d)(24.0 h/d)(3600 s/h) = 2.36x10 6 s 2r v 2 4 2r 2 4 2r 4 2 (3.84x10 8 m) v= a = = 2 = 2 = = 0.00272 m/s 2 6 2 T r Tr T (2.36x10 s) 5-2 Dynamics of Uniform Circular Motion 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: 5-2 Dynamics of Uniform Circular Motion There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome. If the centripetal force vanishes, the object flies off tangent to the circle. Example 5-3 Estimate the force a person must exert on a string attached to a 0.150 kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second (T=0.500 s). v2 (2r/T) 2 (F)R = ma R = m = m r r 4 2 (0.600 m/0.500 s) 2 = (0.150 kg) (0.600 m) = 14 N 5-3 Highway Curves, Banked and Unbanked When a car goes around a curve, there must be a net force towards the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction. 5-3 Highway Curves, Banked and Unbanked If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show. 5-3 Highway Curves, Banked and Unbanked As long as the tires do not slip, the friction is static. If the tires do start to slip, the friction is kinetic, which is bad in two ways: 1. The kinetic frictional force is smaller than the static. 2. The static frictional force can point towards the center of the circle, but the kinetic frictional force opposes the direction of motion, making it very difficult to regain control of the car and continue around the curve. 5-3 Highway Curves, Banked and Unbanked Banking the curve can help keep cars from skidding. In fact, for every banked curve, there is one speed where the entire centripetal force is supplied by the horizontal component of the normal force, and no friction is required. This occurs when: Example 5-6 A 1000. kg car rounds a curve o a flat road of radius 50. m at a speed of 50. km/h (14 m/s). Will the car follow the curve, or it will it skid? Assume (a) the pavement is dry and the coefficient of static friction is s=0.60; (b) the pavement is icy and s=0.25. FN = mg = (1000. kg)(9.8 m/s 2 ) = 9800 N v2 (14 m/s) 2 (F)R = ma R = m = (1000 kg) = 3900 N r (50 m) (a) (Ffr ) max = sFN = (0.60)(9800 N) = 5900 N Since this is greater than 3900 N, the car will follow the curve. (b) (Ffr ) max = sFN = (0.25)(9800 N) = 2500 N Since this is less than 3900 N, the car will not follow the curve, so it will skid. Example 5-7 (a) For a car traveling with speed v around a curve of radius r, determine a formula for the angle at which a road should be banked so that no friction is required. (b) What is the angle for an expressway off-ramp curve of 50 m at a design speed of 50 km/h? mv 2 FR = ma R FNsin = r Fy = ma y = 0 FNcos - mg = 0 FN = mg cos mv 2 mg mv 2 mv 2 FNsin = sin = mgtan = r cos r r v2 tan = rg For r = 50 m and v = 14 m/s, we have (14 m/s) 2 tan = = 0.40 = 22 (50 m)(9.8 m/s 2 ) 5-4 Nonuniform Circular Motion If an object is moving in a circular path but at varying speeds, it must have a tangential component to its acceleration as well as the radial one. 5-4 Nonuniform Circular Motion This concept can be used for an object moving along any curved path, as a small segment of the path will be approximately circular. 5-5 Centrifugation A centrifuge works by spinning very fast. This means there must be a very large centripetal force. The object at A would go in a straight line but for this force; as it is, it winds up at B. Example 5-9 The rotor of an ultracentrifuge rotates at 50,000 rpm (revolutions per minute). The top of a 4.00 cm long test tube is 6.00 cm from the rotation axis and is perpendicular to it. The bottom of the tube is 10.00 cm from the axis of rotation. Calculate the centripetal accelerations in “g’s”, at the top and the bottom of the tube. At top, 2r = (2 )(0.0600 m) = 0.377 m per revolution 50,000 rpm = 833 rev/s so T = 1 = 1.20x10 -3 s/rev 833 rev/s 2r 0.377 m/rev 2 = = 3.14x10 m/s -3 T 1.20x10 s/rev v 2 (3.14x10 2 m/s) 2 aR = = = 1.64x10 6 m/s 2 = 1.67x10 5 g' s r 0.0600 m 2r (2 )(0.1000 m) At the bottom, v = = = 523.6 m/s -3 T 1.20x10 s/rev v 2 (523.6 m/s) 2 aR = = = 2.74x10 6 m/s 2 = 2.80x10 5 g' s r 0.1000 m v=