Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 1 PHYS 1441 – Section 002 Lecture #11 Wednesday, Oct. 13, 2010 Dr. Jaehoon Yu • Force of Friction Review • Example for Motion with friction • Uniform Circular Motion • Centripetal Acceleration • Newton’s Law & Uniform Circular Motion • Unbanked and Banked highways • Newton’s Law of Universal GravitationAnnouncements • Reading Assignments – CH5.4, 5.5, 5.9 and 6.2 • Quiz next Monday, Oct. 18 – Beginning of the class – Covers CH4.5 to CH5.3 • Colloquium today – Our own Dr. Q. Zhang Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 2Wednesday, Oct. 13, 2010 3 PHYS 1441-002, Fall 2010 Dr. Jaehoon YuWednesday, Oct. 13, 2010 Forces of Friction Summary fs≤µsFNResistive force exerted on a moving object due to viscosity or other types frictional property of the medium in or surface on which the object moves. Force of static friction, fs: Force of kinetic friction, fk The resistive force exerted on the object until just before the beginning of its movement The resistive force exerted on the object during its movement These forces are either proportional to the velocity or the normal force. Empirical Formula What does this formula tell you? Frictional force increases till it reaches the limit!! Beyond the limit, the object moves, and there is NO MORE static friction but kinetic friction takes it over. Which direction does kinetic friction apply? Opposite to the motion! PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 4Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 5 Example w/o Friction A crate of mass M is placed on a frictionless inclined plane of angle θ. a) Determine the acceleration of the crate after it is released. F=Free-body Diagram θ"x y Fg n n F= -Mg θ"Supposed the crate was released at the top of the incline, and the length of the incline is d. How long does it take for the crate to reach the bottom and what is its speed at the bottom? d =ax= gsinθ∴ t =2dgsinθx y Fg+ n=May=n − mg cosθ=vixt +12axt2=12gsinθ t2vix+ axt =gsinθ2dgsinθ=2dgsinθ∴ vxf= 2dg sinθWednesday, Oct. 13, 2010 Example w/ Friction Suppose a block is placed on a rough surface inclined relative to the horizontal. The inclination angle is increased till the block starts to move. Show that by measuring this critical angle, θc, one can determine coefficient of static friction, µs. Free-body Diagram θ"x y Fg n n F= -Mg θ"Fs=μsn Net force x comp. y comp. Fg+ n+ fsFgx− fs=PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 6 FsWednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu Uniform circular motion is the motion of an object traveling at a constant speed on a circular path. Definition of the Uniform Circular Motion 7Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu Let T be the period of this motion, the time it takes for the object to travel once around the complete circle whose radius is r. Speed of a uniform circular motion? 2πr T distancetime8Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu The wheel of a car has a radius of 0.29m and is being rotated at 830 revolutions per minute on a tire-balancing machine. Determine the speed at which the outer edge of the wheel is moving. 1830 revolutions min=Ex. : A Tire-Balancing Machine 1.2 × 10−3min revolution 1.2 × 10−3 min = 0.072 s 2πrT= 2π0.29 m( )0.072 s= 25m s9Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu In uniform circular motion, the speed is constant, but the direction of the velocity vector is not constant. Centripetal Acceleration 10 The change of direction of the velocity is the same as the change of the angle in the circular motion!Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu Centripetal Acceleration From the geometry Centripetal Acceleration What is the direction of ac? Always toward the center of circle! sinθ2 =11Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu Newton’s Second Law & Uniform Circular Motion The centripetal * acceleration is always perpendicular to the velocity vector, v, and points to the center of the rotational axis (radial direction) in a uniform circular motion. The force that causes the centripetal acceleration acts toward the center of the circular path and causes the change in the direction of the velocity vector. This force is called the centripetal force. Are there forces in this motion? If so, what do they do? What do you think will happen to the ball if the string that holds the ball breaks? The external force no longer exist. Therefore, based on Newton’s 1st law, the ball will continue its motion without changing its velocity and will fly away following the tangential direction to the circle. *Mirriam Webster: Proceeding or acting in the direction toward the center or rotational axis 12Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu Ex. Effect of Radius on Centripetal Acceleration The bobsled track at the 1994 Olympics in Lillehammer, Norway, contained turns with radii of 33m and 23m. Find the centripetal acceleration at each turn for a speed of 34m/s, a speed that was achieved in the two –man event. Express answers as multiples of g=9.8m/s2. Centripetal acceleration: R=33m R=24m 3.6g13Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu Example of Uniform Circular Motion A ball of mass 0.500kg is attached to the end of a 1.50m long cord. The ball is moving in a horizontal circle. If the string can withstand maximum tension of 50.0 N, what is the maximum speed the ball can attain before the cord breaks? Centripetal acceleration: When does the string break? when the required centripetal force is greater than the sustainable tension. mv2r=Calculate the tension of the cord when speed of the ball is 5.00m/s. Trm= 50.0 × 1.50.500= 12.2 m / s( ) mv2r= 0.500 ×5.00( )21.5= 8.33 N(
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