Extension of Circular Motion & Newton s Laws. Chapter 6 Mrs. Warren Kings High School

Transcription

1 Extension of Circular Motion & Newton s Laws Chapter 6 Mrs. Warren Kings High chool

2 Review from Chapter 4 Uniform Circular Motion Centripetal Acceleration

3 Uniform Circular Motion, Force F r A force is associated with the centripetal acceleration. The force is also directed toward the center of the circle. å 2 F = mac = m r Applying Newton s econd Law along the radial direction gives v ection 6.1

4 Uniform Circular Motion, cont. A force causing a centripetal acceleration acts toward the center of the circle. It causes a change in the direction of the velocity vector. If the force vanishes, the object would move in a straight-line path tangent to the circle. ee various release points in the active figure ection 6.1

5 The Conical Pendulum A small ball of mass, m, is suspended from a string of length, L. The ball revolves with constant speed, v, in a horizontal circle of radius, r. Find an expression for v in terms of the geometry. v = Lg sinq tanq

6 The Conical Pendulum A puck of mass kg is attached to the end of a cord. The puck moves in a horizontal circle of radius 1.50 m. If the cord can withstand a maximum tension of 50.0 N, what is the maximum speed at which the puck can move before the cord breaks? Assume the string remains horizontal during the motion. v = Tr m

7 The Flat Curve Model the car as a particle in uniform circular motion in the horizontal direction. Model the car as a particle in equilibrium in the vertical direction. The force of static friction supplies the centripetal force. The maximum speed at which the car can negotiate the curve is: v = m gr s Note, this does not depend on the mass of the car.

8 The Flat Curve A 1500 kg car moving on a flat, horizontal road negotiates a curve as shown on the previous slide. If the radius of the curve is 35.0 m and the coefficient of static friction between the tires and dry pavement is 0.523, find the maximum speed the car can have and still make the turn successfully.

9 The Banked Roadway These are designed with friction equaling zero. Model the car as a particle in equilibrium in the vertical direction. Model the car as a particle in uniform circular motion in the horizontal direction. There is a component of the normal force that supplies the centripetal force. The angle of bank is found from

10 The Banked Curve A civil engineer wishes to redesign the curved roadway in such a way that a car will not have to rely on friction to round the curve without skidding. In other words, a car moving at the designated speed can negotiate the curve even when the road is covered with ice. uch a road is usually banked, which means that the roadway is tilted toward the inside of the curve. uppose the designated speed for the road is to be 30.0 mi/h and the radius of the curve is 35.0 m. At what angle should the curve be banked?

11 Ferris Wheel A child of mass m rides on a Ferris wheel as shown. The child moves in a vertical circle of radius 10.0 m at a constant speed of 3.00 m/s. Determine the force exerted by the seat on the child at the bottom of the ride. Express your answer in terms of the weight of the child, mg. Do the same when the child is at the top of the ride.

12 Non-Uniform Circular Motion The acceleration and force have tangential components. F r F t produces the centripetal acceleration produces the tangential acceleration The total force is å å å F = Fr + Ft ection 6.2

13 Vertical Circle with Non- Uniform peed 2 v T mg æ ö = ç + cosq è Rg ø A small sphere of mass, m, is attached to the end of a cord of length R and set into motion in a vertical circle about a fixed point O as illustrated. Determine the tangential acceleration of the sphere and the tension in the cord at any instant when the speed of the sphere is v and the cord makes an angle with the vertical. ection 6.2

14 Motion in Accelerated Frames A fictitious force results from an accelerated frame of reference. The fictitious force is due to observations made in an accelerated frame. A fictitious force appears to act on an object in the same way as a real force, but you cannot identify a second object for the fictitious force. Remember that real forces are always interactions between two objects. imple fictitious forces appear to act in the direction opposite that of the acceleration of the non-inertial frame. ection 6.3

15 Centrifugal Force

16 Coriolis Force

17 Motion with Resistive Forces The medium exerts a resistive force, the medium. R, on an object moving through The magnitude of.. The direction of R The magnitude of R is R can depend on the speed in complex ways. We will discuss only two: R is proportional to v R Good approximation for is proportional to v 2 Good approximation for ection 6.4

18 Resistive Force Proportional To peed The resistive force can be expressed as R = -bv b depends on the property of the medium, and on the shape and dimensions of the object. The negative sign indicates is in the opposite direction to v. R ection 6.4

19 Resistive Force Proportional To peed, Example Assume a small sphere of mass m is released from rest in a liquid. Forces acting on it are: Resistive force Gravitational force Analyzing the motion results in dv mg - bv = ma = m dt dv b a = = g - v dt m ection 6.4

20 Resistive Force Proportional To peed, Example, cont. Initially, v = 0 and dv/dt = g As t increases, R increases and a decreases The acceleration approaches 0 when R mg At this point, v approaches the terminal speed of the object.

21 Terminal peed To find the terminal speed, let a = 0 olving the differential equation gives t is the time constant and t = m/b ection 6.4

22 Terminal peed A small sphere of mass 2.00 g is released from rest in a large vessel filled with oil, where it experiences a resistive force proportional to its speed. The sphere reaches a terminal speed of 5.00 cm/s. Determine the time constant and the time at which the sphere reaches 90.0% of its terminal speed.

23 Resistive Force Proportional To v 2 For objects moving at high speeds through air, the resistive force is approximately equal to the square of the speed. R = ½ DrAv 2 D is a dimensionless empirical quantity called the drag coefficient. r is the density of air. A is the cross-sectional area of the object. v is the speed of the object. ection 6.4

24 Resistive Force Proportional To v 2, example Analysis of an object falling through air accounting for air resistance. 1 2 åf = mg - DrAv = ma 2 æ DrA ö 2 a = g - ç v è 2m ø

25 Resistive Force Proportional To v 2, Terminal peed The terminal speed will occur when the acceleration goes to zero. olving the previous equation gives v T = 2 mg DrA

26 Example: kysurfer tep from plane Initial velocity is 0 Gravity causes downward acceleration Downward speed increases, but so does upward resistive force Eventually, downward force of gravity equals upward resistive force Traveling at terminal speed ection 6.4

27 kysurfer, cont. Open parachute ome time after reaching terminal speed, the parachute is opened. Produces a drastic increase in the upward resistive force Net force, and acceleration, are now upward The downward velocity decreases. Eventually a new, smaller, terminal speed is reached. ection 6.4

28 HW Objective Questions: 1, 4, 5, 6, 7 Conceptual Questions: 4, 5, 6, 8 Problems: (ection 6.1): 1, 2, 6, 7, 8, 9/ (ection 6.2): 13, 14, 15, 16, 18/ (ection 6.3): 21, 22/ (ection 6.4): 27, 28, 30, 31