Banked Curves v 2 F = F sinθ = m F cosθ mg c N r N =

Transcription

1 Banked Curves On a frictionless banked curve, the centripetal force is the horizontal component of the normal force. The vertical component of the normal force balances the car s weight.

2 Banked Curves F c v = FN sin θ = m F N cosθ = mg r

3 Banked Curves v F N sin θ = m r F N cosθ = mg tan θ = v rg

4 Banked Curves Example: The Daytona 500 The turns at the Daytona International Speedway have a maximum radius of 316 m and are steeply banked at 31 degrees. Suppose these turns were frictionless. As what speed would the cars have to travel around them? tan θ = v rg v = rg tanθ v = 316 m ( ) 9.8m s ( )tan31! = 43m s 96 mph ( )

5 Vertical Circular Motion In vertical circular motion the gravitational force must also be considered. An example of vertical circular motion is the vertical loop-the-loop motorcycle stunt. Normally, the motorcycle speed will vary around the loop. The normal force, F N, and the weight of the cycle and rider, mg, are shown at four locations around the loop.

6 Vertical Circular Motion 1) ) 3) 4) v1 F N 1 mg = m r v F = N m r v3 F N 3 + mg = m r v F = 4 N 4 m r

7 Vertical Circular Motion There is a minimum speed the rider must have at point 3 in order to stay on the loop. This speed may be found by setting F N3 = 0 in the centripetal force equation for point 3, i.e. in F N3 + mg = mv 3 /r è v 3min = (rg) 1/ e.g. for a track with r = 10 m, v 3min = ((10)(9.8)) 1/ = 9.9 m/s = mph

8 Example: Find the orbital periods of Mercury (~0.4 AU from the Sun) and Neptune (~30 AU from the Sun) and compare with that of the Earth. M = Sun mass, m = planet mass, r = distance to Sun, v = planet speed in orbit F C = mv r 1 AU = distance from Earth to Sun =1.5 x m τ Earth = π τ Mercury = π τ Neptune = π ( ) = G mm r τ = orbital period = πr v τ = π ( )( ) = s ( ) r 3 GM ( )( ) = s = 0.5τ Earth ( ) ( )( ) = s =160τ Earth v r = G M r 3

9 End of material for Midterm 1

10 Chapter 6 Work and Energy

11 Work Done by a Constant Force Work involves force and displacement. W = Fs 1 N m = 1 joule ( J)

12 Work Done by a Constant Force

13 Work Done by a Constant Force More general definition of the work done on an object, W, by a constant force, F, through a displacement, s, with an angle, θ, between F and s: W = ( F cosθ)s cos0 cos90 cos180 = 1 = 0 = 1 Note that W is a scalar!

14 Work Done by a Constant Force Example: Pulling a Suitcase-on-Wheels Find the work done if the force is 45.0 N, the angle is 50.0 degrees, and the displacement is 75.0 m. W = ( F cosθ)s =! "( 45.0 N)cos50.0 # $ 75.0 m ( ) = 170 J

15 Work Done by a Constant Force A weightlifter doing positive and negative work on the barbell: Lifting phase --> positive work. W = F cos0 ( )s = Fs Lowering phase --> negative work. W = F cos180 ( )s = Fs

16 Example. Work done skateboarding. A skateboarder is coasting down a ramp and there are three forces acting on her: her weight W (675 N magnitude), a frictional force f (15 N magnitude) that opposes her motion, and a normal force F N (61 N magnitude). Determine the net work done by the three forces when she coasts for a distance of 9. m. W = (F cos θ)s, work done by each force W = (675 cos 65 o )(9.) = 60 J W = (15 cos 180 o )(9.) = J W = (61 cos 90 o )(9.) = 0 J The net work done by the three forces is: 60 J J +0 J = 1470 J

17 Work Done by a Constant Force Example: Accelerating a Crate The truck is accelerating at a rate of m/s. The mass of the crate is 10 kg and it does not slip. The magnitude of the displacement is 65 m. What is the total work done on the crate by all of the forces acting on it?

18 Work Done by a Constant Force The angle between the displacement and the normal force is 90 degrees. The angle between the displacement and the weight is also 90 degrees. Thus, for F N and W, W = F cos90 ( )s = 0

19 Work Done by a Constant Force The angle between the displacement and the friction force is 0 degrees. f s = ma = ( 10 kg) ( 1.5m s ) =180N W =!" ( 180N)cos0# $ ( 65 m) = J

20 The Work-Energy Theorem and Kinetic Energy Consider a constant net external force acting on an object. The object is displaced a distance s, in the same direction as the net force. F s The work is simply W = (Σ F)s = (ma)s

21 The Work-Energy Theorem and Kinetic Energy W = ( ) ( ) = m 1 v v = 1 mv 1 mv m as f o f o v f = v o + as ( ) constant acceleration kinematics formula as ( ) = 1 v f ( v ) o DEFINITION OF KINETIC ENERGY The kinetic energy KE of and object with mass m and speed v is given by KE = 1 mv SI unit: joule (J)

22 The Work-Energy Theorem and Kinetic Energy THE WORK-ENERGY THEOREM When a net external force does work on an object, the kinetic energy of the object changes according to W = KE KE f o = 1 mv f 1 mv o (also valid for curved paths and non-constant accelerations)

23 The Work-Energy Theorem and Kinetic Energy Example: Deep Space 1 The mass of the space probe is 474 kg and its initial velocity is 75 m/s. If the 56.0 mn force acts on the probe through a displacement of m, what is its final speed?

24 The Work-Energy Theorem and Kinetic Energy W = 1 f mv 1 mv o W = [(Σ F)cos θ]s

25 The Work-Energy Theorem and Kinetic Energy [( ) ] 1 1 F cosθ s = mvf mvo [(Σ F)cos θ]s ( N)cos0 ( m) = 1 ( 474 kg)v f 1 ( 474 kg) ( 75m s) v f = 805m s

26 The Work-Energy Theorem and Kinetic Energy Conceptual Example: Work and Kinetic Energy A satellite is moving about the earth in a circular orbit and an elliptical orbit. For these two orbits, determine whether the kinetic energy of the satellite changes during the motion.

27 Gravitational Potential Energy Find the work done by gravity in accelerating the basketball downward from h o to h f. W = F cosθ ( )s W gravity = mg( h o h ) f

28 Gravitational Potential Energy W gravity = mg( h o h ) f We get the same result for W gravity for any path taken between h o and h f.

29 Gravitational Potential Energy Example: A Gymnast on a Trampoline The gymnast leaves the trampoline at an initial height of 1.0 m and reaches a maximum height of 4.80 m before falling back down. What was the initial speed of the gymnast?

30 Gravitational Potential Energy W = 1 mv f 1 mv o mg ( ) 1 h h = mv o f o W gravity = mg ( ) h o h f v o = g( h o h ) f v o = ( 9.80m s )( 1.0 m 4.80 m) = 8.40m s

31 Gravitational Potential Energy W gravity = mgh o mgh f DEFINITION OF GRAVITATIONAL POTENTIAL ENERGY The gravitational potential energy PE is the energy that an object of mass m has by virtue of its position relative to the surface of the earth. That position is measured by the height h of the object relative to an arbitrary zero level: PE = mgh 1 N m = 1 joule ( J)

32 The Ideal Spring HOOKE S LAW: RESTORING FORCE OF AN IDEAL SPRING The restoring force on an ideal spring is F x = k x SI unit for k: N/m

33 The Ideal Spring Example: A Tire Pressure Gauge The spring constant of the spring is 30 N/m and the bar indicator extends.0 cm. What force does the air in the tire apply to the spring?

34 The Ideal Spring F Applied x = = k x ( 30 N m)( 0.00 m) = 6.4 N

35 Potential energy of an ideal spring A compressed spring can do work.

36 Potential energy of an ideal spring W elastic = ( F cosθ)s = 1 ( kx + kx o f )cos0 ( x o x ) f F 0 F f F = kx W elastic = 1 1 kx o kx f x f x 0 F = 1 ( F + F ) 0 f s F

37 Potential energy of an ideal spring DEFINITION OF ELASTIC POTENTIAL ENERGY The elastic potential energy is the energy that a spring has by virtue of being stretched or compressed. For an ideal spring, the elastic potential energy is PEelastic = 1 kx SI Unit of Elastic Potential Energy: joule (J)