Topic 1: Newtonian Mechanics Energy & Momentum

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1 Work (W) the amount of energy transferred by a force acting through a distance. Scalar but can be positive or negative ΔE = W = F! d = Fdcosθ Units N m or Joules (J)

2

3 Work, Energy & Power Power (P) the rate at which work is done and hence energy is transferred. P = ΔE Δt = W Δt = Fdcosθ = Fv cosθ Δt Units J/s or Watts

4 Is work being performed on the satellite by the force of gravity?

5 Work may be done by individual forces, but there is no net work done on an object when the net force is zero. The object is in equilibrium (at rest or constant velocity)

6 ratio of useful work done by a system to the total work done by the system ratio of useful energy output of a system to the total energy input to the system ratio of useful power output of a system to total power input to the system e = output input x 100

7 K = ½ mv 2 U s = ½ kx 2 U e = Pt = VIt U g = mgδy Q = mcδt = ml

8 Elastic Potential Energy! F = k x! s U s = 1 2 kx2 = F s x F max Force (N) F avg Displacement (m) x

9 Gravitational Potential Energy Path Independence: the change in U g is the same for any path between the same two locations ΔU g = mgδy

10 Conservative & Non-Conservative Forces

11 Conservative & Non-Conservative Forces The total mechanical energy of an object remains constant if the net work done by external non-conservative forces is zero, Wnc = 0 J. E o = E f

12 Conservative & Non-Conservative Forces In the case of non-conservative forces being present, we can state: E o + W nc = E f

13 One of the fastest roller coasters in the world is the Magnum XL-200 at Cedar Point Park in Sandusky, Ohio. The ride includes a vertical drop of 59.4 m. Assume that the coaster has a speed of nearly zero as it crests the top of the hill. Neglect friction and find the speed of the riders at the bottom of the hill.

14 One of the fastest roller coasters in the world is the Magnum XL-200 at Cedar Point Park in Sandusky, Ohio. The ride includes a vertical drop of 59.4 m. Assume that the coaster has a speed of nearly zero as it crests the top of the hill. Neglect friction and find the speed of the riders at the bottom of the hill. E o = E f U g = K mgh = 1/2 mv 2 v = 2gh = (2)(9.8 m/s2)(59.4 m) = 34.1 m/s

15 In the previous example, we ignored non-conservative forces, such as friction. In reality, however, such forces are present when the roller coaster descends the hill. The actual speed of the riders at the bottom is 32.2 m/s. Assuming again that the coaster has a speed of nearly zero at the top of the hill, find the work done by non-conservative forces on a 55.0 kg rider during the descent down the hill.

16 In the previous example, we ignored non-conservative forces, such as friction. In reality, however, such forces are present when the roller coaster descends the hill. The actual speed of the riders at the bottom is 32.2 m/s. Assuming again that the coaster has a speed of nearly zero at the top of the hill, find the work done by non-conservative forces on a 55.0 kg rider during the descent down the hill. E o + W nc = E f U g + W nc = K W nc = K - U g W nc = 1/2 mv 2 - mgh W nc = 1/2 (55.0 kg)(32.2 m/s) 2 - (55.0 kg)(9.8 m/s 2 )(59.4 m) = J W nc = 3.5 x 10 3 J

17 A pendulum is allowed to swing freely from rest. Derive the possible tension in the string as the pendulum bob passes through its lowest position.

18 A pendulum is allowed to swing freely from rest. Derive the possible tension in the string as the pendulum bob passes through its lowest position.

19 A pendulum is allowed to swing freely from rest. Derive the possible tension in the string as the pendulum bob passes through its lowest position. Fc = ma F T F T - mg = mv2 r F T = mg + mv2 L What is v? F g

20 A pendulum is allowed to swing freely from rest. Derive the possible tension in the string as the pendulum bob passes through its lowest position. Fc = ma F T - mg = mv2 r Eo = Ef U g = K F T F T = mg + mv2 L mgh = 1/2 mv 2 gl = 1/2 v 2 v 2 = 2gL F g

21 F T F g A pendulum is allowed to swing freely from rest. Derive the possible tension in the string as the pendulum bob passes through its lowest position. Fc = ma F T - mg = mv2 r F T = mg + mv2 L F T = mg + m(2gl) L F T = mg + 2mg F T = 3mg Eo = Ef U g = K mgh = 1/2 mv 2 gl = 1/2 v 2 v 2 = 2gL

22 A rollercoaster has a loop-de-loop section as shown. Derive the minimum height the car must start from to successfully make it through the loop?

23 A rollercoaster has a loop-de-loop section as shown. Derive the minimum height the car must start from to successfully make it through the loop? E o = E f mgh o = mgh f + 1/2 mv 2 What is v? gh o = g(2r) + 1/2 v 2 Loop height = 2R

24 A rollercoaster has a loop-de-loop section as shown. Derive the minimum height the car must start from to successfully make it through the loop? E o = E f mgh o = mgh f + 1/2 mv 2 gh o = g(2r) + 1/2 v 2 F = ma F N + F g = mv2 r 0 N + mg = mv2 r v 2 = gr

25 A rollercoaster has a loop-de-loop section as shown. Derive the minimum height the car must start from to successfully make it through the loop? E o = E f mgh o = mgh f + 1/2 mv 2 gh o = g(2r) + 1/2 v 2 gh o = g (2r) + 1/2 ( gr) h o = 2.5 r F = ma F N + F g = mv2 r 0 N + mg = mv2 r v 2 = gr

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27 Linear Motion - the quantity of motion of a moving body, measured as a product of its mass and velocity.! p = m v! Which object possesses more momentum? Units kg m/s

28 Linear Motion - the quantity of motion of a moving body, measured as a product of its mass and velocity.! p = m v! Units kg m/s When could granny possess more momentum than the truck?

29 Topic 1: Newtonian Mechanics Impulse (J) the change in momentum of a system Vector

30 Derivation of Impulse J = Δ! p =! FΔt = m! v

31 Calculate the impulse experienced by the 1500 kg car as it bounce off the wall.

32 Calculate the impulse experienced by the 1500 kg car as it bounce off the wall. J = Δp = mδv = m (v - v o ) J = Δp = 1500kg ( +2.60m/s - (-15.0m/s)) J = Δp = 2.64x10 4 kg im/s

33 Compare the forces, impact time, impulse, acceleration, and change in momentum for each vehicle in the impending collision.

34 Compare the forces, impact time, impulse, acceleration, and change in momentum for each vehicle in the impending collision. Newton s 3 rd Law F T = F C

35 Compare the forces, impact time, impulse, acceleration, and change in momentum for each vehicle in the impending collision. F T = F C Newton s 2 nd Law ma = ma So a T < a C

36 Compare the forces, impact time, impulse, acceleration, and change in momentum for each vehicle in the impending collision. F T = F C & Δt T = Δt C So J T = J C Δp T = Δp C

37 Compare the forces, impact time, impulse, acceleration, and change in momentum for each vehicle in the impending collision. Δp T = Δp C mδv = mδv So Δv T < Δv C

38 Conservation of Momentum the total momentum of an isolated system remains constant. p o = p f Isolated System the next external forces acting on the system is zero.

39 Elastic collision - a collision in which the total kinetic energy is conserved p o = p f, Et o = Et f, (mechanical) E o = E f, KE o = KE F Inelastic collision - a collision in which the total kinetic energy is not conserved p o = p f, Et o = Et f, (mechanical) E o E f, KE o KE F

40 A ballistic pendulum consists of a block of wood (m 2 ) suspended by a wire of negligible mass. A bullet (m 1 ) is fired with a speed v 01. Just after the bullet collides with it, the block (with the bullet in it) has a speed v f and then swings to a maximum height (h) above the initial position (see part b of the drawing). Find the speed v 01 of the bullet, assuming that air resistance is negligible. Can we say that the KE of bullet in (a) is equal to the Ug of bullet + block in (b)? Why?

41 A ballistic pendulum consists of a block of wood (m 2 ) suspended by a wire of negligible mass. A bullet (m 1 ) is fired with a speed v 01. Just after the bullet collides with it, the block (with the bullet in it) has a speed v f and then swings to a maximum height (h) above the initial position (see part b of the drawing). Find the speed v 01 of the bullet, assuming that air resistance is negligible. What about using these two points?

42 K bot = ΔU gtop 1/2mv 2 = mgh v = 2gh p o = p m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v m 1 v 1 = (m 1 + m 2 ) 2gh KE Ug v 1 = (m 1 + m 2 ) 2gh m 1

43 Conservation of Momentum in Two Dimensions Momentum is a vector quantity, and is conserved in an isolated system. However, in two dimensions, the x and y components of the total momentum are conserved separately.

44 Rotational Motion Translational Motion all points on a body travel on parallel path (not necessarily straight lines) Rotational Motion all points on a body travel in a circular path with a common axis. can be combined with translational motion

45 Rotational Motion axis of rotation Axis of Rotation an imaginary line with which an object rotates around Angular Displacement (Δθ) the angle through which a rigid object rotates about a fixed axis Δθ = θ - θ o Units radians (rad) or degrees ( ) or revolution (rev) θ = S r

46 Rotational Motion For a full circle the arc length (S) or circumference, is equivalent to 2πr. Therefore: 1 revolution = 360 = 2πr and 1 radian = 360 2π = 180 π = 57.3

47 Rotational Motion Angular Velocity (ω) the rate of change of angular displacement ω = Δθ = 2πf Δt Units radians per second (rad/s) Angular Acceleration (α) the time rate change of angular velocity α = Δω t Units radians per second 2 (rad/s 2 )

48 Rotational Motion If an object changes its angular velocity at a constant rate, the average angular velocity is midway between the initial and final values: ω = ω i + ω f 2

49 Rotational Motion ω = ω o + αt θ = θ o + ω o t +1/2 αt 2 ω 2 = ω o 2 + 2α (θ θ o ) A bicycle wheel rotates with a constant angular acceleration of 3.50 rad/s 2. If the initial angular speed of the wheel is 2.00 rad/s: through what angle does the wheel rotate in 2.00 seconds? what is the angular speed of the wheel after 2.00 s?

50 Rotational Motion ω = ω o + αt ω = (2.00 rad /s) + (3.50 rad/s 2 )(2.00 s) ω = 9.00 rad/s A bicycle wheel rotates with a constant angular acceleration of 3.50 rad/s 2. If the initial angular speed of the wheel is 2.00 rad/s: through what angle does the wheel rotate in 2.00 seconds? what is the angular speed of the wheel after 2.00 s? θ = θ o + ω o t + 1/2αt 2 θ = (2.00 rad/s)(2.00 s) + 1/2(3.50 rad/s 2 )(2.00 s) 2 θ = 4 rad + 7 rad θ = 11 radians 11 radians rad = rev rev 11 rad 2πrad = 1.75 revolutions = 1.75 revolutions

51 Rotational Motion In the familiar ice-skating stunt known as crack the whip, a number of skaters attempt to maintain a straight line as they skate around a central person who remains in place. Compare the angular velocity and the tangential velocity of the skaters as shown.

52 Rotational Motion Uniform Circular Motion constant speed and constant radius Acceleration present only centripetal acceleration Non-Uniform Circular Motion non-constant speed Acceleration present 1. Centripetal Acceleration rate of change of direction of linear (tangential) velocity 2. Tangential Acceleration - rate of change of magnitude of linear (tangential) velocity 3. Angular Acceleration rate of change of angular velocity

53 Rotational Motion

54 Rotational Motion When the tires roll, there is a relationship between the angular speed at which the tires rotate and the linear speed (assumed constant) at which the car moves forward. Similar reasoning reveals that:

55 Moment of Inertia Moment of Inertia (I) rotational inertia - a quantity expressing a body s tendency to resist angular acceleration. depends on mass and distribution of mass Units kg m 2 I = m r 2

56 Moment of Inertia

57 Moment of Inertia A solid circular disk has a mass of 1.2 kg and radius of 0.16 m. Each of three identical thin rods has a mass of 0.15 kg. The rods are attached perpendicularly to the plane of the disk at its outer edge to for a three-legged stool (see the drawing). Find the moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center.

58 Moment of Inertia A solid circular disk has a mass of 1.2 kg and radius of 0.16 m. Each of three identical thin rods has a mass of 0.15 kg. The rods are attached perpendicularly to the plane of the disk at its outer edge to for a three-legged stool (see the drawing). Find the moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center. I stool = I disk + 3I rod I stool = 1 2 M disk R2 + 3M rod R 2 I stool = 1 2 ( 1.2 kg) ( 0.16 m) 2 + 3( 0.15 kg) ( 0.16 m) 2 I stool = kg m 2

59 Torque of a Rotating Body τ = r F = r F sinθ A bicycle wheel can be accelerated by either pulling on the chain that is on the gear or by pulling on a string wrapped around the tire. The wheel s radius is 0.38 meter, while the radius of the gear is 0.14 meter. If you obtained the needed acceleration with a force of 15-Newtons on the chain, what force would you need to exert on the strap?

60 Torque of a Rotating Body τ = r F = r F sinθ A bicycle wheel can be accelerated by either pulling on the chain that is on the gear or by pulling on a string wrapped around the tire. The wheel s radius is 0.38 meter, while the radius of the gear is 0.14 meter. If you obtained the needed acceleration with a force of 15-Newtons on the chain, what force would you need to exert on the strap? τ chain = τ wheel F chain r gear = F string r wheel F string = F r chain gear r wheel (15 N)(0.14 m) F string = 0.38 m F string = 5.5 N

61 Torque of a Rotating Body Newton s 2 nd Law can be rewritten to represent rotational motion as: The advantage of using τ = I α is that it can be applied to any rigid body rotating about a fixed axis, and not just to a particle.

62 Torque of a Rotating Body A merry-go-round made of a solid steel wheel has a mass of 150 kilograms and a diameter of 4.4 meters. A person pushes the handlebar of the stationary merry-goround for 15.0 seconds to make it rotate to a final speed of 8.0 rev/s. Determine: the torque applied to the merry-go-round. how much force was exerted on the handlebar.

63 Torque of a Rotating Body A merry-go-round made of a solid steel wheel has a mass of 150 kilograms and a diameter of 4.4 meters. A person pushes the handlebar of the stationary merry-goround for 15.0 seconds to make it rotate to a final speed of 8.0 rev/s. Determine: the torque applied to the merry-go-round. how much force was exerted on the handlebar. τ = I α τ = (1/2 mr 2 ) ( Δω t ) τ = (1/2 mr 2 ) ( 2πf ) t τ = 1/2 (150 kg)(2.2 m)2 2π (8.0 rev/s) 15.0 s τ = Nim = 1.2 x 10 3 Nim F = τ r = 1.2 x 103 Nim 2.2 m = 550 N

64 Rotational Work A rope is wrapped around a wheel and is under a constant tension F. If the rope is pulled out a distance s, the wheel rotates through an angle of: The rotational work can be determined since: θ = s r W = F d W = F s W = F r θ W R = τ θ

65 Rotational Kinetic Energy The rotational work done by a net external torque causes the rotational kinetic energy to change. The kinetic energy of the entire rotating body is the sum of the kinetic energies of the particles. K = 1 2 m v 2 T K = 1 2 m r2 ω 2 K = ( 1 2 I) ω 2 K R = 1 2 I ω 2

66 Total Mechanical Energy E T! = 1 Total Mechanical Energy 2 mv2 "#$ Translational kinetic energy Iω 2 + mgh!! Rotational kinetic energy Gravitational potential energy Which cylinder will reach the bottom of the incline first?

67 The solid cylinder wins!!! E o = E f 1 2 mv o 2 Iω 2 + mgh o o = 1 2 mv f 2 Iω 2 + mgh f f mgh o = 1 2 mv f 2 Iω f2 mgh o = 1 2 mv f 2 I v f r 2 vf = 2 mgh o m + I r 2 Hoop Solid Cylinder vf = 2 mgh o m + mr2 r 2 vf = 2 mgh o m + 1 / 2 mr2 r 2 vf = gh o vf = 4gh o 3 = 1.15 gh o

68 Angular Momentum

69 Conservation of Angular Momentum Conservation of Angular Momentum the total angular momentum of a system remains constant (is conserved) if the net external torque acting on the system is zero.

70 Conservation of Angular Momentum An artificial satellite is placed into an elliptical orbit about the earth, as shown in the figure. As the satellite travels from perigee to apogee determine how the following quantities will change: gravitational force: tangential speed: kinetic energy: linear momentum: angular momentum:

71 Conservation of Angular Momentum Can also use L o = L f. An artificial satellite is placed into an elliptical orbit about the earth, as shown in the figure. As the satellite travels from perigee to apogee determine how the following quantities will change: gravitational force: tangential speed: kinetic energy: linear momentum: angular momentum: v = F g = Gm E m s r 2 = ( )( )( ) F m E r = ( )(-) ( ) = F ( ) 2 g = v ( K = 1/2 mv 2 = (-)(-)( ) 2 = K) ( p = mv = (-)( ) = p) m s v 2 F c = F g ( L = mvr = (-)( )( ) = same L) r = Gm E m s r 2

72 Conservation of Angular Momentum A 34.0 kg child runs with a speed of 2.80 m/s tangential to the rim of a stationary merry-go-round. The merry-go-round has a moment of inertia of 510 kg m 2 and a radius of 2.31 m. When the child jumps onto the merry-go-round, the entire system begins to rotate. What is the angular speed of the system?

73 Conservation of Angular Momentum A 34.0 kg child runs with a speed of 2.80 m/s tangential to the rim of a stationary merry-go-round. The merry-go-round has a moment of inertia of 510 kg m 2 and a radius of 2.31 m. When the child jumps onto the merry-go-round, the entire system begins to rotate. What is the angular speed of the system? L o = L f m c vr = (I + m c r 2 )ω ω = m c vr I + m c r 2 ω = (34.0 kg)(2.80 m/s)(2.31 m) (510 kg m 2 + (34.0 kg)(2.31 m) 2 ω = 3.18 rad/s

74 Angular Impulse Angular Impulse Angular Momentum theorem The angular impulse on an object is equal to the object s final angular momentum minus the object s initial angular momentum. τ net = I α τ net = I ω f - ω i t τ net t = I (ω f - ω i ) τ net t = L f - L i ΔL = τ Δt

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