2 Rotational Coordinates Ridged objects require six numbers to describe their position and orientation: 3 coordinates 3 axes of rotation
3 Rotational Coordinates Use an angle θ to describe rotations about a single fixed axis. Always measure θ in radians. θ f Δθ θ 0
4 Rotational Coordinates The angle θ takes the place of a translational coordinate, x, y, or z. We define the change in Δθ as: Δθ = θ f θ 0
5 Rotational Velocity Just like we defined velocity for translational motion, we can define velocity for rotations as well. ω avg = Δθ Δt Instantaneous angular velocity is defined as: ω = dθ dt
6 Rotational Acceleration We also define rotational acceleration as: α avg = Δω Δt And instantaneous angular acceleration as: α = dω dt = d2 θ dt 2
7 Rotational-Translational Analog v avg = Δx Δt a avg = Δv Δt ω avg = Δθ Δt α avg = Δω Δt The definitions of angular velocity and acceleration are analogous to the definitions for translational velocity and acceleration.
8 Rotational Kinematics Just like with translational motion, we can relate Δθ, ω, and α. Δθ = ω avg Δt Δω = α avg Δt For constant angular acceleration ω avg = 1 ω ω f Δθ = 1 α Δt 2 + ω 2 0 Δt ω 2 f = ω α Δθ
9 Rotational-Translational Analog v avg = Δx Δt a = Δv Δt v avg = 1 2 v 0 + v f ω avg = Δθ Δt α = Δω Δt ω avg = 1 2 ω 0 + ω f Δx = 1 2 a Δt 2 + v 0 Δt Δθ = 1 2 α Δt 2 v f 2 = v a Δx ω f 2 = ω α Δθ The rotational kinematic equations are analogous to the translational kinematic equations.
10 Rotational-Translational Analog Δx = t 0 t fv t dt Δθ = t 0 t fω t dt Δv = t 0 t fa t dt Δω = t 0 t fα t dt The same analogs extend to non-constant acceleration problems as well.
11 Rotational-Translational Analogy The rotational and translational equations are analogous. Replace: Δx Δθ v ω a α
12 Right Hand Rule #1 Like their translational counterparts Δθ, ω, and α are vectors. How do we describe the direction of ω? Use the Right Hand Rule
13 Right Hand Rule #1 Right Hand Rule #1: Curl the fingers on your right hand in the direction of the rotation, your thumb will point in the direction of Δ θ and ω. ω
14 Direction of Angular Acceleration The direction of α is given by: In general: α = ω f ω 0 Δt If ω is increasing, α points in the same direction as ω. If ω is decreasing, α points in the opposite direction from ω.
15 Rolling Without Slipping Δx
16 Rolling Without Slipping Whenever a wheel or pulley rolls without slipping: Δx = r Δθ v = r ω a = r α
17 Translations From Rotation s =? What about the velocity and acceleration of a point on a rotating wheel?
18 Translations From Rotations The displacement and speed of a point on a spinning object is found by: s = r Δθ v = r ω Note that this does not work for acceleration: a r α
19 Rotational Accelerations Recall that it is the component of the acceleration that points parallel to v that is responsible for changing an object s speed. Therefore, a t = r α Where a t is the component of a that points tangent to the circular path.
20 Rotational Accelerations When an object travels around a circular path the acceleration is broken into two parts: Tangent Causes the change in the object s speed a a c a t Centripetal Causes the change in the object s direction
21 Acceleration Uniform Circular Motion Uniform Circular Motion ω is constant Although speed is constant, acceleration is non-zero The acceleration responsible for uniform circular motion is called centripetal acceleration We can calculate a c by relating Δ v to Δt
22 v f r f v 0 r 0
23 v f Δ v v 0 r f Δθ r 0 v 0
24 Centripetal Acceleration When an object travels with uniform circular motion, the acceleration always points towards the center of the circular path with: a c = v2 r = r ω2
25 Centripetal Acceleration - Example A centrifuge starts from rest and experiences a constant rotational acceleration of 4 rad.. The average radius s 2 of the arm of the centrifuge is r = 5 cm. What is the acceleration of the centrifuge after completing one quarter of a rotation?
26 a c = 6250 g = m s 2 v =? r = 0.05 m ω =?
27 Centripetal Acceleration - Example A circular track has a radius of 500 m. When the race starts, the diver hits the gas peddle, causing the car s speed to increase at a rate of 6 m s 2. What is the magnitude of the car s acceleration after 10 s have passed?
28 a c = 6250 g = m s 2 v =? r = 0.05 m ω =?
29 Centripetal Acceleration - Example A car drives around a level turn. The tires have a coefficient of friction of μ s = 0.8, and the turn has a radius of 900 m. How fast can the car go around the turn without sliding?
30 N r v =? F c =? F c W
31 Centripetal Acceleration - Example A car drives around an banked turn with a radius of 900 m. The turn is designed so that a car traveling 10 m/s will be able go around the turn even when the coefficient of friction is reduced to μ s = 0. What angle is the turn banked at?
32 N y N θ N x r v =? F c θ W
33 Newton s Law of Universal Gravity Every particle exerts an attractive force all other particles The force is given by: F G = Gm 1m 2 r 2 G is the universal gravitational constant: G = N m2 kg 2
34 Newton s Law of Universal Gravity F G = Gm 1m 2 r 2 Note that r, is the distance between the centers of the two masses Therefore, when standing on the surface of the Earth: r = R E + h R E
35 Orbit Any force can supply the centripetal force required to keep an object in uniform circular motion. When a satellite obits the Earth, the centripetal force is supplied by the gravitational force. F g
36 Orbit An important problem in orbital motion is determining how fast a satellite must move to stay in orbit. Suppose a satellite is a distance r from the center of the Earth, how fast must it move?
37 Kepler s Third Law of Planetary Motion Kepler s Third Law relates the period T to the radius of the orbit R
38 Kepler s Third Law of Planetary Motion Kepler s Third Law of Planetary Motion Says: T 2 R 3 = const.
39 Cross-Product To discuss the effect that a force has on a rotating object, we need to introduce a new type of vector operation: The Cross-Product
40 Cross-Product The Cross-Product, also known as a vector product, is a product between two vectors: A B = C
41 Cross-Product The Cross-Product A B = C is defined as: C = A y B z A z B y + A z B x A x B z i j + A x B y A y B x k
42 Cross-Product An easy way to remember the formula for A B is: i j k A B = det A x A y A z B x B y B z
43 Cross-Product In general, the magnitude of the crossproduct is given by: A B = A B sin θ How do we find the direction of A B?
44 Right Hand Rule #2 A B A B
45 Torque Toque is denoted by the Greek letter tau, τ Torque is the rotational analog of a force The torque describes the tendency for a force to cause rotational acceleration.
46 Torque F Applying a force further from the pivot increases the torque. The distance between the force and the axis of rotation is called the lever arm.
47 Torque F r r F F t Only the component of the force that is perpendicular to the lever arm generates a torque.
48 Torque In general: τ = F t r = F r sin θ
49 Torque Clearly τ is a cross-product between F and r. But is it: F r or r F We want τ to point in the direction of α.
50 Torque In general: τ = r F
51 Torque - Example A rope is wrapped around a circular disk with a radius of 0.1 m. A 50 N force is applied to the rope as shown in the figure. What is the torque generated by the tension in the rope? 100 N R
52 Torque - Example Now the force is applied at an angle. What is the torque generated by the tension in the rope? R N
53 Rotational Dynamics What does torque tell us? Consider a force applied to a single point mass: F = m a To find the effect of torque, cross both sides with r: r F = m r a
54 In general: Rotational Dynamics τ = m r 2 α I Where I describes the rotational inertia of the point mass. I is called the moment of inertia.
55 Rotational-Translational Analog F = m a τ = I α Translational to Rotational: F τ a α m I
56 Moment of Inertia Formulas for the moment of inertia of various shapes can be found in tables like this.
57 Torque - Example A box of mass m hangs from a massless string which winds around a circular disk of mass M and radius R. What is the acceleration of the mass?
58 R M m
59 Torque - Example Tension can not be the same M on opposite sides of a massive pulley. Since, τ = T 1 T 2 R = I α T 1 T 2
60 Torque - Example R Calculate the acceleration of M an Atwood Machine with for masses m 1 > m 2 and a pulley of mass M and radius R. m 1 m 2
61 R M m 1 m 2
62 Moment of Inertia Moment of Inertia is rotational inertia For a collection of rigidly attached masses: I = n m n r n 2
63 Moment of Inertia For a solid continuous object, the moment of inertia is calculated by: I = r 2 dm Where dm is the mass of an infinitesimal piece of the object.
64 Moment of Inertia I = r 2 dm For a: Rod dm = λ dx; λ = mass length Disk dm = σ da; σ = mass area Volume dm = ρ dv; ρ = mass volume
65 Moment of Inertia Calculate the moment of inertia of a metal rod of length L and mass M rotated about one end.
66 x = 0 M L x = L
67 Moment of Inertia Calculate the moment of inertia of a metal rod of length L and mass M rotated about its center.
68 M x = L 2 L x = + L 2
69 Parallel Axis Theorem The moment of inertia of an object depends on the location of axis of rotation. The moment of inertia about any axis can be related to the moment of inertia about the center of mass using the parallel axis theorem.
70 Parallel Axis Theorem The Parallel Axis Theorem states: The moment of inertia of an object of mass M rotated about an axis a distance R from the center of mass, is given by: I = I cm + M R 2
71 Parallel Axis Theorem M L 2 L I I= cm I= 1 cm + M L L
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