Rotational Motion Rotational Kinematics

Transcription

1 Rotational Motion Rotational Kinematics Lana Sheridan De Anza College Nov 16, 2017

2 Last time 3D center of mass example systems of many particles deforming systems

3 Overview rotation relating rotational and translational quantities rotational kinematics torque

4 Rotation of Rigid Objects Now we understand that while we can treat a collection of particles as a single point particle at the center of mass, we do not have to do that. This will allow us to describe another important kind of motion: rotation.

5 Rotation of Rigid Objects Now we understand that while we can treat a collection of particles as a single point particle at the center of mass, we do not have to do that. This will allow us to describe another important kind of motion: rotation. Begin with rotational kinematics.

6 Pitfall Prevention 10.1 Remember the Radian In rota- for the disc, a fixed reference here As the for disc rotational rotates, a particle motion. at line is chosen. A particle at P P Figure moves through 10.1 illustrates an arc length an ove is located at a distance r from To begin, consider a rotating disc. disc s on rotates a circular about path of a radius fixed r. axis p the rotation axis through O. through The angular the position center of of P is the u. disc ticle at P is at a fixed distance r radius r. (In Pfact, every element o convenient r to represent s the pos r u the distance from the origin to P O P Reference O ence line fixed in space Reference as show line line changes in time while r remai cle from the reference line, wh a b length s as in Figure 10.1b. The s = rθ ; θ As the disc rotates, a particle at relationship = s Figure r 10.1 A compact disc P r moves is constant through in an time arc for length a rigid object. rotating about a fixed axis Rotation of Rigid Objects s on a circular path of radius r. The angular position of P is u. through O perpendicular to the plane of the figure. P

7 s on a circular path of radius r. through O perpendicular to the Units The angular for θ: position radians. of Often P is u. written as plane rad. of the But figure. notice, that a dimensional analysis gives [m] [m] = 1, unitless! The radian is an artificial unit. In fact, angles given inpitfall radians Prevention are dimensionless P Remember the Radian In rota- for the disc, a fixed reference here As the for disc rotational rotates, a particle motion. at line is chosen. A particle at P P Figure moves through 10.1 illustrates an arc length an ove is located at a distance r from To begin, consider a rotating disc. disc s on rotates a circular about path of a radius fixed r. axis p the rotation axis through O. through The angular the position center of of P is the u. disc ticle at P is at a fixed distance r radius r. (In Pfact, every element o convenient r to represent s the pos r u the distance from the origin to P O P Reference O ence line fixed in space Reference as show line line changes in time while r remai cle from the reference line, wh a b length s as in Figure 10.1b. The s = rθ ; θ As the disc rotates, a particle at relationship = s Figure r 10.1 A compact disc P r moves is constant through in an time arc for length a rigid object. rotating about a fixed axis Rotation of Rigid Objects

8 Don t fall into the trap of using angles measured in degrees in rotational equations. How does the angle advance in time? Rotation of Rigid Objects O y r,t f u i u f,t i Figure 10.2 A particle on a rotating rigid object moves from to along the θ arc = of θ f a circle. θ i In the time interval Dt 5 t f 2 t i, the radial x position of a rigi the object, such angular position the object and th Such identificati lational motion which is the orig motion that the As the particl tion in a time sweeps out an an placement of the The rate at whic spins rapidly, th slowly, this displ rates can be qua

9 Angular Speed Rate at which the angle advances is a speed: the angular speed, ω. Average angular speed: ω avg = θ f θ i t f t i = θ t Instantaneous angular speed: ω = dθ dt

10 Angular Acceleration Rate at which the angular speed changes: the angular acceleration, α. Average angular acceleration: α avg = ω f ω i t f t i = ω t Instantaneous angular acceleration: α = dω dt

11 Rotation of Rigid Objects and Vector Quantities We can also define these quantities as vectors! This might seem a bit strange because the direction of the motion any point (other than the axis) on a rotating disc is always changing.

12 Rotation of Rigid Objects and Vector Quantities We can also define these quantities as vectors! This might seem a bit strange because the direction of the motion any point (other than the axis) on a rotating disc is always changing. However, the angle can be positive or negative depending on whether it is clockwise or counterclockwise from the reference point. This is the same way that a 1-dimensional displacement x can be positive or negative based on whether it is on the left or right of the origin.

13 Rotation of Rigid Objects and Vector Quantities We can also define these quantities as vectors! This might seem a bit strange because the direction of the motion any point (other than the axis) on a rotating disc is always changing. However, the angle can be positive or negative depending on whether it is clockwise or counterclockwise from the reference point. This is the same way that a 1-dimensional displacement x can be positive or negative based on whether it is on the left or right of the origin. In that case we could write x, and the direction of the vector was indicated by the sign of the quantity x.

14 Rotation of Rigid Objects and Vector Quantities By convention, we define the counterclockwise direction to be 296 positive. The vector Chapter itself10 is drawn Rotation along of a the Rigid axis Object of rotation. About a Fixed Axis Then we can write: θ = s r ˆn ; v S ω = dθ direction of v S for the particle is out o is counterclockwise and into the pla wise. To illustrate this convention, it strated in Figure When the fou direction of rotation, the extended r direction v S of S a follows from its definit v S if the angular speed is increasing in speed Figure is decreasing 10.3 The in right-hand time. rule for determining the direction of the angular velocity vector ; α Analysis = dω Model: R dt dt Constant Angular where v S ˆn is a unit vector perpendicular to the plane of rotation. In our study of translational motion,

15 Comparison of Linear and Rotational quantities Linear Quantities x v = dx dt a = dv dt Rotational Quantities θ ω = dθ dt α = dω dt

16 Rotational Kinematics If α is constant, we have basically the same kinematics equations as before, but the relations are between the new quantities. ω f = ω i + αt θ f = θ i + ω i t αt2 ω 2 f = ω 2 i + 2α θ ω avg = 1 2 (ω i + ω f ) θ f = θ i (ω i + ω f )t

17 Kinematics Comparison Linear Quantities v f = v i + at x f = x i + v i t at2 Rotational Quantities ω f = ω i + αt θ f = θ i + ω i t αt2 v 2 f = v 2 i + 2a x ω 2 f = ω 2 i + 2α θ v avg = 1 2 (v i + v f ) ω avg = 1 2 (ω i + ω f ) x f = x i (v i + v f )t θ f = θ i (ω i + ω f )t

18 and for the velocity, Relating Rotational Quantities to Translation of Points There is no translational analog to part (B) because transl Consider a point on the rotating object. How does its speed relate to the angular speed? O y S v r u 10.3 Angular a In this section, we derive acceleration of a rotating of a point in the object. rotates about a fixed axi circle whose center is on Because point P in Figu is always tangent to the ci nitude of the tangential v v 5 ds/dt, where s is the d path. Recalling that s 5 r Figure 10.4 As a rigid object We know s = rθ, so rotates sinceabout the object s the fixed axis speed (the is its speed along the path s, z axis) through O, the point P has a tangential velocity S v that is always tangent v = ds to the circular path of radius r. dt = r dθ dt P s x v f 5 v i 1 at m/s 1 (3

19 Relating Rotational Quantities to Translation of Points Since ω = dθ dt, that gives us and expression for (tangential) speed v = rω And differentiating both sides with respect to t again: a t = rα

20 Relating Rotational Quantities to Translation of Points Since ω = dθ dt, that gives us and expression for (tangential) speed v = rω And differentiating both sides with respect to t again: a t = rα Notice that the above equation gives the rate of change of speed, which is the tangential acceleration.

21 Centripetal Acceleration Remember: a t = dv dt where v is the speed, not velocity. So, a t = rα But of course, in order for a mass at that point, radius r, to continue moving in a circle, there must be a centripetal component of acceleration also. a c = v 2 = ω 2 r r

22 Centripetal Acceleration Remember: a t = dv dt where v is the speed, not velocity. So, a t = rα But of course, in order for a mass at that point, radius r, to continue moving in a circle, there must be a centripetal component of acceleration also. a c = v 2 = ω 2 r r For a rigid object, the force that supplies this acceleration will be some internal forces between the mass at the rotating point and the other masses in the object. Those are the forces that hold the object together.

23 Linear and Rotational Kinematics To get an idea of another way to confirm the relation between the linear and rotational kinematics equations: v f = v i + at v f r = v i r + a r t ω f = ω i + αt

24 Linear and Rotational Kinematics To get an idea of another way to confirm the relation between the linear and rotational kinematics equations: v f = v i + at v f r = v i r + a r t ω f = ω i + αt The scalar versions of the equations can all be recovered this way.

25 out e of 3.00 s. Example 1 5. A wheel starts from rest and rotates with constant W angular acceleration to reach an angular speed of 12.0 rad/s in 3.00 s. Find (a) the magnitude of the angular acceleration of the wheel and (b) the angle in radians through which it rotates in this time interval. guage nd. ing ed? ion t 1 ter- 4. A bar on a hinge starts from rest and rotates with an angular acceleration a t, where a is in rad/s 2 and t is in seconds. Determine the angle in radians through which the bar turns in the first 4.00 s. Page 325, #5 Section 10.2 Analysis Model: Rigid Object Under Constant Angular Acceleration 6. A centrifuge in a medical laboratory rotates at an angular speed of rev/min. When switched off, it rotates through 50.0 revolutions before coming to rest. Find the constant angular acceleration of the centrifuge. 7. An electric motor rotating a workshop grinding wheel M at rev/min is switched off. Assume the wheel has a constant negative angular acceleration of magni- w ef a fr e o Secti 15. A

26 Example 1 Known: ω i, ω f, t (a) Angular accel., α?

27 Example 1 Known: ω i, ω f, t (a) Angular accel., α? ω f = ω i + αt

28 Example 1 Known: ω i, ω f, t (a) Angular accel., α? ω f = ω i + αt α = ω f ω i t α = 12.0 rad/s s α = 4.00 rad s 2

29 Example 1 Known: ω i, ω f, t, α (b) Angle in radians, θ?

30 Example 1 Known: ω i, ω f, t, α (b) Angle in radians, θ? Either use θ = ω i t αt2 = (4)(3)2 = 18 radians or use θ = 1 2 (ω i + ω f )t θ = 1 ( )(3.00) 2 = 18.0 radians

31 Example 2 the constant angular acceleration of the centrifuge. 7. An electric motor rotating a workshop grinding wheel M at rev/min is switched off. Assume the wheel has a constant negative angular acceleration of magnitude 2.00 rad/s 2. (a) How long does it take the grinding wheel to stop? (b) Through how many radians has the wheel turned during the time interval found in part (a)? Page 325, #8 8. A machine part rotates at an angular speed of Q/C rad/s; its speed is then increased to 2.2 rad/s at an angular acceleration of 0.70 rad/s 2. (a) Find the angle through which the part rotates before reaching this final speed. (b) If both the initial and final angular speeds are doubled and the angular acceleration remains the same, by what factor is the angular displacement changed? Why? 9. A dentist s drill starts from rest. After 3.20 s of constant angular acceleration, it turns at a rate of rev/min. (a) Find the drill s angular acceleration. (b) Determine the angle (in radians) through which the drill rotates during this period. Secti 15. A A 4 tu 16. M o tu e 17. A W d th o la o to fo 18. F h

32 Example 2 Known: ω i, ω f, α (a) θ?

33 Example 2 Known: ω i, ω f, α (a) θ?

34 Example 2 Known: ω i, ω f, α (a) θ? ω 2 f = ω 2 i + 2α θ

35 Example 2 Known: ω i, ω f, α (a) θ? ω 2 f = ω 2 i + 2α θ θ = ω2 f ω2 i 2α = (2.2)2 (0.060) = 3.5 rad

36 Example 2 (b) If both ω i and ω f are doubled, α kept constant, what happens to θ? θ = (2 ω f ) 2 (2 ω i ) 2 2α = 4 ω2 f ω2 i 2α = 4 θ

37 Torque Torque is a measure of force-causing-rotation. It is not a force, but it is related. It depends on a force vector and its point of application relative to an axis of rotation. Torque is given by: τ = r F That is: the cross product between a vector r, the displacement of the point of application of the force from the axis of rotation, and an the force vector F

38 Torque Torque is a measure of force-causing-rotation. It is not a force, but it is related. It depends on a force vector and its point of application relative to an axis of rotation. Torque is given by: τ = r F That is: the cross product between a vector r, the displacement of the point of application of the force from the axis of rotation, and an the force vector F Units: N m Newton-meters. These are not Joules!

39 Torque O The component F sin f tends to rotate the wrench about an axis through O. d r S r f F sin f Figure 10.7 The force F S has a f S F F cos f Line of action greater rotating tendency about an axis through τ = r O as F = F increases rf sin φand ˆn as the moment arm d increases. where φ is the angle between r and F, and ˆn is the unit vector perpendicular to r and F, as determined by the right-hand rule Torqu In our study of tra we studied the cau What is the cause Imagine trying ular to the door s hinges. You will ac force near the doo When a force is to rotate about th axis is measured b but we will consid Chapter 11. Consider the wr perpendicular to t

40 Vectors Properties and Operations Multiplication by a vector: The Cross Product Let A = A x i + A y j B = B x i + B y j, A B = (A x B y A y B x )k The output of this operation is a vector. Equivalently, Pitfall Prevention 11.1 The Vector Product Is a Vector Remember that the result of taktion vector r and the applied force vector S F. In the situation shown, S r and S F lie in the xy plane, so the torque is along the z axis. u Properties of the vector product S The direction of C is perpendicular S S to the plane formed by A and B, and its direction is determined by the right-hand rule. A S S B C S A S S B S C S B A S Figure 11.2 The vector product S S S A 3 B is a third vector C having a magnitude AB sin u equal to the area of the parallelogram shown. follows: 1. Unl the imp The mus han 2. If A S that 3. If A S 4. The 5. The whe the It is left from the d and k^ obey A B = AB sin θ ˆn AB Cross products of unit vectors where ˆn AB is a unit vector perpendicular to A and B. Signs are i and i^ 3 12 The cro ing determ

41 Vectors Properties and Operations (See page 336 in Serway and Jewett.) The Cross Product - with k components In general: A = A x i + A y j + A z k B = B x i + B y j + B z k, A B = (A y B z A z B y )i + (A z B x A x B z )j + (A x B y A y B x )k How do we usually implement this formula? Via the determinant of a matrix: i j k A B = A x A y A z B x B y B z

42 Vector Operations: Cross Product Practice Try it yourself! Find A B when: A = 1i + 2j + 3k ; B = 1i 4j + 5k

43 Vector Operations: Cross Product Practice Try it yourself! Find A B when: A = 1i + 2j + 3k ; B = 1i 4j + 5k Now find B A...

44 Vector Operations: Cross Product Practice Try it yourself! Find A B when: A = 1i + 2j + 3k ; B = 1i 4j + 5k Now find B A... First A B: A B = 22i 8j 2k

45 Vector Operations: Cross Product Practice Try it yourself! Find A B when: A = 1i + 2j + 3k ; B = 1i 4j + 5k Now find B A... First A B: A B = 22i 8j 2k B A = 22i + 8j + 2k

46 Vectors Properties and Operations (See page 336 in Serway and Jewett.) The Cross Product - with k components A B = AB sin θ ˆn AB

47 Vectors Properties and Operations (See page 336 in Serway and Jewett.) The Cross Product - with k components Properties A B = AB sin θ ˆn AB The cross product is not commutative: A B B A. In fact, it is anticommutative because A B = (B A). If A B, A B = 0. If A B, A B = AB ˆn AB.

48 Summary rotation rotational kinematics torque 3rd Collected Homework due Monday, Nov 20. Next test Monday, Nov 27. (Uncollected) Homework Serway & Jewett, Read ahead in Chapter 10. Ch 10, onward from page 288. Probs: 3 (set yesterday) Ch 10, onward from page 288. Probs: 7, 11, 15, 17, 19, 21, 25