2D Kinematics Relative Motion Circular Motion Lana heridan De Anza College Oct 5, 2017
Last Time range of a projectile trajectory equation projectile example began relative motion
Overview relative motion uniform circular motion nonuniform circular motion
on P of the particle relative ative Relative to observer Motion B with the we see One that the veryvectors useful rtechnique P A for physical reasoning is considering n other frames of reference. (4.22) A reference frame is a coordinate system that an observer adopts. e, noting that v BA is con- Different observers may have different perspectives: different frames of reference. Consider a pair of observers, one stationary (A), one moving with constant velocity v BA. Both observe a (4.23) W W Galilean velocity particle P. by observer A and transformation up B is its icle velocity rather than v, o reference frames.) Equation equations. They relate A B P bservers in relative motion. When relative velocities are rpa rpb uter ones (P, A) match the t velocities for the particle, We can verify that by taking A vbat B vba Figure 4.20 A particle located x
Frames of Reference to the other. Q uick Quiz 39.1 How do we relate coordinates in different frames of reference? observer in the Two frames and y O vt x y x O x v P (event) Figure 39.2 An event occurs at Galilean transformations: a point P. The event is seen by two observers in inertial frames and x = x 9, + where vt, 9 y moves = y with, z a = velocity z, t = t v relative to. x uppose some observed by an o frame means tha The event s locati nates (x, y, z, t). W of an observer in with uniform rela Consider two in stant velocity v al We assume the or space at some inst and the observer frames in Figure fixed location in
Relative Motion (4.23) by observer A and u P B is its icle velocity rather than v, o reference frames.) Equation equations. They relate bservers in relative motion. When relative velocities are uter ones (P, A) match the t velocities for the particle, We can verify that by taking A W W Galilean velocity transformation A vbat rpa B B rpb vba P x Figure 4.20 A particle located at P is described by two observers, one in the fixed frame of reference r PA A and = the r PB other + vin BA the t conclude that a P A 5 frame a B, which moves to the right P B he acceleration Differentiating, of the partis the same as that measured vector relative to A, and r P B is its notingwith that a constant v velocity v BA. The vector r BA is a constant: PA is the particle s position tive to the first frame. position dr PA vector relative to B. = dr PB +v BA dt dt u PA = u PB + v BA
vrelative BA Motion Example dt A boat crossing a wide river with moves a constant with velocity a vspeed of 10.0 km/h BA. The relative to the water. The vector water r PA inis the particle s river position has a uniform speed of vector relative to A, and 5.00 km/h due east relative to the Earth. If r P B is its position vector relative to the B. boat heads due north, determine the velocity of the boat relative to an observer standing on either bank. 1 ore, we conclude that a P A 5 a P B hat is, the acceleration of the partirence is the same as that measured city relative to the first frame. r peed of ver has a e Earth. locity of bank. s a river will not end up W a N vbr E you relau vre vbe Figure 4.20 A particle located at P is described by two observers, one in the fixed frame of reference A and the other in the frame B, which moves to the right W b N v br = 10.0 km/h v re = 5.00 km/h E vbr v re 2 Page Figure 97, 4.21 erway (Example & Jewett 4.8) (a) A boat aims directly across a u vbe
Relative Motion and Rip Currents Life and death application: rip currents. In shallow ocean water, a rip current is a strong flow of water away from the shore.
Relative Motion and Rip Currents If you are caught in one, which way should you swim? 1 Diagram from Wikipedia.
Relative Motion and Rip Currents If you are caught in one, which way should you swim? 1 Diagram from Wikipedia.
Relative Motion and Rip Currents If you are caught in one, which way should you swim? 1 Diagram from Wikipedia.
Relative Motion Example, #49, pg 206 A bolt drops from the ceiling of a moving train car that is accelerating northward at a rate of 2.50 m/s 2. (a) What is the acceleration of the bolt relative to the train car? (b) What is the acceleration of the bolt relative to the Earth? (c) Describe the trajectory of the bolt as seen by an observer inside the train car. (d) Describe the trajectory of the bolt as seen by an observer fixed on the Earth. Let the x-axis point north, and the y-axis point up.
Another Example, #53, pg106 A science student is riding on a flatcar of a train traveling along a straight, horizontal track at a constant speed of 10.0 m/s. The student throws a ball into the air along a path that he judges to make an initial angle of 60.0 with the horizontal and to be in line with the track. The student s professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does she see the ball rise?
Another Example, #53, pg106 A science student is riding on a flatcar of a train traveling along a straight, horizontal track at a constant speed of 10.0 m/s. The student throws a ball into the air along a path that he judges to make an initial angle of 60.0 with the horizontal and to be in line with the track. The student s professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does she see the ball rise? v x = 10.0 m/s and v i = v x cos θ h = v 2 i sin 2 θ 2g or v 2 f,y = v 2 i,y 2gh
Another Example, #53, pg106 A science student is riding on a flatcar of a train traveling along a straight, horizontal track at a constant speed of 10.0 m/s. The student throws a ball into the air along a path that he judges to make an initial angle of 60.0 with the horizontal and to be in line with the track. The student s professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does she see the ball rise? v x = 10.0 m/s and v i = v x cos θ h = v 2 i sin 2 θ 2g or v 2 f,y = v 2 i,y 2gh h = 15.3 m
it. From the right triangle in Figure 3.2b, we find that sin u y/r and that cos x/r. (A review of trigonometric functions is given in Appendix B.4.) Therefore, Circular Motion: Radial Coordinates y (x, y) r sin u = y r cos u = x r r y O u x y tan u = x u x a e 3.2 (a) The plane polar coordinates of a point are represented by the distance r and the u, where u is measured counterclockwise from the positive x axis. (b) The right triangle used to (x, y) to (r, u). b
it. From the right triangle in Figure 3.2b, we find that sin u y/r and that cos x/r. (A review of trigonometric functions is given in Appendix B.4.) Therefore, Circular Motion: Radial Coordinates s conceptualize the problem. We wish to find r and u. We expect r to a y lem and re simply r coordisubstitunot have tution of alize step O r (x, y) Figure 3.3 (Example 3.1) sin u = y r cos u = x r Finding u polar coordinates when y u x tan u = ( 3.50, 2.50) x Cartesian coordinates are given. x e 3.2 (a) The plane polar coordinates of a point are represented by the distance r and the u, where u is measured counterclockwise from the positive x axis. (b) The right triangle used to (x, y) to (r, u). b u r r y (m) y x (m)
it. From the right triangle in Figure 3.2b, we find that sin u y/r and that cos x/r. (A review of trigonometric functions is given in Appendix B.4.) Therefore, Circular Motion: Radial Coordinates s conceptualize the problem. We wish to find r and u. We expect r to a y lem and re simply r coordisubstitunot have tution of alize step O r (x, y) Figure 3.3 (Example 3.1) sin u = y r cos u = x r Finding u polar coordinates when y u x tan u = ( 3.50, 2.50) x Cartesian coordinates are given. x e 3.2 Give (a) The position plane polar coordinates of ina terms point are ofrepresented r and θ. by Tothe transform distance r and from the r, θ to x, y: (x, y) to (r, u). x = r cos θ y = r sin θ u, where u is measured counterclockwise from the positive x axis. (b) The right triangle used to b u r r y (m) y x (m)
Radial Coordinates ˆ θ ˆ r ˆ θ ˆ r We can define perpendicular radial and tangential unit vectors, but their direction changes with the motion of a particle around a circle.
the center of the circle. Let us now find the magnitude of the acceleration of the particle. C diagram of the position and velocity vectors in Figure 4.15b. The figur the vector representing the change in position D r for an arbitrary ti As a car moves around a circular path its velocity changes, if not in The particle follows a circular path of radius r, part of which is shown by magnitude, then in direction. Circular Motion r O Top view v a b c r i v i r qu r f v f v i v f u If the radius remains constant and the speed of the car does as well, then v points toward the center of the circle. v Figure stant s particl, its v tion fo Dv, wh That means the acceleration vector does, too.
be modeled as a particle. If it moves Uniform Circular Motion onstant speed v, the magnitude of its The velocity vector points along a tangent to the circle (4.14) otion is given by is (4.15) (4.16) For uniform circular motion: the radius is constant the speed is constant the magnitude of the acceleration is constant ipetal Acceleration of the Earth AM r a c v Examples: of constan fectly circu form magn nucleus in hydrogen
Uniform Circular Motion The magnitude of the acceleration is given by a c = v 2 r If the constant speed is v, then the time period for one complete orbit is T = 2πr v
Uniform Circular Motion Pitfall Prevention 10.1 Remember the Radian In rotational equations, you must use angles expressed in radians. angle u from the reference line, ev the same angle u. Therefore, we object as well as with an individua We can also consider the Don t rate fall into atthe which trap of using the angular coordinate is angles measured in degrees in changing: rotational equations. O y r,t f ui u f,t i Figure 10.2 A particle on a rotating rigid object moves from to along the arc of a circle. In the time interval Dt 5 t 2 t i, the radial line of θ length = r moves θ f through θ i an angular displacement Du 5 u f 2 u i. Then we can define the angular speed, ω, as Average angular speed ω = dθ dt x position of a rigid object in its rot the object, such as a line connect angular position of the rigid obje the object and the fixed reference uch identification is similar to the lational motion as the distance x which is the origin, x 5 0. Therefo motion that the position x does in As the particle in question on o tion in a time interval Dt as in F sweeps out an angle Du 5 u f 2 u i. placement of the rigid object: The rate at which this angular di spins rapidly, this displacement c slowly, this displacement occurs in rates can be quantified by definin omega) as the ratio of the angular val Dt during which the displaceme v avg
Uniform Circular Motion ω gives the amount by which the angle θ advances in radians, per unit time. Therefore, ω = 2π T where T is the period (time for one revolution).
Uniform Circular Motion ω gives the amount by which the angle θ advances in radians, per unit time. Therefore, ω = 2π T where T is the period (time for one revolution). Putting in the expression for T (T = 2πr v ): ω = 2π v 2πr ω = v r
Uniform Circular Motion ω gives the amount by which the angle θ advances in radians, per unit time. Therefore, ω = 2π T where T is the period (time for one revolution). Putting in the expression for T (T = 2πr v ): ω = 2π v 2πr ω = v r This gives us another expression for the centripetal acceleration: a c = ω 2 r
Radial and Tangential Accelerations Quick Quiz 4.4 2 A particle moves in a circular path of radius r with speed v. It then increases its speed to 2v while traveling along the same circular path. (i) The centripetal acceleration of the particle has changed by what factor? Choose one: A 0.25 B 0.5 C 2 D 4 1 Page 93, erway & Jewett
Radial and Tangential Accelerations Quick Quiz 4.4 2 A particle moves in a circular path of radius r with speed v. It then increases its speed to 2v while traveling along the same circular path. (ii) From the same choices, by what factor has the period of the particle changed? A 0.25 B 0.5 C 2 D 4 1 Page 93, erway & Jewett
be modeled as a particle. If it moves Uniform Circular Motion onstant speed v, the magnitude of its The velocity vector points along a tangent to the circle (4.14) otion is given by is (4.15) (4.16) For uniform circular motion: the radius is constant the speed is constant ipetal Acceleration of the Earth AM r a c v Examples: of constan fectly circu form magn nucleus in hydrogen the magnitude of the acceleration is constant, a c = v 2 = ω 2 r r
ngential acceleration component causes a change in the speed v of the part omponent is parallel to the instantaneous velocity, and its magnitude is give Non-Uniform Circular Motion a t 5 ` dv dt ` (4 Path of particle a t a r a a a r a r a t a t a
Radial and Tangential Accelerations a t 5 ` dt ` Path of particle a t ved e nt on ts - ial a a r a a t a r a = a t + a r a = a t ˆθ acˆr Let ˆθ(t) be a unit vector in the direction of the velocity. Note that its direction changes with time! v(t) = v(t) ˆθ(t)
Radial and Tangential Accelerations a = dv dt ; v(t) = v(t) ˆθ(t) Find the acceleration using the product rule: a = dv dt ˆθ + v dˆθ dt
Radial and Tangential Accelerations a = dv dt ; v(t) = v(t) ˆθ(t) Find the acceleration using the product rule: a = dv dt ˆθ + v dˆθ dt ( The term dv ˆθ ) dt is all in the tangential component of the acceleration. ( ) But how to find what is? We need to find how ˆθ changes v dˆθ dt with time. (It rotates, but at what rate?)
Radial and Tangential Accelerations: How do the perpendicular axes change? Let s find out! ˆθ is changing, so let us say that ˆθ i is the initial tangential unit vector and ˆθ f is the final tangential unit vector.
Radial and Tangential Accelerations: How do the perpendicular axes change? ˆθ is changing, so let us say that ˆθ i is the initial tangential unit vector and ˆθ f is the final tangential unit vector. Both vectors are 1 unit long (the same length). Both remain perpendicular to the radial direction. Therefore we have two similar triangles! r i vθ ˆ ii r qu r f vθ ˆ f f vθ ˆ i u ˆ vθ ff v Δθˆ
Radial and Tangential Accelerations: How do the perpendicular axes change? ˆθ is changing, so let us say that ˆθ i is the initial tangential unit vector and ˆθ f is the final tangential unit vector. Both vectors are 1 unit long (the same length). Both remain perpendicular to the radial direction. Therefore we have two similar triangles! r i vθ ˆ ii r qu r f vθ ˆ f f vθ ˆ i u ˆ vθ ff v Δθˆ (ˆθ) = r ˆθ i r d dt ˆθ = 1 r dr dt = v r This tells us how fast the tangential unit vector changes in direction.
Radial and Tangential Accelerations: How do the perpendicular axes change? d dt ˆθ = v r This tells us how fast the tangential unit vector changes in direction. Now consider that the direction of change must be radial! d dt ˆθ = v r ˆr
Radial and Tangential Accelerations a = dv dt = d (v ˆθ) dt Find the acceleration using the product rule: We said a = a t ˆθ acˆr so, a = dv dt ˆθ + v dˆθ dt = dv dt ˆθ + v ( v ) r ˆr = dv dt ˆθ v 2 r ˆr tangen. radial a c = v 2 r
Radial and Tangential Accelerations pg 105, #41 Problems 105 0 cm inate cenbe at e the te of mall dge). arch ntal, Figin a away. ond, 41. A train slows down as it rounds a sharp horizontal M turn, going from 90.0 km/h to 50.0 km/h in the 15.0 s it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50.0 km/h. Assume the train continues to slow down at this time at the same rate. 42. A ball swings counterclockwise in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9 past the lowest point on its way up, its total acceleration is 1222.5 i^ 1 20.2 j^2 m/s 2. For that instant, (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball. 1 43. Page(a) 93, Can erway a & particle Jewett moving with instantaneous speed
ummary relative motion circular motion Quiz start of class tomorrow. Collected Homework! due tomorrow, Oct 6. (Uncollected) Homework erway & Jewett, Ch 4, onward from page 104. Problems: 40, 43, 45, 51 (relative and circular motion probs, set last time)