PHYS-2010: General Physics I Course Lecture Notes Section IV

Transcription

1 PHYS-010: General Phsics I Course Lecture Notes Section IV Dr. Donald G. Luttermoser East Tennessee State Universit Edition.3

2 Abstract These class notes are designed for use of the instructor and students of the course PHYS-010: General Phsics I taught b Dr. Donald Luttermoser at East Tennessee State Universit. These notes make reference to the College Phsics, 7th Edition (005) tetbook b Serwa and Faughn.

3 IV. Motion in Two Dimensions A. Vector Arithmetic. 1. We now will be working with both the and aes. Vectors are represented on an - graph as an arrow with a distinct direction indicated b the arrowhead: A Vector A θ a) Note that vector A makes an angle θ with the -ais. b) The -ais is alwas the reference line which vector direction angles are measured.. Vector addition. a) Two or more vectors can be added graphicall b placing the beginning of the nd vector at the tail of the 1st vector. IV 1

4 IV PHYS-010: General Phsics I Eample IV 1. diagrams below. Graphicall add vectors A and B in the B A 5 R = A + B θ R A R is 7 units long in the direction and is units long in the direction. 5 B 7

5 Donald G. Luttermoser, ETSU IV 3 b) Two or more vectors can be added algebraicall b rewriting the vectors into components, &, then adding the respective components. Eample IV. the Eample IV-1. Algebraicall add vectors A and B from R = A + B where R = R + R and R = A + B R = A + B A = A + A = 5ˆ +0ŷ B = B + B = ˆ +ŷ R = A + B = (A + B )ˆ +(A + B )ŷ = (5 + ) ˆ +(0+)ŷ = 7ˆ +ŷ R is 7 units long in the direction and is units long in the direction. c) The actual length of a vector is determined from the Pthagorean theorem: R = R + R (IV-1)

6 IV 4 PHYS-010: General Phsics I = note that the length of a vector is sometimes written as R = R, (IV-) and that the length is alwas taken to be the positive root of the square root in Equation (IV-1). d) The angle that a vector makes with the -ais is determined b R = R R = R θ R = R R R = R cos θ = R sin θ (IV-3) or tan θ = sin θ cos θ = or tan θ = R R. ( ) θ = tan 1 R R. (IV-4)

7 Donald G. Luttermoser, ETSU IV 5 Eample IV 3. What are the lengths and angles of the 3 vectors, A, B, and R, in Eample IV-? A =5ˆ : A =5, A =0 B =ˆ +ŷ : B =, B = R =7ˆ +ŷ : R =7, R = A = 5 +0 = 5 = 5 B = + = 4+4= 8= 4 = R = 7 + = = θ A = tan 1 ( 0 7 θ B = tan 1 ( ) ) ) = tan 1 (0) = 0 = tan 1 (1) = 45 ( θ R = tan 1 tan 1 (0.857) , 7 though here I would also have accepted 0 as the answer for θ R. Note that from this point forward, we will use the standard equals = sign even when the answer is not eact (as indicated above with the sign). 3. Vector subtraction. a) Graphicall: Flip the nd vector so that it is pointing in the opposite direction, then follow the rules for vector addition: R = A B R = A +( B)

8 IV 6 PHYS-010: General Phsics I Eample IV 4. Subtract B from A in Eample IV-1. B A 5 - B A 5 -

9 Donald G. Luttermoser, ETSU IV 7 R = A - B A θ R 3 B 5 R is 3 units in the direction and is units in the direction. b) Subtracting one vector from another algebraicall is done b subtracting the respective components: R = A B =(A B )ˆ +(A B )ŷ. Algebraicall subtract B from A in E- Eample IV 5. ample VI-4. R = (5 ) ˆ +(0 ) ŷ = 3ˆ ŷ = i.e., 3 units over in + direction, units down in direction.

10 IV 8 PHYS-010: General Phsics I R = R = 3 +( ) = 9+4 = 13 = = 4 ( ( θ = tan 1 R = tan 1 R 3 = 33.7= 30 ) ) = tan 1 ( ) (remember to round-off as per the significant digits of the input parameters). B. Velocit and Acceleration in D. 1. Displacement: r = r f r i. (IV-5) r r i r f

11 Donald G. Luttermoser, ETSU IV 9. Velocit: a) Average Velocit: v = r t. (IV-6) b) Instantaneous Velocit Velocit: r v lim t0 t d r dt. (IV-7) 3. Acceleration: a) Average Acceleration: a = v t. (IV-8) b) Instantaneous Acceleration Acceleration: v a lim t0 t d v dt. (IV-9) c) Note that a = a in this course = acceleration is constant. C. Projectile Motion. 1. In this section, we will make use of the following assumptions: a) The free-fall acceleration, g, has a magnitude of 9.80 m/s, is constant over the range of motion, and is directed downward. b) The effect of air resistance is negligible (hence the size of the object is relativel small and its surface is smooth). c) The rotation of the Earth does not affect the motion (hence the distance traveled is small with respect to the Earth s radius).

12 IV 10 PHYS-010: General Phsics I. For such projectile motion, the path is alwas a parabola. v v = v o v v o v o v o g v v v o θ ο v o v o -v o θ ο v a) The acceleration in the direction is g, just as in free fall. b) The acceleration in the direction is 0 (air friction is neglected). c) θ is called the projection angle. 3. Breaking motion into components. a) The vectors in our equations of motion can be broken up into and components: v = v cos θ v = v sin θ. b) We can just use the 1 D equations of motion and set them up for each component:

13 Donald G. Luttermoser, ETSU IV 11 -direction (horizontal) -direction (vertical) = + v (t t ) = + v (t t ) 1 g(t t ) (a) v = v v = v g(t t ) (b) v = v g( ) (c) (IV-10) c) The speed, v, of the projectile can be computed at an instant b v = v + v. (IV-11) d) The angle with respect to the ground at which the projectile is moving can be found with ( ) θ = tan 1 v v. (IV-1) 4. Problem-solving strateg for D motion: a) Select a coordinate sstem (tpicall Cartesian). b) Resolve the initial velocit vector into and components. c) Treat the horizontal motion and the vertical motion independentl. d) Horizontal motion has constant velocit. e) Vertical motion has constant acceleration.

14 IV 1 PHYS-010: General Phsics I Eample IV 6. Problem 3.56 (Page 77) from the Serwa & Faughn tetbook: A ball is thrown straight upward and returns to the thrower s hand after 3.00 s in the air. A second ball is thrown at an angle 30.0 with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown verticall? ma θ ο = 30.0 ο BALL 1 BALL Ball 1 v (Ball ) =? t =0 t f =3.00 s = 0 Step 1: Get ma first b realizing = ma =? that v = 0 at that position: =0 v =0 v = v gt v = 0 (at top) t = 1 (3.00 s) = 1.50 s v = v + gt =0+(9.80 m/s 1.50 s) v =14.7 m/s Step : ma now is determined with v = v g g = v v

15 Donald G. Luttermoser, ETSU IV 13 = v v g ma = (14.7 m/s) (0 m/s) (9.80 m/s ) = 11.0 m ma Step 3: Now solve for v for the ball thrown at 30 up to ma (remember ma =11.0 m and v = 0 at the top of the parabola): v = v g v = v ma +g ma v = 0 + (9.80 m/s )(11.0 m) v v = 15.6 m s = 14.7 m/s, which is eactl the initial velocit for the ball thrown straight up (as it should be). Then to find the total initial velocit, v, we just use a little trig: v = v sin θ v = v sin θ = 14.7 m/s sin 30 = v =9.4 m/s m/s 0.5 D. Relative Velocit. 1. Observers in different frames of reference ma measure different displacements or velocities for an object in motion.. For measuring motion in one frame of reference with respect to a different frame of reference, we must add the motion of the other frame to our equations.

16 IV 14 PHYS-010: General Phsics I a) Displacement: r = r + u t (IV-13) r displacement in our frame r displacement in the other frame u velocit of the other reference frame (wrt our frame) t time b) Velocit: v = v + u (IV-14) v velocit in our frame v velocit in the other frame Eample IV 7. Problem 3.63 (Page 78) from the Serwa & Faughn tetbook: A hunter wishes to cross a river that is 1.5 km wide and flows with a speed of 5.0 km/h parallel to its banks. The hunter uses a small powerboat that moves at a maimum speed of 1 km/h with respect to the water. What is the minimum time necessar for crossing? Let v = v br, where br means boat wrt river and u = v re, where re means river wrt Earth, then v re = 5.0 km/hr v = v + u v = v br = 1 km/h ˆ u = v re =5.0 km/h ŷ v br 1.5 km

17 Donald G. Luttermoser, ETSU IV 15 In velocit space v v v = v re v = v br θ v Determine v & θ: The actual velocit of the boat wrt the Earth is v = v be = v + v = km/h = 169 km/h = 13. km/h and the angle that the boat s makes wrt the -ais is ( ) θ = tan 1 v = tan 1 v 5.0 km/h 1 km/h = tan 1 =.6=3. The time it will take to cross the river is derived from v = = r t r r t t (set r =0,t =0) or t = r v

18 IV 16 PHYS-010: General Phsics I r is now the path length of our trip: cos θ = 1.5 km r r r = = 1.5 km cos θ 1.5 km cos.6 θ =1.6 km 1.5 km t = r v = 1.6 km 13. km/h =0.15 h = 0.15 h 60 min 1h t =7.5 min = min