PHYS 1114, Lecture 9, February 6 Contents:

Transcription

1 PHYS 4, Lecture 9, February 6 Contents: Continued with projectile motion: The kicko problem in football was treated analytically, obtainin formulas for maimum heiht and rane in terms of initial speed and kicko anle. Inorin air resistance, the larest rane is reached when the kicko anle is 45 = /4 rad at iven kicko speed. The same formulas hold enerally, for eample when throwin a ball or shootin a cannon. 2 Four representations of motion usin the eample of a man walkin at constant speed in one direction were iven. 3 The inclined plane was introduced and an eample was worked out. 4 An eample of projectile motion was treated numerically: Dive of 5 m plank. i

2 Equations for Projectile Motion The main equations of Chapter 2 (Motion on a Line) are = o + v o t + 2a t 2 v = v o + a t v 2 v 2 o = 2a ( o ) where we have added subscripts to indicate that we are now oin to apply these to the horizontal motion components. The equations for the vertical motion components are simply y = y o + v yo t + 2a y t 2 v y = v yo + a y t v 2 y v 2 yo = 2a y (y y o ) replacin every by a y. Nothin really new! To obtain the equations for projectile motion, we only have to substitute a = 0 and a y =, ivin 8 = o + v o t, v = v o = constant, >< y = y o + v yo t 2 t2, v y = v yo t, >: vy 2 vyo 2 = 2(y y o ). (The second and third eqation for the -components both just become v = v o when a = 0.)

3 Typical eample of projectile motion The horizontal velocity component is constant if we can inore air resistance. But, because of the ravitational pull of the Earth ivin a constant downward acceleration, the vertical velocity component may at first be upward but decreasin till it becomes zero, at the maimal heiht H. After this the vertical velocity component starts rowin in the downward direction. Finally the projectile hits the round (or makes its first bounce) havin covered distance R. y v 0 H θ R In the above picture the vertical motion is slowed down at first as the vertical velocity and acceleration components point in opposite directions. Later, when they point in the same direction speedin up occurs. y-ais v 0y = v 0 sinθ v 0 θ v 0 = v 0 cosθ -ais The initial velocity components v o, v yo are also determined also by speed v o and anle : 2

4 y v 0 θ H R v o = v o cos, v yo = v o sin. Also, let us assume o = 0, y o = 0, so that we may rewrite our main equations as = v o t, y = v yo t 2 t 2, v y = v yo t. Heiht H At the top we have v y = 0, y = H, and, say, t = t. Then, v y = 0 = v yo t, y = H = v yo t 2 t 2. Solvin the first identity for t and substitutin that in the second, we find, (also usin 2 = 2), t = v yo, H = v yo v yo 2 vyo 2 vyo 2 = 2 = (v o sin )

5 Time of fliht T At the end of the fliht we have = R, y = 0, and, say, t = T. Then, = R = v o T, y = 0 = v yo T 2 T 2 = T (v yo 2 T ). The last equation has both the trivial solution T = 0, the moment of kick-o, and the desired solution v yo 2 T = 0, or T = 2v yo = 2t. This result shows the symmetry, that both the upward and downward portions of the motion take equal time. Finally, Rane R versus initial anle iven v o y 67 o 23 o 45 o R = v o T = v o 2v yo = 2v ov yo = 2(v o cos ) (v o sin ) =) R = 2 sin cos v2 o = sin(2 ) v2 o. In eamples you can substitute iven numerical variables early and that may make the equations look simpler to you. But you will have less calculator manipulations if you work out alebraic answers first. If needed, the trionometric identity sin(2 ) = 2 sin cos will be provided at the eam. 4

6 Four representations of motion Consider a man walkin at constant speed. We can draw a picture of his motion, by drawin him a few times, indicatin his proress after fied time intervals. To describe his proress mathematically (or physically), we use the point mass model, replacin the man s body by a point mass. This is ad hoc for now, but can be justified with the concept of center of mass introduced later. Havin the point mass description, we can now also plot the motion in a raph, position versus time t, or ive the correspondin formula. The four representations: Pictures Point masses t Graph = o + v o t Formula 5

7 Gravity s acceleration = 9.80 m/s 2 a y = down! Inclined plane q cosq q Acceleration a = sin, induced by ravity,down incline. Note: When is small, sin small and a small. Note: Both pairs of anles with value marked have les that are pairwise orthoonal (makin anles of 90 derees). Therefore, the two anles are equal by a well-known theorem in eometry. 6

8 v=0 sin 30 = 2 L=5.0m cosq q=30 o 30 o o 90 o 8 a = m/s2 >< o = 0 = 5.0 m >: v o = 0 t unknown! v, t unknown! v unknown! o 0 v? v o m a 4.90 m/s 2 t? 8 >< = o + v o t + 2 at2 v = v o + at >: v 2 = vo 2 + 2a( o ) So we can solve the first and the third equation: 5.0 m = = (4.90 m/s2 ) t 2 =) t 2 = m 4.90 m/s2 =) t 2 = s2 =) t = p 2.04 s =.95 s =) t =.2 s v 2 = (4.90 m/s 2 )(5.0 m) = 49.0 m 2 /s 2 =) v = 7.0 m/s 7

9 Projectile motion a = 0 o = v o t y y o = v yo t 2 t2 v y = v yo t v 2 y = v 2 yo 2(y y o ) Dive of 5m plank v o =5.00m/s 5.00m q? variable value o 0? y o 5.00 m y 0 v o = v 5.00 m/s v yo 0 v y, v?,? a y = 9.80 m/s 2 t?? Check the unknowns in the four equations of motion: We can solve time t from the first y motion equation and, knowin t, we can solve and v y. 8

10 y = y o + v yo t 2t 2 =) 0 = 5.00 m m/s2 =) t 2 = m 9.80 m/s 2 =.020 s2 =) t =.00 s =.0 s = o + v o t = 0 + (5.00 m/s)(.00 s) = 5.05 m v y = v yo t = 0 (9.80 m/s 2 )(.00 s) = 9.90 m/s v The speed is iven by Pythaoras and the anle by an inverse tan: q v y v = q v 2 + v 2 y = p (5.00 m/s) 2 + ( 9.90 m/s) 2 = p 23.0 m/s =. m/s tan = v v y = =) = 26.8 = rad (The final answers are rounded to three places, keepin more places in the intermediate steps.) 9