Physics 111: Lecture 2 Today’s Agenda

Physics 111: Lecture 2Today’s Agenda • Recap of 1-D motion with constant acceleration • 1-D free fall • example • Review of Vectors • 3-D Kinematics • Shoot the monkey • Baseball • Independence of x and y components

Review: • For constant acceleration we found: x t v • From which we derived: t a t

12 22 32 42 Recall what you saw:

1-D Free-Fall • This is a nice example of constant acceleration (gravity): • In this case, acceleration is caused by the force of gravity: • Usually pick y-axis “upward” • Acceleration of gravity is “down”: y t v t a y t ay =  g

Ball w/ cup Gravity facts: Penny & feather • g does not depend on the nature of the material! • Galileo (1564-1642) figured this out without fancy clocks & rulers! • demo - feather & penny in vacuum • Nominally, g= 9.81 m/s2 • At the equator g = 9.78 m/s2 • At the North pole g = 9.83 m/s2 • More on gravity in a few lectures!

1000 m Problem: • The pilot of a hovering helicopter drops a lead brick from a height of 1000 m. How long does it take to reach the ground and how fast is it moving when it gets there? (neglect air resistance)

y Problem: • First choose coordinate system. • Origin and y-direction. • Next write down position equation: • Realize that v0y = 0. 1000 m y = 0

y Problem: • Solve for time t when y = 0 given that y0 = 1000 m. • Recall: • Solve for vy: y0= 1000 m y = 0

Lecture 2, Act 11D free fall • Alice and Bill are standing at the top of a cliff of heightH. Both throw a ball with initial speedv0, Alice straightdownand Bill straightup. The speed of the balls when they hit the ground arevAandvBrespectively.Which of the following is true: (a)vA < vB(b) vA = vB (c) vA > vB Alice v0 Bill v0 H vA vB

Lecture 2, Act 11D Free fall • Since the motion up and back down is symmetric, intuition should tell you that v = v0 • We can prove that your intuition is correct: Equation: This looks just like Bill threw the ball down with speed v0, sothe speed at the bottom shouldbe the same as Alice’s ball. • Bill v0 v= v0 H y = 0

Lecture 2, Act 11D Free fall • We can also just use the equation directly: Alice: same !! Bill: Alice v0 Bill v0 y = 0

Chicago r Urbana Vectors (review): • In 1 dimension, we could specify direction with a + or - sign. For example, in the previous problem ay = -g etc. • In 2 or 3 dimensions, we need more than a sign to specify the direction of something: • To illustrate this, consider the position vectorr in 2 dimensions. • Example: Where is Chicago? • Choose origin at Urbana • Choose coordinates of distance (miles), and direction (N,S,E,W) • In this case r is a vector that points 120 milesnorth.

Vectors... • There are two common ways of indicating that something is a vector quantity: • Boldface notation: A • “Arrow” notation: A =

Vectors... • The components of r are its (x,y,z) coordinates • r= (rx ,ry ,rz ) = (x,y,z) • Consider this in 2-D (since it’s easier to draw): • rx = x = r cos • ry = y = r sin where r = |r| (x,y) y arctan( y / x ) r  x

Vectors... • The magnitude (length) of r is found using the Pythagorean theorem: r y x • The length of a vector clearly does not depend on its direction.

û Unit Vectors: • A Unit Vector is a vector having length 1 and no units • It is used to specify a direction • Unit vector u points in the direction of U • Often denoted with a “hat”: u = û • Useful examples are the Cartesian unit vectors [i, j, k] • point in the direction of the x, yand z axes U y j x i k z

Vector addition: • Consider the vectors A and B. Find A+B. B B A A A C=A+B B • We can arrange the vectors as we want, as long as we maintain their length and direction!!

Vector addition using components: • Consider C=A + B. (a) C = (Ax i + Ayj) + (Bx i + By j) = (Ax + Bx)i + (Ay + By)j (b)C = (Cx i+ Cy j) • Comparing components of (a) and (b): • Cx = Ax + Bx • Cy = Ay + By By C B Bx A Ay Ax

Lecture 2, Act 2Vectors • Vector A = {0,2,1} • Vector B = {3,0,2} • Vector C = {1,-4,2} What is the resultant vector, D, from adding A+B+C? (a){3,5,-1}(b){4,-2,5}(c){5,-2,4}

Lecture 2, Act 2Solution D = (AXi+ AYj+ AZk) + (BXi+ BYj+ BZk) + (CXi+ CYj + CZk) = (AX + BX + CX)i+ (AY + BY+ CY)j+ (AZ + BZ + CZ)k = (0 + 3 + 1)i + (2 + 0 - 4)j+ (1 + 2 + 2)k = {4,-2,5}

3-D Kinematics • The position, velocity, and acceleration of a particle in 3 dimensions can be expressed as: r = x i + y j + z k v = vx i + vy j + vz k(i,j,kunit vectors ) a = ax i + ay j + az k • We have already seen the 1-D kinematics equations:

3-D Kinematics • For 3-D, we simply apply the 1-D equations to each of the component equations. • Which can be combined into the vector equations: r = r(t) v = dr / dt a = d2r / dt2

3-D Kinematics • So for constant acceleration we can integrate to get: • a = const • v = v0 + a t • r = r0 + v0 t + 1/2 a t2 (where a, v, v0, r, r0, are all vectors)

2-D Kinematics lost marbles • Most 3-D problems can be reduced to 2-D problems when acceleration is constant: • Choose y axis to be along direction of acceleration • Choose x axis to be along the “other” direction of motion • Example: Throwing a baseball (neglecting air resistance) • Acceleration is constant (gravity) • Choose y axis up: ay = -g • Choose x axis along the ground in the direction of the throw

“x” and “y” components of motion are independent. Cart • A man on a train tosses a ball straight up in the air. • View this from two reference frames: Reference frame on the moving train. Reference frame on the ground.

Problem: • Mark McGwire clobbers a fastball toward center-field. The ball is hit 1 m (yo ) above the plate, and its initial velocity is 36.5 m/s(v ) at an angle of 30o () above horizontal. The center-field wall is 113 m (D) from the plate and is 3 m (h) high. • What time does the ball reach the fence? • Does Mark get a home run? v h  y0 D

Problem... • Choose y axis up. • Choose x axis along the ground in the direction of the hit. • Choose the origin (0,0) to be at the plate. • Say that the ball is hit at t = 0, x = x0 = 0 • Equations of motion are: vx = v0xvy = v0y - gt x = vxt y = y0 + v0y t - 1/ 2 gt2

Problem... • Use geometry to figure out v0x and v0y : g Find v0x = |v| cos . and v0y = |v| sin . y v v0y  y0 v0x x

Problem... • The time to reach the wall is: t = D / vx (easy!) • We have an equation that tell us y(t) = y0 + v0y t + a t2/ 2 • So, we’re done....now we just plug in the numbers: • Find: • vx = 36.5 cos(30) m/s = 31.6 m/s • vy = 36.5 sin(30) m/s = 18.25 m/s • t = (113 m) / (31.6 m/s) = 3.58 s • y(t) = (1.0 m) + (18.25 m/s)(3.58 s) - (0.5)(9.8 m/s2)(3.58 s)2 = (1.0 + 65.3 - 62.8) m = 3.5m • Since the wall is 3 m high, Mark gets the homer!!

Lecture 2, Act 3Motion in 2D • Two footballs are thrown from the same point on a flat field. Both are thrown at an angle of 30o above the horizontal. Ball 2 has twice the initial speed of ball 1. If ball 1 is caught a distance D1 from the thrower, how far away from the thrower D2 will the receiver of ball 2 be when he catches it?(a) D2 = 2D1 (b) D2 = 4D1 (c) D2 = 8D1

Lecture 2, Act 3Solution • The distance a ball will go is simply x= (horizontal speed) x (time in air) = v0x t • To figure out “time in air”, consider the equation for the height of the ball: • When the ball is caught, y = y0 (time of catch) two solutions (time of throw)

v0,2 v0y ,2 v0,1 ball 2 ball 1 v0y ,1 v0x ,1 v0x ,2 Lecture 2, Act 3Solution x = v0x t • So the time spent in the air is proportional to v0y: • Since the angles are the same, both v0y and v0x for ball 2are twice those of ball 1. • Ball 2 is in the air twice as long as ball 1, but it also has twicethe horizontal speed, so it will go 4 times as far!!

Shooting the Monkey(tranquilizer gun) • Where does the zookeeper aim if he wants to hit the monkey? • ( He knows the monkey willlet go as soon as he shoots ! )

Shooting the Monkey... r = r0 • If there were no gravity, simply aim at the monkey r =v0t

Dart hits the monkey! Shooting the Monkey... r= r0 - 1/2 g t2 • With gravity, still aim at the monkey! r= v0t - 1/2 g t2

Recap:Shooting the monkey... x= v0 t y= -1/2 g t2 • This may be easier to think about. • It’s exactly the same idea!! x = x0 y= -1/2 g t2

Recap of Lecture 2 • Recap of 1-D motion with constant acceleration. (Text: 2-3) • 1-D Free-Fall (Text: 2-3) • example • Review of Vectors (Text: 3-1 & 3-2) • 3-D Kinematics (Text: 3-3 & 3-4) • Shoot the monkey (Ex. 3-11) • Baseball problem • Independence of x and y components • Look at textbook problemsChapter 3: # 35, 39, 69, 97

Physics 111: Lecture 2 Today’s Agenda