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Unformatted text preview: Physics 112: Lecture 4 Today's Agenda More on Vectors Motion in 2D/3D How to define A - B? Vector Subtraction Define the difference as the vector sum of A and -B: A - B = A + (-B) Connect the tail of -B with the head of A A A C=A-B -B -B B A Vectors obey these rules: A+B=B+A c ( A + B) = cA + cB A + (-A) = 0 Vector Algebra A short problem:
B C A Vector C = which of the following? ( 1 ) A + B ( 2 ) B - A ( 3 ) (4) A o ther B T Vector Components Often easier to break vectors into components along reference directions (less trig) E.g. in 2 dimensions:
y x x Products of Vectors Scalar product ("dot product") Will use this later, e.g. to describe work done by a force Vector product ("cross product") Will need this later for torque and angular momentum Scalar product AB Defined as the magnitude of A multiplied by the component of B parallel to A Result is not a vector! r r A " B # A B cos $ B B cos A ! Components of scalar product
Scalar product in terms of the components? In 2D: AB = (Axi+Ayj)(Bxi+Byj) =AxiBxi +AxiByj + AyjBxi + AyjByj ii=1, ij=0 = AxBx + AyBy i.e. the scalar product of two vectors is just the sum of the product of the components! Another short question
Vectors A and B: A = (3.0 m)i + (4.0 m) j B = (2.0 m)i + (-2.0 m)j What's the angle (in degrees) between A and B? T Another short question
Vectors A and B: A = (3.0 m)i + (4.0 m) j B = (2.0 m)i + (-2.0 m)j What's the angle (in degrees) between A and B? cos " = Ax Bx + Ay By 3m # (2m) + 4m # ($2m) = AB (3m) 2 + (4m) 2 (2m) 2 + (2m) 2 $2m 2 1 cos " = =$ = $0.14 % " = 98.1o 5m # 2 2m 5 2 ! Vector product C=AxB Defined as a vector quantity that is perpendicular to both A and B. Direction given by the "right hand rule". Magnitude: |C| = |A| |B| sin
C B A 3-D Kinematics !!! The position, velocity, and acceleration of a particle in 3 dimensions can be expressed as: r= xi+yj+zk v = vx i + vy j + vz k a = ax i + ay j + az k (i , j , k unit vectors ) We have already seen the 1-D kinematics equations: x = x(t) dx v= dt dv d 2 x a= = 2 dt dt 3-D Kinematics
d ^ ^ v(t) = (x(t)i + y(t) ^ + z(t) k ) j dt dx ^ dy ^ dz ^ v(t) = i+ j+ k dt dt dt dx dy dz " v x = ,v y = ,v z = dt dt dt d2x d2y d 2z " ax = 2 ,ay = 2 ,az = 2 dt dt dt Which can be combined into the vector definitions: r v = dr / dt a = d2r / dt2 ! = r(t)
Calculus with vectors: USE COMPONENTS 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, v0, and r0, are all constant vectors and v(t) and r(t) are vectors that are functions of time) 2-D Kinematics 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. A man on a train tosses a ball straight up in the air. View this from two reference frames: (more on reference frames later)
Reference frame on the moving train. Cart Reference frame on the ground. Problem: A baseball player hits a fastball toward centerfield. 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 centerfield wall is 113 m (D) from the plate and is 3 m (h) high. At what time does the ball reach the fence? Is this a home run?
v y0 D h 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 = v0x x = v xt vy = v0y - gt y = y0 + v0y t - 1/ 2 gt2 Problem... Use geometry to figure out v0x and v0y :
g Find and v0x v0y v0x = |v| cos . v0y = |v| sin . y v y0 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: a = -g 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 = 3.5 m Since the wall is 3 m high, it's a homer!! Recap of Lecture 4 Vector Kinematics (Text: 3-1 & 3-2) 3-D Kinematics (Text: 3-3) Baseball problem Independence of x and y components ...
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