This preview shows page 1. Sign up to view the full content.
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=AB 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 3D 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 1D kinematics equations: x = x(t) dx v= dt dv d 2 x a= = 2 dt dt 3D 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 3D 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) 2D Kinematics Most 3D problems can be reduced to 2D 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: 31 & 32) 3D Kinematics (Text: 33) Baseball problem Independence of x and y components ...
View
Full
Document
This note was uploaded on 09/29/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell University (Engineering School).
 Fall '07
 LECLAIR,A
 Physics, mechanics

Click to edit the document details