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Physics 207: Lecture 4, Pg 1
Lecture 4
Goals for Chapter 3 & 4
Perform vector algebra
- ( addition & subtraction) graphically or by xyz components
- Interconvert between Cartesian and Polar coordinates Work with 2D motion
- Distinguish position-time graphs from particle trajectory plots
- Trajectories Obtain velocities Acceleration: Deduce components parallel and perpendicular to the trajectory path Solve classic problems with acceleration(s) in 2D (including linear, projectile and circular motion) Discern different reference frames and understand how they relate to (^) motion in stationary and moving framesmotion in stationary and moving frames
Assignment: Read thru Chapter 5.4Assignment: Read thru Chapter 5. MP Problem Set 2MP Problem Set 2 due this Wednesdaydue this Wednesday
Physics 207: Lecture 4, Pg 2
Example of a 1D motion problem
A cart is initially traveling East at a constant speed of 20 m/s. When it is halfway (in distance) to its destination its speed suddenly increases and thereafter remains constant. All told the cart spends a total of 10 s in transit with an average speed of 25 m/s. What is the speed of the cart during the 2nd^ half of the trip? Dynamical relationships (only if constant acceleration):
v x = v x 0 +a x ∆ t
2 2
1 x = x 0 +v (^) x 0 ∆ t + a x ∆ t
a x = const
(v v ) 2
1 v
v v 2a (x x )
x(avg) x x
x 0
2 x
2 x
0
0
= +
− = −
( )
(totaltime)
(displacement )
v average velocity
t
x
And^ =
Physics 207: Lecture 4, Pg 3
The picture
Plus the average velocity Knowns: x 0 = 0 m t 0 = 0 s v 0 = 20 m/s t 2 = 10 s vavg = 25 m/s relationship between x 1 and x 2 Four unknowns x 1 v 1 t 1 & x 2 and must find v 1 in terms of knowns
t 0 t (^1) t 2
v 1 ( > v 0 ) a 1 =0 m/s^2
v 0 a 0 =0 m/s^2
x 0 x (^1) x 2
2 0
v 2 0
t t
x x
t
x
v (^) x = v x 0
x = x 0 +v (^) x (^) 0 ∆ t
a x = 0
Physics 207: Lecture 4, Pg 4
Using
Four
unknowns
Four
relationships
t 0 t (^1) t 2
v 1 ( > v 0 ) a 1 =0 m/s^2
v 0 a 0 =0 m/s^2
x 0 x (^1) x 2
( )
v
2 2 0
1 1
2 0
2 0
x x x
t t
x x
t
x
= −
−
−
∆
∆
x = x 0 +v (^) x 0 ∆ t
x 1 (^) = x 0 +v 0 ( t 1 − t 0 ) x 2 (^) = x 1 +v 1 ( t 2 − t 1 )
Physics 207: Lecture 4, Pg 7
Vectors and 2D vector addition
The sum of two vectors is another vector.
A = B + C
B
C A
B
C
D = B + 2 C ???
Physics 207: Lecture 4, Pg 8
2D Vector subtraction
Vector subtraction can be defined in terms of addition.
B - C
B
C
B
- C
B - C
= B + (-1) C
A Different directionand magnitude!
A = B + C
Physics 207: Lecture 4, Pg 9
Reference vectors: Unit Vectors
A Unit VectorUnit Vector is a vector having length 1 and no units
It is used to specify a direction.
Unit vector (^) uu points in the direction of (^) UU
Often denoted with a “hat”: (^) uu = û
U = |U|U = |U|^ ûû
ûû
x
y
z
ii
jj
kk
Useful examples are the cartesian unit vectors [ (^) i, j, ki, j, k ] Point in the direction of the x , y and z axes. R = rx i + ry j + rz k
Physics 207: Lecture 4, Pg 10
Vector addition using components:
Consider, in 2D, CC = AA + BB. (a) (^) CC = (Ax ii + Ay jj ) + (Bx ii + By jj ) = (Ax + Bx ) ii + (Ay + By ) (b) C = (CC x ii + Cy jj )
Comparing components of (a) and (b):
Cx = Ax + Bx Cy = Ay + By |C| = [ ( Cx )^2 + ( Cy )^2 ]1/
CC
Bx A A
By BB
Ax
Ay
Physics 207: Lecture 4, Pg 13
Converting Coordinate Systems In polar coordinates the vector R = (r,θ) In Cartesian the vector R = (rx,ry) = (x,y) We can convert between the two as follows:
- In 3D cylindrical coordinates ( r ,θ, z ), r is the same as the magnitude of the vector in the x-y plane [ sqrt (x^2 +y^2 )]
θ = tan-1^ ( y / x )
2 2 r = x + y
y
x
(x,y)
rr
ry
rx
ˆi ˆ j
cos
cos
R x y
r y r
r x r
y
x
θ
θ
Physics 207: Lecture 4, Pg 14
Resolving vectors into components
A mass on a frictionless inclined plane
A block of mass m slides down a frictionless ramp
that makes angle θ with respect to horizontal.
What is its acceleration a?
θ
m a
Physics 207: Lecture 4, Pg 15
Resolving vectors, little g & the inclined plane
g (bold face, vector) can be resolved into its x,y or x ’ ,y ’ components g = - g j g = - g cos θ j ’ + g sin θ i ’
The bigger the tilt the faster the acceleration….. along the incline
x ’
y ’
x
g y
Physics 207: Lecture 4, Pg 16
Dynamics II: Motion along a line but with a twist (2D dimensional motion, magnitude and directions) Particle motions involve a path or trajectory
Recall instantaneous velocity and acceleration
These are vector expressions reflecting x, y & z motion
rr^ =^ rr (t)^ vv^ = d rr^ / dt^ aa^ = d^2 rr^ / dt^2
Physics 207: Lecture 4, Pg 19
Instantaneous Acceleration
The instantaneous acceleration is the limit of the average acceleration as ∆ v /∆ t approaches zero
The instantaneous acceleration is a vector with components parallel (tangential) and/or perpendicular (radial) to the tangent of the path
Changes in a particle’s path may produce an acceleration
The magnitude of the velocity vector may change The direction of the velocity vector may change (Even if the magnitude remains constant) Both may change simultaneously (depends: path vs time)
Physics 207: Lecture 4, Pg 20
Generalized motion with non-zero acceleration:
need both path & time
Two possible options: Change in the magnitude of v
Change in the direction of v
a = 0
a (^) = 0
a =^ a +^ a
Animation
0 with^22
a ≠ a = ar + at
r r
a (^) r ≡ a ⊥
r r
a (^) t a ||
r r ≡
a
r
v
r
Physics 207: Lecture 4, Pg 21
Kinematics
The position, velocity, and acceleration of a particle in 3-dimensions can be expressed as: r r = x ii + y jj + z kk v v = vx ii + vy jj + vz kk ( ii , jj , kk unit vectors ) a a = ax ii + ay jj + az kk
All this complexity is hidden away in r r = (^) rr ( ∆ t) vv = d rr / dt aa = d^2 rr / dt^2
2
2
dt
d x ax = (^2)
2
dt
d y ay = (^2)
2
dt
d z az =
dt
dx vx = dt
dy vy = dt
dz vz =
x = x ( ∆ t ) y = y ( ∆ t ) z = z ( ∆ t )
2 2
1
with, if constant accel.,e.g. x (∆ t )= x 0 + vx 0 ∆ t + ax ∆ t
Physics 207: Lecture 4, Pg 22
Special Case
Throwing an object with x along the horizontal and y along the vertical.
x and y motion both coexist and t is common to both
Let g act in the – y direction, v0x= v 0 and v0y= 0
y
0 4 t x
t = 0
y
t
x
x vs t y vs t x^ vs^ y
Physics 207: Lecture 4, Pg 25
Exercise 1 & 2
Trajectories with acceleration
A rocket is drifting sideways (from left to right) in deep space, with its engine off, from A to B. It is not near any stars or planets or other outside forces.
Its “constant thrust” engine (i.e., acceleration is constant) is fired at point B and left on for 2 seconds in which time the rocket travels from point B to some point C Sketch the shape of the path from B to C.
At point C the engine is turned off.
Sketch the shape of the path after point C
Physics 207: Lecture 4, Pg 26
Exercise 1
Trajectories with acceleration
A. A
B. B
C. C
D. D
E. None of these
B
C
B
C
B
C
B
C
A
C
B
D
From B to C?
Physics 207: Lecture 4, Pg 27
Exercise 3
Trajectories with acceleration
A. A
B. B
C. C
D. D
E. None of these
C
C
C
C
A
C
B
D
After C?
Physics 207: Lecture 4, Pg 28
Lecture 4
Assignment: Read through Chapter 5.4Assignment: Read through Chapter 5.
MP Problem Set 2MP Problem Set 2 due Wednesdaydue Wednesday
