chapter 3 motion in 2 dimensions. 1) displacement, velocity and acceleration displacement is the...
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Chapter 3Motion in 2 dimensions
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1) Displacement, velocity and acceleration
• displacement is the vector from initial to final position
Δrr =
rr −
rr0
Δrx = x - component of Δrr = Δx
Δry = y - component of Δrr = Δy
Δy
Δx
Δx = rx − r0 x = x − x0
Δy = ry − r0 y = y − y0
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1) Displacement, velocity and acceleration
• average velocity
rv =
Δrr
Δt
rv
vx =ΔxΔt
=x−x0
Δt
vy =ΔyΔt
=y−y0Δt
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1) Displacement, velocity and acceleration
• instantaneous velocity
rv = lim
Δt→ 0
Δrr
Δt
vx = limΔt→ 0
ΔxΔt
vy = limΔt→ 0
ΔyΔt
rv0
rv
v can change even if v is constant
v is tangent to the path
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1) Displacement, velocity and acceleration
• average acceleration
ra =
Δrv
Δt=
rv−
rv0
Δt
rv0
rv
rv
ra
ax =Δvx
Δt
ay =Δvy
Δt
not in general parallelto velocity
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1) Displacement, velocity and acceleration
• instantaneous acceleration
rv0
rv
ax = limΔt→ 0
Δvx
Δt
ay = limΔt→ 0
Δvy
Δt
ra = lim
Δt→ 0
Δrv
Δt
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1) Displacement, velocity and acceleration
• instantaneous acceleration
rv0
rv
ra = lim
Δt→ 0
Δrv
Δt
- object speeding up in a straight line
rv0
rv
Δrv
acceleration parallel to velocity
- object at constant speed but changing direction
rv Δ
rv
acceleration perp. to velocity
ra
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2) Equations of kinematics in 2d
• Superposition (Galileo): If an object is subjected to two separate influences, each producing a characteristic type of motion, it responds to each without modifying its response to the other.
• That is, we consider x and y motion separately
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2) Equations of kinematics in 2d
A bullet fired vertically in a car moving with constant velocity, in the absence of air resistance (and ignoringCoriolis forces and the curvature of the earth), will fall back into the barrel of the gun. That is, the bullet’s x-velocity is not affected by the acceleration in the y-direction.
vx
vy
vy
vx
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2) Equations of kinematics in 2d
• That is, we can consider x and y motion separately
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2) Equations of kinematics in 2d
• That is, we can consider x and y motion separatelyDisplacement : x, y (x0 =0,y0 =0)Velocity: vx,vy v0x,v0y
Acceleration: ax,ay (a0x =ax,a0y =ay)
Time: t (t0 =0)
x =v0xt+12 axt
2
vx =v0x + axt
vx2 =v0x
2 + 2axx
x= 12 (v0x + vx)t
x=vxt−12 axt
2
y =v0yt+12 ayt
2
vy =v0y + ayt
vy2 =v0y
2 + 2ayy
y= 12 (v0y + vy)t
y=vyt−12 ayt
2
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Example
rv0
ra
x
y v0 x =22 m/s; v0y =14 m/s
ax = 24 m/s2; ay =12 m/s2
t=7.0 s
Find x, y,rv at t =7.0 s
x =v0xt+12 axt
2 =740 m
y=v0yt+12 ayt
2 =390 m
vx =v0x + axt=190 m/svy =v0y + ayt= 98 m/s
v = vx2 + vy
2 = 210 m/s
tanθ =vy
vx
=0.516 → θ =27º
rv
v0 = v0x2 + v0y
2 = 26 m/s
tanθ =vy
vx
=0.516 → θ =32.5º
rv
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3) Projectile Motion (no friction)
a) EquationsConsider horizontal (x) and vertical (y) motion separately (but with the same time)
Horizontal motion: No acceleration ==> ax=0
x =v0xt+12 axt
2 → x=v0xtvx =v0x + axt→ vx =v0x
Vertical motion: Acceleration due to gravity ==> ay= ±g
- usual equations for constant acceleration
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3) Projectile Motion (no friction)
Example: Falling care package
x
Find x.
Step 1: Find t from vertical motion
Step 2: Find x from horizontal motion
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3) Projectile Motion (no friction)
Example: Falling care package
x
Step 1:
Given ay, y, v0y
Solve
y =v0yt+12 ayt
2 =−12 gt2
t=2yg
=14.6 s
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3) Projectile Motion (no friction)
Example: Falling care package
x
Step 2:
x =v0xt=1680 m
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3) Projectile Motion (no friction)b) Nature of the motion: What is y(x)?
Eliminate t from y(t) and x(t):
x =v0xt ; y=v0yt−12 gt2
t=x
v0x
x
y
Substitute:
y =v0y
v0x
⎛
⎝⎜⎞
⎠⎟x−
g2v0x
2
⎛
⎝⎜⎞
⎠⎟x2
y(x) is a parabola
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3) Projectile Motion (no friction)
c) Cannonball physics
Initial velocity rv is given:
Either (v0 x ,v0 y ) or (v0 ,θ)
Find (i) height, (ii) time-of-flight, (iii) range
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3) Projectile Motion (no friction)
c) Cannonball physics
Initial velocity rv is given:
Either (v0 x ,v0 y ) or (v0 ,θ)
(i) Height: Consider only y-motion: v0y given, ay=-g knownThird quantity from condition for max height: vy=0
Use v2 =v0y2 + 2ayy
When y=H ,v=0, so
0 =v0y2 −2gH → H =
v0y2
2g
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Use y =v0yt+12 ayt
2
When y=0,
0 =v0yt−12 gt2 (Trivial solution: t=0)
Non-trivial solution: t=2v0y
g
3) Projectile Motion (no friction)
c) Cannonball physics
Initial velocity rv is given:
Either (v0 x ,v0 y ) or (v0 ,θ)
(ii) Time-of-flight: Consider only y-motion: v0y given, ay=-g knownThird quantity from condition for end of flight; y=0
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For y =0, x=R and t=2v0y
g, so
from x=v0xt, R=2v0xv0y
g
3) Projectile Motion (no friction)
c) Cannonball physics
Initial velocity rv is given:
Either (v0 x ,v0 y ) or (v0 ,θ)
(iii) Range: Consider x-motion using time-of-flight: x=v0xt
or, R =v0
2 (2cosθ sinθ)g
=v0
2 sin(2θ)g