Transcript 1 PHYSICS 231 Lecture 6: Motion in 2 dimensions

PHYSICS 231
Lecture 6: Motion in 2 dimensions
Remco Zegers
Walk-in hour: Thursday 11:30-13:30
HELPROOM
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Displacement in 2D
Often, we replace motion
in 2D into horizontal and
vertical components.
In vector notation:
r= x+ y
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Velocity and acceleration
The definitions made in 1D remain the same in 2D:


v   r /  t  average velocity in 2D


r
v  lim
 instantane ous velocity in 2D
t  0  t


a   v /  t  average accelerati on in 2D


v
a  lim
 instantane ous velocity in 2D
t  0  t
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While studying motion in 2D one almost
always makes a decomposition into
horizontal and vertical components of the
motion, which are both described in 1D
•Remember that the object can accelerate in one
direction, but remain at the same speed in the other
direction.
•Remember that after decomposition of 2D motion into
horizontal and vertical components, you should should
investigate both components to understand the complete
motion of a particle.
•After decomposition into horizontal and vertical
directions, treat the two directions independently.
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Parabolic motion: a catapult
vx=v0cos
vy=v0sin-2g=0
V=v0
vx=v0cos
vy=v0sin
vx=v0cos
vy=v0sin-1g
Vt=v0+at
vx=v0cos
vy=v0sin-3g
vx=v0cos
vy=v0sin-4g

vx=v0cos
vy=v0sin-5g
t=0
t=1
t=2
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t=3
t=4
t=5
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Question
 A hunter aims at a bird that is some distance away
and flying very high (i.e. consider the vertical
position of the hunter to be 0), but he misses. If
the bullet leaves the gun with a speed of v0 and
friction by air is negligible, with what speed vf
does the bullet hit the ground after completing its
parabolic path?
v0
vf
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Answer
 First consider the horizontal direction:
V0x=V0cos()
Since there is no friction, there is no change in
the horizontal component: Vfx=V0cos()=V0x
 Next the vertical direction:
V0y=V0sin()
Vy(t)=V0y-gt
xy(t)=Voyt-0.5gt2 (g=9.81 m/s2)
Boundary condition: bullet hits the ground:
0= Voyt-0.5gt2
t=0 or t=2V0y/g
So, Vfy(t)=V0y-(2V0y/g)g=-V0y
 Total velocity=(V0x2+(-V0y)2)=V0!!!!
 The speed of the bullet has not changed, but the
vertical component of the velocity has changed
sign.
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A
B
Pop and Drop
For A: Vy=-0.5gt2
Vx=0
For B: Vy=-0.5gt2
Vx=V0
For A: Xy=X0-0.5gt2
Xx=0
For B: Xy=X0-0.5gt2
Xx=V0t
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Shoot the monkey
The hunter aims his arrow
exactly at the monkey
At the moment he fires, the monkey
drops off the branch. What happens?
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The hor. Position of the arrow is:
x(t)=d-v0cos()t
x(t)=0 at t=d/v0cos()=tx=0
Its vertical position is:
y(t)=v0sin()t-0.5gt2
y(tx=0)=dtan()-0.5gt2=h-0.5gt2
Always hit!!
h
The vertical position of the
monkey is: y(t)=h-0.5gt2
The horizontal position is 0
d

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Another example
 A football player throws a ball with initial velocity
of 30 m/s at an angle of 30o degrees w.r.t. the
ground. How far will the ball fly before hitting the
ground? And what about 60o? And at what angle is
the distance thrown maximum?
X(t)=30cos()t
Y(t)=30sin()t-0.5gt2
=0 if t(30sin()-0.5gt)=0
t=0 or t=30sin()/(0.5g)
X(t=30sin()/(0.5g))=900cos()sin()/(0.5g)
=900sin(2)/g
if =30o X=79.5 m
if =60o X=79.5 m !!
Maximum if sin(2) is maximum, so =45o
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X(=45o)=91.7 m
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