Transcript Physics 141 Mechanics Yongli Gao Lecture 4 Motion in 3-D

Physics 141
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Mechanics
Lecture 4
Motion in 3-D
Motion in 2-dimensions or 3-dimensions has to be
described by vectors. However, what we have learnt from
1-dimensional kinematics is of great value since we can
treat each dimension separately as a 1-D problem.
The motion of a particle is described by the position vector
r=r(t)=x(t)i+y(t)j+z(t)k, a vector pointing from the origin to
(x,y,z), the coordinates of the particle. The components x(t),
y(t), z(t) are functions of time t and can have forms very
different from each other.
The displacement of the particle is described by the
displacement vector
Dr=r(t2)-r(t1)=(x(t2)-x(t1))i+(y(t2)-y(t1))j+(z(t2)-z(t1))k.
Note here the unit vectors are not changing with time.
Velocity in Higher Dimensions
• The average velocity is again defined as the displacement
divided by the time interval taken for the displacement
Dr r(t2 )  r(t1 )
v

Dt
t2  t1
• The instantaneous velocity, or velocity in short, is the time
derivative of the position vector
r(t  Dt)  r(t) dr dx dy
dz
v  lim

 i j k
Dt0
Dt
dt dt
dt
dt
dx
dy
dz
or, vx  , vy  ,vz 
dt
dt
dt
• Speed
v | v | vx 2  vy2  vz 2
Acceleration in Higher Dimensions
• The average acceleration is defined as the velocity change
divided by the time interval taken for the velocity change
Dv v(t2 )  v(t1 )
a

Dt
t2  t1
• The instantaneous acceleration, or acceleration in short, is
the time derivative of the velocity vector
dvy
v(t  Dt)  v(t) dv dv x
dvz
d 2r
a  lim


i
j
k 2,
Dt0
Dt
dt
dt
dt
dt
dt
dvy d 2 y
dvx d 2 x
dvz d 2 z
or, ax 
 2 , ay 
 2 , az 
 2
dt
dt
dt dt
dt dt
• Magnitude a | a | ax 2  ay 2  az2
Motion of Constant Acceleration
• If the acceleration a is a constant vector, then a  a
Note that both the magnitude and direction are constant.
• Similar to the 1-D case, we have then
1 2
v  v0  at,r  r0  v0t  at
2
Each of the above equations are in fact three equations
vx  v0 x  a x t, vy  v0 y  a y t,vz  v0z  azt
1
x  x 0  v 0 x t  ax t 2
2
1 2
y  y0  v0 y t  ay t
2
1 2
z  z0  v0z t  azt
2
Projectile Motion
• One common and useful motion of constant acceleration in
higher dimensions is that of a projectile. We have discussed
projectile motion confined in 1-D: the free fall body. What
if the initial velocity is not vertical? Let’s formulate the
motion of projectile.
• If we ignore the air friction, the only acceleration is due to
the gravity in the vertical direction. The projectile will
move within the plane defiled by the initial velocity and the
vertical direction. Let’s set up the coordinate system with
XOY in this plane. There is no motion in the third direction
and we don’t have to worry about it. We further define x to
be horizontal and y vertical.
• Assume the initial velocity
y
of the projectile is v0 at an
v0
g
angle q to the horizontal.
We choose the coordinate
q
system shown on the right:
0
x
• The initial velocity is
v0 = v0 cosq i + v0 sinq j and the constant acceleration is a=gj. The initial position is the origin.
The motion along horizontal direction is
x(t) = v0 t cosq
In the vertical direction,
1
y(t)  v0t sinq  gt 2
2 v (t ) = 0, or
The maximum height is reached when
y max
v sin q
v0 sin q  gtmax  0  tmax  0
g
•
• The maximum height
1
v0 2 sin2 q
2
ymax  v0 sin q  tmax  gtmax 
2
2g
• To get the range of the projectile, we just have to let y(t)=0
1 2
1
y(t)  v0t sin q  gt  0  t(v0 sin q  gt)  0
2
2
2v0 sin q
or,tflight 
g
xmax  v0 cosq  tflight
2v0 sin q v0 2 sin 2q
 v0 cosq

g
g
Example:
A basket ball is thrown with an angle q at the basket
distance D away and height H above the ground. What
should be the speed v0 of the ball?
Solution:
This is a projectile problem with the final y position at H.
1 2
y(t)  v0t sinq  gt  H
x(t)  v0t cosq  D
2
Eliminate t we get
2
D
1  D 

H  v0
sin q  g
v0 cosq
2 v0 cosq 
gD2
D tan q  H 
2v0 2 cos2 q
gD2
v0 
2 cos2 q (D tan q  H)
Demonstration: Monkey and Cannon
Suppose you are in a jungle with a huge cannon. You see a
monkey and decided to shoot it with your cannon. At the
moment you fires, the monkey falls down. How would you
have to aim to get a hit, disregard the moral issues and air
friction.
It seems a complicated problem since you have to consider
a moving target. In reality, all you have to do is just to aim
straight at the monkey. If there were no gravity, it’s
obviously right. With the gravity turned on, though, both
the monkey and the cannon ball fall the same amount in the
same time if you aimed right.
Uniform Circular Motion
• If the position of a particle is
r(t)  Rcoswti  Rsin wtj
y
r
its trajectory is a circle with
wt
radius R. The angle of the
x
position vector with the x-axis,
q(t)=wt, increases linearly with time.
dq
The angular velocity
described the rate of angular
change. In this case, dt dq
 w  constant
dt
Such motion is called uniform circular motion. The period
of uniform circular motion is the time taken for one full
revolution, or Dq2p, and it is T=2p/w.
Velocity of Uniform Circular Motion
• The velocity of a uniform circular motion is
dr dx
dy
dz
d
v
 i  j  k  (Rcoswti  Rsin wtj)
dt dt
dt
dt
dt
 wRsin wti  wRcoswtj
The magnitude of the velocity is constant
v | v | w 2 R2 sin2 wt  w 2 R2 cos2 wt  wR
Also v is always perpendicular to r since
r • v  (Rcoswti  Rsin wtj) • (Rsinwti  Rcoswtj)  0
Acceleration of Uniform Circular Motion
• The acceleration of a uniform circular motion is
dv
dv dvx
dv
d
a

i  y j  z k  (wRsin wti  wRcoswtj)
dt
dt
dt
dt
dt
 w 2 Rcos wti  w 2 Rsin wtj
The magnitude of the acceleration is also constant 2
v
4 2
2
4 2
2
2
a | a | w R cos wt  w R sin wt  w R 
R
Also a is always perpendicular to v since
a • v  (w 2 Rcoswti  w 2 Rsinwtj) • (Rsinwti  Rcoswtj)  0
And a is always antiparallel to r, pointing to the center of
the motion. It is therefore named centripetal acceleration.