Outline  Two Dimensional Motion


 Vector and Scalar Quantities
 Scalar quantity possesses magnitude only. Examples:
speed, mass and temperature. When you state a magnitude
you give a number, e.g. 50 miles/hr
 Vector quantities possess both magnitude and direction.
Examples: displacement, velocity, acceleration, and force.
Magnitude and direction means, for example, 50 miles/hr, south
 Two vectors are equal if they have the same magnitude and
direction. Moving a vector to a different location does
not change the vector as long as its direction and magnitude
remain constant.
 Usefulness of Vectors
 Principle of superposition: 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.
 A vector analysis allows you to separate twodimensional
motion into two onedimensional motions and then combine
them at the end of the problem. Example: projectile
motion
 Rules for vector addition
 For vectors along a straight line use a positive sign for
a vector to the right and a negative sign for a vector to
the left.
 For vectors at an angle to each other use:
 The polygon method. In Fig. 1a below, the vectors
are placed head to tail. The resultant vector goes
from the tail of the first vector A to the head
of the second vector B to give C = A
+ B
 The parallelogram method. Draw the vectors with
their tails at one point. Complete the parallelogram
and draw the diagonal to find the resultant. See Fig.
1b below.
 Cartesian coordinates
 Unit vectors. The unit vector i, j,
and k are unit vectors along the X, Y, and Zaxes,
respectively. When multiplied by a number or a symbol
that represents a quantity it becomes a vector with the
magnitude of the quantity (or symbol). For example,
5m i is a vector of length 5 m along the Xaxis.
 Components of vectors
 If a vector lies in the XY plane, it can be written
as a component in the Xdirection added vectorally
to a component in the Ydirection. A = A_{x}i
+ A_{y}j, where A_{x} and
A_{y} are the X and Y components of the vector
A (Fig. 2a below).
 From Fig. 2b above, we see that A_{x}
= A cos Θ A_{y}
= A sin Θ
and A = (A_{x}^{2 } +
A_{y}^{2})^{1/2
}
 Utility of components. If two vectors are
equal, their components along any chosen axis are
equal. If C = D, then
C_{x }= D_{x
} and C_{y }= D_{y
}
 Example (Fig. 3 below)
A = A_{x}i + A_{x}j
B = B_{x}i + B_{x}j
A + B = (A_{x}i + A_{y}j)
+ (B_{x}i + B_{y}j)
=
(A_{x} + B_{x)})i + (A_{y}+
B_{y})j
C = C_{x}i + C_{x}j.
If C =A + B, C_{x} =(A_{x}
+ B_{x)}), and
C_{y} =(A_{y}+ B_{y})
[Fig. 3 above]
 Sample problems in 107
Problem Set for Two Dimensional Motion: 14.
 Constant Acceleration
 For one dimensional motion
 In the Xdirection
v_{x} = v_{ox} + a_{x}t
and
x = x_{o }+ v_{ox }t + 1/2 a_{x}t^{2}
 In the Ydirection
v_{y} = v_{oy} + a_{y}t
and
y = y_{o }+ v_{oy }t + 1/2 a_{y}t^{2}
 For two dimensional motion
v = v_{o} + at
and r = r_{o } + v_{o}t
+ 1/2 at^{2},
where r_{o } = x_{o }i +
y_{o}j, v_{o } = v_{ox}i
+ v_{oy }j, and a = (a_{x}i
+ a_{y}j)
r = xi + yj. From A we
see we may write:
r = (x_{o }+ v_{ox }t + 1/2 a_{x}t^{2})i
+ (y_{o }+ v_{oy }t + 1/2 a_{y}t^{2})j,
or
r = (x_{o}i + y_{o}j)
+ (v_{ox}i + v_{oy }j)t
+ 1/2(a_{x}i + a_{y}j)t^{2,
} or
r = r_{o
}+ v_{o}t
+ 1/2 at^{2}
 Motion In TwoDimensions
 Projectile Motion
 Sample problem. An object is given an initial
velocity of 25 m/s at an angle of 53^{0} with
the horizontal. Find the initial position of the
object along (a) the Xaxis, x_{o}, (b)
the Y axis, y_{o}, the initial velocities
along (c) the Xaxis, v_{ox}, and (d) the
Yaxis, v_{oy}. (e) Indicate the directions
of v_{ox} and v_{oy} on the figure. Find
(f) the maximum vertical height y_{max}, (g) the
time for object to hit the ground, (h) the distance x
the projectile travels in a horizontal direction, (i)
v_{x}, v_{y}, and v just before the object
hits the ground, and (j) the angle v makes with
the horizontal. Take g = 10 m/s^{2}.
 Approach to solution
 Choose an appropriate X and Y axis. For a
projectile problem, the appropriate Xaxis is the
horizontal direction and the Yaxis the vertical direction.
Draw the figure, as shown in Fig. 4b below.
 Set up a chart for the X and Y components of the
descriptive quantities. Identify the quantities asked
for in (a) through (d) of the statement of the problem
along with the components of the acceleration and
enter them in the chart. Then write appropriate formulae
for the components of motion.
X

Y

(a) x_{o} = 0

(b) y_{o} =25
m

(c) v_{ox} =
25 m/s cos 53^{o
} =15 m/s

(d) v_{oy} =
25 m/s sin 53^{o}
= 20 m/s

a_{x} = 0

a_{y} = 10 m/s^{2}^{}

x(t) = x_{o}
+ v_{ox}t + 1/2 a_{x}t^{2}^{}

v_{y}^{2}(y)
= (v_{oy})^{2} + 2a_{y}(y
 y_{o})

(e) x(t) = 0 + 15 m/s
t + 0

(f) At y_{max},
v_{y}^{2}(y) = 0 =
(20m/s)^{2} 20m/s^{2}(y_{max}25m)
(y_{max} 25 m) = 20 m, or
y_{max }= 45 m


(g) y(t) = y_{o}
+ v_{oy}t + 1/2 a_{y}t^{2}
when object hits ground,
0 = 25 m + 20 m/s t  5 m/s^{2} t^{2},
or t^{2}
4 s t  5 s^{2} = 0
(t  5 s)(t +1 s) = 0
t = 5 s. t cannot be negative.

(h) x(5s) = (15m/s)(5s)
= 75m


(i) v_{x}(t)
= v_{ox} + a_{x}t
v_{x}(5 s) = v_{ox} = 15m/s

(i) v_{y}(t)
= v_{oy} + a_{y}t
v_{y}(5 s) = 20 m/s (10 m/s)(5s)
= 30 m/s

v = (v_{x}^{2}
+ v_{y}^{2})^{1/2}
= [15^{2} + (30)^{2}]^{1/2}
m/s = 33.5m/s

(j) tan Θ
= v_{y}/v_{x} = 30/15 = 
2.
Θ =  63^{o}

 Sample problems in 107
Problem Set for Two Dimensional Motion: 614.
 Uniform Circular Motion
 Definition: A body travelling in a circle with a constant
speed.
In Fig. 6 below, v = v + Δv
 Instantaneous velocity tangent to the circle at any
instant.
 Displacement vector sweeps through same angle as the
velocity vector.
 Centripetal acceleration
 Look at the triangles of Fig. 6b above.
 Δv/v = Δr/r
or
Δv = v Δr/v.
 To find magnitude of acceleration consider:
Δv/ Δt
= (v/r)( Δr/ Δt)
a = lim Δv/ Δt
= (v/r)(v) = v^{2}/r
^{Δ}^{t
}^{
o}
 Since a = lim of Δv/ Δt
as Δt
0, find the direction of the acceleration by looking at
the direction of Δv
as Δt approaches
0. Δv and
a point into the center of the circle.
 Descriptive Terms
 Period T is the time for one complete revolution
 Frequency f is the number of revolutions per second = 1/T
 The speed of the object = 2 πr/T
= 2 πrf is a CONSTANT.
It's velocity is not constant because it changes direction.
 The magnitude of the acceleration = v^{2}/r
is a constant, but the direction of the acceleration
is not a constant since it continually changes direction.
The direction changes so this is not a case of constant
acceleration and you cannot use the formulas for constant
acceleration.
 Angular velocity ω
is the time rate of change of angular displacement =
dΘ/dt
 The angle ΔΘ subtends
an arc whose length is Δs.
 By definition, the angle in radians, ΔΘ=
Δs/r.
 Since v = Δs/ Δt,
ΔΘ/ Δt = ( Δs/r)/ Δt = v/r. The limit as Δt
approaches zero of ΔΘ/ Δt is ω.
So ω = v/r.
 The direction of ω
is given by the right hand rule: When the fingers
of your right hand are curled in the direction of the
motion, the direction of your thumb points in the direction
of ω.
For Fig. 6a above, ω
is out of the page.
 While v is not constant for uniform circular
motion, ω
is constant.
Θ = ωt.
 Sample problems in 107
Problem Set for Two Dimensional Motion: 1518.

