Chapter

Two Dimensional Kinematics

Blinn College - Physics 1401 - Terry Honan

. - Kinematical Variables in 2D

The General Problem

Position as a Function of Time

There are two equivalent ways of describing position in two dimensions.  One is by labeling the coordinates (x,y).  The other is by giving the position vector (OverVector[r]) of the position.  The two are related; the coordinates are the components of the position vector.  To label position as a function of time we can consider x and y as separate functions of time or as (OverVector[r]) as a function of time.

x(t)  and  y(t)   ⟺   (OverVector[r])(t)=⟨x(t),y(t)⟩

The actual path followed by the body is called the trajectory.  It is represented by a plot of a path in the xy plane.

Average Velocity

The average velocity, as we saw in the one dimensional case, refers to two times. At (Subscript[t, i]) the position vector is (Subscript[(OverVector[r]), i])=(OverVector[r])((Subscript[t, i])) and at (Subscript[t, f]) it is (Subscript[(OverVector[r]), f])=(OverVector[r])((Subscript[t, f])).  The displacement is the difference of these two positions

Δ (OverVector[r])=(Subscript[(OverVector[r]), f])-(Subscript[(OverVector[r]), i])=⟨(Subscript[x, f])-(Subscript[x, i]),(Subscript[y, f])-(Subscript[y, i])⟩.

The average velocity vector is then defined as

(Subscript[(OverVector[v]), ave])=((Δ (OverVector[r]))/(Δ t))=⟨((Δ x)/(Δ t)),((Δ y)/(Δ t))⟩=⟨(Subscript[v, ave,x]) ,(Subscript[v, ave,y])⟩.

Instantaneous Velocity

The instantaneous velocity is defined as the limit of the average velocity as Δ t approaches zero; it is the time derivative of the position vector.

(OverVector[v])=(lim_(Δ t->0))((Δ(OverVector[r]))/(Δ t))=⟨(lim_(Δ t->0))((Δ x)/(Δ t)),(lim_(Δ t->0))((Δ y)/(Δ t))⟩=⟨(Subscript[v, x]),(Subscript[v, y])⟩.

The magnitude of the velocity is called the speed.

v=‖ (OverVector[v]) ‖=(Sqrt[(Subscript[v, x]^2)+(Subscript[v, y]^2)])=speed

Average Acceleration

As with the one dimensional case acceleration is to velocity as the velocity is to position.

(Subscript[(OverVector[a]), ave])=((Δ (OverVector[v]))/(Δ t))=⟨((Δ (Subscript[v, x]))/(Δ t)),((Δ (Subscript[v, y]))/(Δ t))⟩=⟨(Subscript[a, ave,x]),(Subscript[a, ave,y])⟩.

Instantaneous Acceleration

The instantaneous acceleration is similarly defined as a limit of the average acceleration or simply as the time derivative of the velocity.

(OverVector[a])=(lim_(Δ t->0))((Δ (OverVector[v]))/(Δ t))=⟨(lim_(Δ t->0))((Δ (Subscript[v, x]))/(Δ t)),(lim_(Δ t->0))((Δ (Subscript[v, y]))/(Δ t))⟩=⟨(Subscript[a, x]),(Subscript[a, y])⟩.

Acceleration with Constant Speed

The acceleration is the time derivative of the velocity. Constant velocity implies zero acceleration. Speed is the magnitude of the acceleration so there can be an acceleration if the speed is constant, because the direction can change. If the speed is constant then the acceleration must be perpendicular to the direction of motion, which is the direction of the velocity.

constant speed    ⟺    (OverVector[a]) ⊥ (OverVector[v])

If a car is driving at a constant speed and turning left then the acceleration is toward the left.

Parallel and Perpendicular Components of Acceleration

If the speed is changing then there is a forward or backward component of the acceleration, where an increasing speed gives a forward component and a decreasing speed gives a backward component. If the speed and direction are changing then you have both components parallel and perpendicular to the direction of motion. Suppose a car is braking while turning left. There are then two components of the acceleration, the parallel component points backward and the perpendicular component points to the right.

. - Constant Velocity and Acceleration

The cases of constant velocity and acceleration follows the one dimensional case.  Since we are now dealing with vectors the initial position and initial velocity, introduced previously, become vector quantities.

const (OverVector[v])    ⟹    (OverVector[a])=(OverVector[0])  and  (OverVector[r])(t)=(Subscript[(OverVector[r]), 0])+(OverVector[v]) t

const (OverVector[a])    ⟹    (OverVector[v])(t)=(Subscript[(OverVector[v]), 0])+(OverVector[a]) t  and  (OverVector[r])(t)=(Subscript[(OverVector[r]), 0])+(Subscript[(OverVector[v]), 0]) t+(1/2)(OverVector[a]) (t^2)

For 2D motion with constant acceleration we can modify the four constant acceleration equations from Chapter 2 into two sets of four equations.

({{x Equations,          y  Equations}, {(Subscript[v, x])=(Subscript[v, 0,x])+(Subscript[a, x]) t, (Subscript[v, y])=(Subscript[v, 0,y])+(Subscript[a, y]) t}, {Δ x=(1/2) ((Subscript[v, x])+(Subscript[v, 0,x])) t, Δ y=(1/2) ((Subscript[v, y])+(Subscript[v, 0,y])) t}, {Δ x=(Subscript[v, 0,x]) t +(1/2) (Subscript[a, x]) (t^2), Δ y=(Subscript[v, 0,y])t +(1/2) (Subscript[a, y]) (t^2)}, {(Subscript[v, x]^2)=(Subscript[v, 0,x]^2)+2 (Subscript[a, x]) Δ x, (Subscript[v, y]^2)=(Subscript[v, 0,y]^2)+2 (Subscript[a, y]) Δ y.}})

Projectile Motion

An important case of motion with constant acceleration is that of projectile motion. Projectile motion is two dimensional motion of body under the influence of only gravity. If gravity is all that acts means that we ignore air resistance. Spin effects like a curving baseball or a slicing golf ball are associated with air resistance and will also be ignored.

For projectile motion we will take x as the horizontal direction and y as the vertical direction. The acceleration due to gravity is downward and of magnitude g.  Writing this as a vector gives

(OverVector[a])=-g (OverHat[y])=⟨0,-g⟩    or    (Subscript[a, x])=0  and  (Subscript[a, y])=-g

Inserting these components into the two sets of equations above gives:

({{Horizontal, Vertical}, {Equations, Equations}, {(Subscript[v, x])=(Subscript[v, 0,x]), (Subscript[v, y])=(Subscript[v, 0,y])-g  t}, {Δ x=(Subscript[v, 0,x]) t , Δ y=(1/2) ((Subscript[v, y])+(Subscript[v, 0,y])) t}, {, Δ y=(Subscript[v, 0,y]) t -(1/2) g (t^2)}, {, (Subscript[v, y]^2)=(Subscript[v, 0,y]^2)-2 g Δ y.}})

The horizontal motion is simple.  Since there is no horizontal (the x direction) acceleration, the x component of the velocity is constant.  The vertical part (the y direction) of the motion is equivalent to free fall. The key to solving projectile motion problems is keeping the two parts separate.

If a projectile is launched at an initial angle of θ with an initial speed (Subscript[v, 0]) then the components of the initial velocity are given by

(Subscript[v, 0,x])=(Subscript[v, 0]) cos θ  and  (Subscript[v, 0,y])=(Subscript[v, 0]) sin θ.

Example . - An Initial Horizontal Velocity

Solution

Solution

h=1.35(QuantityUnit) ,  Δ x=2.42(QuantityUnit)  and  g=9.80(QuantityUnit)/(((QuantityUnit))^2)

(Subscript[v, 0])=(Sqrt[(Subscript[v, 0,x]^2)+(Subscript[v, 0,y]^2)])=(Sqrt[(Subscript[v, 0,x]^2)+(0^2)])=(Subscript[v, 0,x])

Δ y=(Subscript[v, 0,y]) t -(1/2) g (t^2)  ⟹  -h=0-(1/2) g (t^2)  ⟹  h=(1/2) g (t^2)  ⟹  t=(Sqrt[((2 h)/g)])=0.52489(QuantityUnit)

Δ x=(Subscript[v, 0,x]) t   ⟹  (Subscript[v, 0,x])=Δ x/t=4.67(QuantityUnit)/(QuantityUnit)

Solution

v=(Sqrt[(Subscript[v, x]^2)+(Subscript[v, y]^2)])=(Sqrt[(Subscript[v, 0,x]^2)+(Subscript[v, y]^2)])

(Subscript[v, y])=(Subscript[v, 0,y])-g  t=-g t=-5.1438(QuantityUnit)/(QuantityUnit)

v=(Sqrt[(Subscript[v, 0,x]^2)+(Subscript[v, y]^2)])=6.95(QuantityUnit)/(QuantityUnit)

Solution

θ=(tan^(-1))((Subscript[v, y])/(Subscript[v, 0,x]))=-47.7°

Example . - Punted Football

Solution

(Subscript[v, 0])=27(QuantityUnit)/(QuantityUnit)   and   θ=63°

(Subscript[v, 0,x])=(Subscript[v, 0]) cos θ=12.258(QuantityUnit)/(QuantityUnit)  and  (Subscript[v, 0,y])=(Subscript[v, 0]) sin θ=24.057(QuantityUnit)/(QuantityUnit)

(Subscript[v, y]^2)=(Subscript[v, 0,y]^2)-2 g Δ y  ⟹  0=(Subscript[v, 0,y]^2)-2 g (Subscript[y, max])  ⟹  (Subscript[y, max])=((Subscript[v, 0,y]^2)/(2 g))=29.5(QuantityUnit)

Solution

Δ y=(Subscript[v, 0,y]) t -(1/2) g (t^2)  ⟹  0=(Subscript[v, 0,y]) t -(1/2) g (t^2)=t ((Subscript[v, 0,y])-(1/2) g t)

0=(Subscript[v, 0,y])-(1/2) g T  ⟹  T=((2 (Subscript[v, 0,y]))/g)=4.9096(QuantityUnit)=4.91(QuantityUnit)

Solution

R=Δ x=(Subscript[v, 0,x])t=(Subscript[v, 0,x])T=60.2(QuantityUnit)

The range R of a projectile is the total horizontal distance traveled in the air when it returns to its original level, Δ y=0. In the punted football example above we saw the the total time to return to the same level was T=2 (Subscript[v, 0,y])/g and the range R was R=Δ x=(Subscript[v, 0,x])t=(Subscript[v, 0,x])T.  Combining these expression we get a general expression for the range.

T=((2 (Subscript[v, 0,y]))/g)  and  R=Δ x=(Subscript[v, 0,x])t=(Subscript[v, 0,x])T  ⟹  R=(Subscript[v, 0,x])((2  (Subscript[v, 0,y]))/g)=((2 (Subscript[v, 0,x])(Subscript[v, 0,y]))/g)=((2 (Subscript[v, 0,])cos θ (Subscript[v, 0,]) sin θ)/g)=(((Subscript[v, 0,]^2) )/g)2 cos θ sin θ.

There is a trig identity for the sine of twice an angle: sin(2 θ)=2 sin θ cos θ. This gives an expression for the range.

R=((Subscript[v, 0]^2)/g) sin(2 θ)

Remember that this only applies when Δ y=0.

Example . - Punted Football (continued)

Solution

R=((Subscript[v, 0]^2)/g) sin 2 θ=60.2(QuantityUnit)