Mod 5 Flashcards
Important equations to remember
V=Vx+Vy
Vx=Vcosθ
Vy=Vsinθ
V=(V^2x+V^2y)^1/2
θ=tan^-1Vy/Vx
Theory behind projectile motion
A projectile is the name given to an object which is unpowered and the only force acting on the object is force due to the gravity.
Assumptions:
There is no such thing as air, If there was air, then the projectile would also experience a changing air resistance force. In the absence of this air resistance force, we will see that an object undergoing projectile motion will trace out a parabolic path.
The acceleration due to gravity is a constant value and has a constant direction (this way we can use our existing equations of uniform acceleration for analysis). This is a reasonably safe assumption if the height the projectile reaches is relatively low. We will see in a separate focus area how acceleration due to gravity changes as height changes.
We ignore the impact of the curvature of the Earth. Another relatively safe assumption (#FES for life). The range of our projectiles are usually small enough to assume that the Earth is relatively flat over the distance.
We also ignore any impact that the rotational motion of the Earth may have on velocity and/or acceleration
Describe the trajectory of an ideal projectile accounting air resistance
It has a decreased max height and range as well as a loss of symmetry. This is due to the Initial air force opposing the motion of the object and acts both horizontally and vertically. The maximum vertical displacement is lower as the air force opposes the upwards vertical motion. The range will decrease as the air force will cause the initial horizontal velocity to decrease.
Vy^2=uy^2+2aysy
sy=-uy^2/2ay
What affects the components of projectile motion
Projectile motion is two dimensional motion and can be analysed by considering its horizontal (Vx) and vertical (Vy) components of velocity.
The vertical component of velocity is affected by acceleration due to gravity and hence undergoes a constant acceleration due to this gravity (ay=-9.8 ms^-2)
The horizontal component of velocity isn’t affected by any external forces hence maintaining a constant value throughout the entire time of flight. Provided air resistance and the curvature of earth are ignored, it maintains a parabolic path. Since these components of velocity are perpendicular to each other, one has no impact on the other, they can therefore be considered as isolated or independent and hence be solved individually.
Conditions of type 2 projectile motion
An object is projected horizontally with a velocity of u from a cliff or some other object at height h.
Initial vertical velocity is 0
Initial horizontal velocity is ux
Overall vertical displacement is -h
Conditions of type 1 projectile motion
A projectile launched with an initial velocity at an angle θ above the horizontal. This projectile then strikes another point such that the change in vertical displacement from the beginning to the end of its journey = 0
Initial vertical velocity is usinθ
Initial horizontal velocity is ucosθ
Vertical velocity at max height is 0
Horizontal velocity at max height is ucosθ
Total velocity at max height is ucosθ. V=(Vx^2+Vy^2)^1/2. Vy=0
Final vertical displacement is 0
Final vertical velocity is -usinθ
Time to max height is t=(vy-uy)/ay. t=-uy/ay=-usinθ/ay
Time from max height to ground is t=-uy/ay
Total time of flight is T.O.F=-2uy/ay
Sy=max when tmax =T.O.F/2 and vy=o which is at max height
Sy=0 when t=o or t=T.O.F and when vy=+-uy
T.O.F time of flight
Range equation is sx=(u^2sin2θ)/g
Maximum range will always occur when the launch angle is 45°
Derivation of range equation
Sx=uxt
Sy=uyt+1/2ayt^2
o=uyt +1/2ayt^2
=uyt-g/2 x t^2
g/2 x t^2=uyt
t=2uy/y=(2usinθ)/g=T.O.F
Sx=uxt =ucosθ x (2usinθ)/g
=u^2x 2sinθ cosθ
(u^2 x sin2θ)/g
Conditions of type 3 projectile motion
Type 3 is a projectile launched with an initial velocity at an angle θ above or below the horizontal. This projectile then strikes another point such that there is a change in vertical displacement from the beginning to the end of its journey. The projectile will finish above or below where it started. Sy doesn’t equal 0.
Initial vertical velocity is +- usinθ
Overall vertical displacement is +-h
What is a projectile
What is gravity
Gravitational force effect on projectile
An unpowered object and the only force acting on it is gravitational force
A force occurring between two objects with mass.
It is constant force acting vertically downwards
How to do type 4 projectile motion
What is the question asking for
Like when sx=25m, is sy>20, want to get one solution before and after sx=25m
Then solve for t or any other variable
How to find time for stroboscopic images
Divide the time by the no of gaps
To get time use suvat equation or use frequency or period
How to describe acceleration and net force direction in circular motion
Radially inward or outward
Source of centripetal force in
Ball on a string whirled in a circle
Car driving around a corner
Satellite orbiting the Earth
Tension
Friction
Gravity
Note: If force is removed, then the object will fly off at a tangent.
Centripetal acceleration formula
Centripetal force formula
How to find velocity in circular motion
ac=v^2/r where centripetal acceleration acts radially inwards
Fc=mv^2/r
v=d/t = 2πr/T =2πrf
Angular velocity formula
Angular displacement formula
ω=Δθ/Δt = v/r
Δθ: angular displacement
ω: rotational velocity
θ=arc length/radius = s/r
Note: If angular displacement is 360, then the arc length is 2πr. The angular displacement would be 2πr/r=2π
