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Free-fall & air resistancePowerPoint Presentation

Free-fall & air resistance

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Free-fall & air resistance. AP Physics C. The Elephant and Feather—Free-Fall.

Free-fall & air resistance

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Free-fall & air resistance

AP Physics C

- Suppose that an elephant and a feather are dropped off a very tall building from the same height at the same time. Suppose also that air resistance could somehow be eliminated such that neither the elephant nor the feather would experience any air drag during the course of their fall. Which object - the elephant or the feather - will hit the ground first?

- The elephant and the feather each have the same force of gravity.
- The elephant has more mass, yet both elephant and feather experience the same force of gravity.
- The elephant experiences a greater force of gravity, yet both the elephant and the feather have the same mass.
- On earth, all objects (whether an elephant or a feather) have the same force of gravity.
- The elephant weighs more than the feather, yet they each have the same mass.
- The elephant clearly has more mass than the feather, yet they each weigh the same.
- The elephant clearly has more mass than the feather, yet the amount of gravity (force) is the same for each.
- The elephant has the greatest acceleration, yet the amount of gravity is the same for each.

- http://www.physicsclassroom.com/mmedia/newtlaws/efff.cfm

- Suppose that an elephant and a feather are dropped off a very tall building from the same height at the same time. We will assume the realistic situation that both feather and elephant encounter air resistance. Which object - the elephant or the feather - will hit the ground first?

- The elephant encounters a smaller force of air resistance than the feather and therefore falls faster.
- The elephant has a greater acceleration of gravity than the feather and therefore falls faster.
- Both elephant and feather have the same force of gravity, yet the acceleration of gravity is greatest for the elephant.
- Both elephant and feather have the same force of gravity, yet the feather experiences a greater air resistance.
- Each object experiences the same amount of air resistance, yet the elephant experiences the greatest force of gravity.
- Each object experiences the same amount of air resistance, yet the feather experiences the greatest force of gravity.
- The feather weighs more than the elephant, and therefore will not accelerate as rapidly as the elephant.
- Both elephant and feather weigh the same amount, yet the greater mass of the feather leads to a smaller acceleration.
- The elephant experiences less air resistance and than the feather and thus reaches a larger terminal velocity.
- The feather experiences more air resistance than the elephant and thus reaches a smaller terminal velocity.
- The elephant and the feather encounter the same amount of air resistance, yet the elephant has a greater terminal velocity

- http://www.physicsclassroom.com/mmedia/newtlaws/efar.cfm
- But why then does the elephant, which encounters more air resistance than the feather, fall faster? After all doesn't air resistance act to slow an object down? Wouldn't the object with greater air resistance fall slower?

- In the diagrams below, free-body diagrams showing the forces acting upon an 85-kg skydiver (equipment included) are shown. For each case, use the diagrams to determine the net force and acceleration of the skydiver at each instant in time.

- When does terminal velocity occur?
- What is the net force equal to?
- What is the acceleration equal to?

- its speed ____________.
- the air resistance force on it _______________.
- the net force on it _______________.
- its acceleration _______________.
- until it reaches ________________, where the net force on it and its acceleration are ___________. Terminal velocity occurs when the air resistance force ___________ the weight of the falling object. This means that:
- the object is falling with a _______________ - its acceleration is ________________.
- heavy objects will have a ______________ terminal velocity than light objects.

- Air resistance often called "drag" force is the result of collisions of the object's leading surface with air molecules. The actual amount of air resistance encountered by the object is dependent upon a variety of factors but for the purposes of this article only the following two are discussed. The two most common factors which have a effect upon the amount of air resistance are the speed of the object and the cross-sectional area of the object. Thus:
- Increased speeds result in an increased amount of resistance.
- Increased cross-sectional areas result in an increased amount of air resistance.

- Matt drops an object of mass m into a fluid. The viscous force of the fluid is given by : f = bv, where b is the constant that depends on the properties of the fluid and the object and v is the velocity of the object.
- Draw a FBD.
- Apply Newton’s 2nd Law.
- What is the net force and acceleration of the object when it reaches its terminal velocity?
- What is its terminal velocity in terms of m, b, and g?

x

v

a

t

t

t

Position vs

time

Acceleration

vs

time

Velocity

vs

time

- What is the velocity and acceleration at t = 0?
- Sketch the x-t, v-t and a-t graphs from t = 0 to the time that the object reaches terminal velocity.

- Draw a FBD for a mass m that is dropped in a fluid. The mass experiences a viscous force of F = bv.
- Apply Newton’s 2nd law to the FBD.
- Circle those variables that vary over time from t = 0 until the mass reaches its terminal velocity.

- Draw the FBD.
- Apply Newton’s 2nd Law.
- F = ma
- mg – bv = ma

- mg – bv = m dv
dt

- mg – v = m dv
b bdt

- vt – v = m dv
b dt

- v – vt = - m dv
b dt

- dv = - bdt
v – vt m

- ∫ dv = - b ∫ dt
v – vt m

- What are the limits of integration?

- ln(v – vt) = -bt
m

- ln(v – vt ) –
ln (– vt) = -bt

m

- ln (v – vt) = -b t
– vt m

- v – vt = e-b/m t
– vt

- v = vt(1 - e-b/m t)

- How would you use the velocity-time function to determine the acceleration?
- v = vt(1 - e-b/m t)
- a = dv = d(vt(1 - e-b/m t))
dtdt

- a = vt(-b)( - e-b/m t)
m

- a =mg(-b)( - e-b/m t)
b m

- a = g e-b/m t
- Go back to the velocity-time and acceleration-time graphs and compare to the velocity and acceleration functions.

- How would you use the velocity-time function to determine the position?
- dy = v = vt(1 - e-b/m t)
dt

- dy = v = vt(1 - e-b/m t)
- ∫dy = ∫(vt(1 - e-b/m t)) dt

- y = vtt – vtm(1 - e-b/m t)
b