8. Motion in One Dimension with Air Drag

The purpose of this lab is to analyze how changing force affects motion in one dimension.

Introduction: If a changing force (which causes a changing acceleration) is applied to an object then the object Fnet will be different at every time.

Ø  An object under the influence of Drag is one example. Two basic relationships can be used to analyze the object.              Vnew = Vbold + aavg

1.       By unit analysis we can show that the equation above is valid.

2.       We use aavg  in equation 1 because a is changing therefore taking the average is gives a more accurate value.

3.       We come up with an analogous equation relating Ynew to Yold.

                                       Vnew=VoldΔt+ aavg Δt

                                                                               2

4.       The benefit of choosing a small Δt is that the average of acceleration is closer to the actual acceleration problem.

 

Ø  Since the acceleration depedends on the forces acting on the oblject, we must specify precisely what these forces are. I our problem the drag force will be assumed to depened on the first power of the velocity, and can be written as  F_D = - Kv

Where K is proportionally constant.

 

1.       Motion diagram of the object falling down.

 

a= F / m = D - G = (-kv-mg) / m = - (g + kv / m )


b. Give the condition for the object at terminal velocity (a=0). Using this condition solve for k.

a= 0, So net force is 0, mg = kv_t , k= mg / v_t

c. Substitute k into your expression for the acceleration from part a.

a = - (g + kv / m )
k = mg / v_t So, a = -g (1+ v / v_t )


2. Open the spreadsheet in the Physics Apps file called air drag. Describe what the spreadsheet is calculating.

a. What are the assumption? (i.e. initial values? )

t_o = 0s , g = -9.8m/s^2 , initial velocity = 20m/s , terminal velocity = -40m/s , △t = 0.1s

b. What is v-halfstep?

V-halfstep is the average velocity in 0.1s.

c. What is the a-halfstep?

A-halfstep is the average acceleration in 0.1s.


3. Using the graph paper provided, draw scales graphs of position vs. time, velocity vs. time and acceleration vs. time for the object no air drag.

position vs. time graph when there is no drag

p= -4.9t^2 + 20t +1000

velocity vs. time graph when there is no drag

v= -9.8t + 20

Make predictions on another sheet of graph paper about how position and velocity graphs would change if you include air drag (D= -kv).

position vs. time graph when drag is equal to -kv

velocity vs. time graph when drag is equal to -kv

Now look at a drag force that is dependent on the square of the velocity. Assuming a drag force,
F_D= | kv^2 |, find the new formula for the acceleration.

a= F / m = D - G = (kv^2 - mg) / m
mg = k v_t^2 , k = mg / v_t ^2
So, a= [(v^2/ v_t^2) -1] g

position vs. time graph when drag is equal to |kv^2|

velocity vs. time graph when drag is equal to |kv^2|

Conclusion:

In this lab, we analyzed how changing force affects motion in one dimension. We plotted v-t and p-t graphs in three cases: with no drag; drag is equal to -kv and drag is equal to | kv^2 |.

    

7. Centripetal Force

Purpose: To verify Newton's second law of motion for the case of uniform circular motion.

Equipment: Centripetal force apparatus, metric scale, vernier caliper, stop watch, slotted weight set, weight hanger, triple beam balance.

Introduction: Then centripetal force apparatus is designed to rotate a known mass through a circular path of known radius. By timing the motion for a definite number of revolutions and knowing the total distance that the mass has traveled, the velocity can be calculated. Thus the centripetal force, F, necessary to cause the mass to follow its circular path can be determined from Newton's second law.
F=mv^2/r
Where m is the mass, v is the velocity, and r is the radius of the circular path.
Here we have used the fact that for uniform circular motion, the acceleration, a, is given by:
a=v^2/r


Procedure:
1. For each trail the position of the horizontal crossarm and the verticle indicator post must be such that the mass hangs freely over the post when the spring is detached. After making this adjustment, connect the spring to the mass and practice aligning the bottom of the hanging mass with the indicator post while rotating the assembly.

2.Measure the time for 50 revolutions of the apparatus. Keep the velocity as constant as possible by keeping the pointer on the bottom of the mass aligned with the indicator post. A while sheet of paper placed as a background behind the apparatus can be helpful in getting the alignment as close as possible. Using the same mass and radius, measure the time for three different trials. Record all data in a neat excel table.

3.Using the average time obtained above, calculate the velocity of the mass. From this calculation the centripetal force exerted on the mass during its motion.

4.Independently determine the centripetal force by attaching a hanging a weight to the mass until it once again is positioned over the indicator post (this time at rest). Since the spring is being stretched by the same amount as when the apparatus was rotating, the force stretching the spring should be the same in each case.


Data:
mass= 475grams
radius= 16.5cm
v= 2πr·f
f=50/t
a=v^2/r
F=mv^2/r
(r= radius, f= frequency, t= time of 50 revolutions, m= mass, a= acceleration, F=calculated centripetal force)

The force diagram for the hanging weight:
The average calculated force: 7.36N

The average measured force: 7.066N

Percent difference: 4.44%

5.Add 100g to the mass and repeat steps 2,3,4 above.

Data:

mass= 473grams
radius= 16.5cm

 

The average calculated force: 7.53N

The average measured force: 7.252N

Percent difference: 3.83%

Conclusion:

The sum total of all interactions is directly proportional to the acceleration of the object.