Lab 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.

Centripetal apparatus

    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 apparatus to find the measured force 

 
 The force diagram for the hanging weight:

Our five trials when m=475grams, r=16.5cm

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= 575grams
    radius= 16.5cm

The average calculated force: 7.53N

The average measured force: 7.252N

Percent difference: 3.83%

Conclusion:

      According to our data, we find that when radius is constant, the greater mass, the lower centripetal force. Our average calculated force is really close to measured force, that's means our data is convincing. Because the centripetal force is from the spring. The radius is constant means the length of spring doesn't change, so when the mass changes, the centripetal force is still constant. The formula: F=mv^2/r, F and r are constant, when mass increases, the velocity will decrease.
      In this lab, we verify Newton's second law of motion for the case of uniform circular motion. That the acceleration a of a body is parallel and directly proportional to the net force F acting on the body, is in the direction of the net force, and is inversely proportional to the mass m of the body, F = ma. In this case, the direction of net force points to the center of circle, F= ma, where a=v^2/r. 
      The causes of error:
      (1) When we measure the time for 50 revolutions of apparatus we can't keep the speed exactly constant. So, our frequency is not accurate.
      (2) Our experiment is influenced by air resistance, so our centripetal force actually is not only from the spring.
     

Lab 6: Drag Force on a coffee Filter

Purpose: To study relationship between air drag forces and the velocity of a falling body.


Equipment: Computer with Logger Pro software, lab pro, motion detector, nine coffee filters, meter stick

Introduction: When an object moves through a fluid, such as air, it experiences a drag force that opposes its motion. This force generally increases with velocity of the object. In this lab we are going to investigate the velocity dependence of the drag force. We still start by assuming the drag force,  Fd, has a simple power law dependence on the speed given by Fd=  k |v| ^n, where the power n is to be determined by the experiment.
This lab will investigate drag forces acting on a falling coffee filter. Because of the large surface area and low mass of these filters, they reach terminal speed soon after being released.

Procedure:

NOTE: You will be given a packet of nine nested coffee filters. It is important that the shape of this packet stays the same throughout the experiment so do not take the filters apart or otherwise alter the shape of the packet. Why is it important for the shape to stay the same? Explain and use a diagram.

1. Login to your computer with username and password. Start the Logger Pro software, open the

Mechanics folder and the graphlab file. Don’t forget to label the axes of the graph and create an appropriate title for the graph. Set the data collection rate to 30 Hz.

2. Place the motion detector on the floor facing upward and hold the packet of nine filters at a minimum height of 1.5 m directly above the motion detector. (Be aware other of nearby objects which can cause reflections.) Start the computer collecting data, and then release the packet. What should the position vs time graph look like? Explain.

Verify that the data are consistent. If not, repeat the trial. Examine the graph and using the mouse, select (click and drag) a small range of data points near the end of the motion where the packet moved with constant speed. Exclude any early or late points where the motion is not uniform.

3. Use the curve fitting option from the analysis menu to fit a linear curve (y = mx + b) to the selected data. Record the slope (m) of the curve from this fit. What should this slope represent? Explain.

Repeat this measurement at least four more times, and calculate the average velocity. Record all data in an excel data table.

4. Carefully remove one filter from the packet and repeat the procedure in parts 2 and 3 for the remaining packet of eight filters. Keep removing filters one at a time and repeating the above steps until you finish with a single coffee filter. Print a copy of one of your best x vs t graphs that show the motion and the linear curve fit to the data for everyone in your group (Do not include the data table; graph only please).

5. In Graphical Analysis, create a two column data table with packet weight (number of filters) in one column and average terminal speed (|v|) in the other. Make a plot of packet weight (y-axis) vs. terminal speed not velocity (x-axis). Choose appropriate labels and scales for the axes of your graph. Be sure to remove the “connecting lines” from the plot. Perform a power law fit of the  data and record the power, n, given by the computer. Obtain a printout of your graph for each member of your group. (Check the % error between your experimentally determined n and the theoretical value before you make a printout – you may need to repeat trials if the error is too large.)

6. Since the drag force is equal to the packet weight, we have found the dependence of drag force on speed. Write equation 1 above with the value of n obtained from your experiment. Put a box around this equation. Look in the section on drag forces in your text and write down the equation given there for the drag force on an object moving through a fluid. How does your value of n compare with the value given in the text? What does the other fit parameter represent? Explain.


Our data:

Time vs. Position graph when there is 5 filters

The slope of  Time vs. Position graph represents the terminal velocity (positive direction is upward). So on this graph, the terminal velocity is -1.66m/s. The terminal speed is 1.66m/s.

Time vs. Position graph when there is 8 filters

 On this graph, the terminal velocity is -1.99m/s. The terminal speed is 1.99m/s. 

The average terminal speed of 1-9 trails

We find that when the number of filter decreases, the average terminal speed also decreases. 

                The number of filter vs. average speed graph

We use the power law fit to find the best fitting curve that  Y=2.14X^1.95 (X= terminal speed; Y= number of coffee filters). We find n is 1.95, which is almost 2. That means it matches to the value that given in the text that n=2.

Question:

(1)Why is it important for the shape to stay the same?
     ---Because the drag relates to objects' cross section area. If filter's shape changes, then it's drag will also       change. 

(2)What should the position vs time graph look like?
     ---The curve decreases faster and faster at first. Then, at a moment, the curve turns to be linear, decreases with constant slope.

(3) What should this slope represent (y = mx + b)?
     ---The slope m represents the terminal velocity.

(4)How does your value of n compare with the value given in the text? What does the other fit parameter represent?
    ---We find n is 1.95, which is almost 2. That means it matches to the value that given in the text that n=2. 
         Because Mg= (1/4)Av^2 (A= cross section area). We found that Y=2.14X^1.95 (X= terminal speed; Y= number of coffee filters). So. Y× mg= (1/4)A×(X^2) (m= mass of each coffee filter). Also, Y= [A/ (4gm)]× (X^2). So 2.14 represents A/ (4gm): the cross section area of coffee filter divides four times a coffee filter's gravity force.

Conclusion:

      In this lab, we study the relationship between air drag forces and the velocity of a falling body.  According to our graph and data, we notice that when  an object begins to fall, its speed gradually increases at first. Then, at a moment, the velocity will be constant, and this object will maintain this condition until it toughs down. The reason is that when the speed is increasing, the Drag is also increasing, that Drag=  k |v| ^2. The net force of object is Mg-Drag, so acceleration= (Mg-Drag)/M. The Drag is increasing, the acceleration is decreasing. When Drag is equal to mass of the object, the acceleration turns to be zero. So the object will keep falling will constant velocity.

     The causes of error:
      (1) When filters are falling, their shapes are easy to change. Because drag relates to objects' cross section area, so if filters' shapes change, the drag will also change.
      (2) The filters in the air may be influenced by wind, so they may have extra horizontal velocities. So, the velocity we got may be not correct.  
      (3) We use rounding values to get the "number of filter vs. average speed graph", so it can't be exactly correct.