12.Balanced Torques and Center of Gravity

Purpose: To investigate the conditions for rotational equilibrium of a rigid bar and to determine the center of gravity of a system of masses.

Equipment: Meter stick, meter stick clamps (knife edge clamp), balance support, mass set, weight hangers, unknown masses, balance.

Introduction: The condition for rotational equilibrium is that the net torque on an object about some point in the body, O, is zero. Remember that the torque is defined as the force times the lever arm of the force with respect to the chosen point O. The lever arm is the perpendicular distance from O to the line of action of the force.

Procedure:

Note: In each of the following steps, where appropriate, make a careful sketch showing the meter stick with the applied forces and mark their locations. Also, show the point, O, about which you are calculating torque.

1. Balance the meter stick in the knife edge clamp and record the position of the balance point. What point in the meter stick does this correspond to?

48.5cm

 

2. Select two different masses (100 grams or more each) and using the meter stick clamps and weight hangers, suspend one on each side of the meter stick support at different distances from the support. Adjust the positions so the system is balanced. Record the masses and positions. Is it necessary to include the mass of the clamps in your caculation? EXPLAIN! Sum the torques about your pivot point O and compare with the expected value.

 

mass1: 150g position1: 0.1535m
mass2: 50g position2: 0.1098m

mass of meter stick with clamp: 0.1807 kg

We should include the mass of the clamps, because now the clamp with the mass can be seen as a system, so the mass should be all together.

torqueleft = W1 × Ll = [(50+ 20)/1000]kg × 9.8m/s^2 × [(48.5- 24.1)/100]m = 0.500 N·m
torqueright = W2 × L2 = [(150+ 20)/1000]kg × 9.8m/s^2 × [(58.5- 48.5)/100]m = 0.509 N·m
difference of torque = 0.167-0.166= 0.001 N·m

expected value of L2:
W1·Ll = W2·L2
[(50+ 20)/1000]kg × 9.8m/s^2 × [(48.5- 24.1)/100]m= [(150+ 20)/1000]kg × 9.8m/s^2 × [L2/100]m
L2= 10.047cm
experimental value= 58.5cm-48.5cm= 10cm

percent of error= [(.16971651)/.1651]× 100%= 2.79%



3. Place the same two masses used above at different locations on the same side of the support and balance the system with a third mass on the opposite side. Record the masses and positions. Calculate the net torque on this system about the point support and compare with the expected value.

mass1: 150g position1: 68.5cm
mass2: 50g position2: 58.5cm
mass3: 100g position2: 15.3cm
mass of clamp: 20g


torqueright = W1·Ll+ W2·L2
[(50+ 20)/1000]kg × 9.8m/s^2 × [(58.5- 48.5)/100]m+ [(150+ 20)/1000]kg × 9.8m/s^2 × [(68.5-48.5)/100]m = 0.4018 N·m

torqueleft = W3·L3
[(100+ 20)/1000]kg × 9.8m/s^2 × [(48.5-15.3)/100]m = 0.3904 N·m

difference of torque = 0.4018-0.3904= 0.0114 N·m

expected value of L3:
W3·L3= W1·Ll+ W2·L2
[(100+ 20)/1000]kg × 9.8m/s^2 × L3= 0.4018 N·m
L3= 34.17cm

experimental value of L3= 48.5-15.3= 33.2cm


percent of difference= [(34.17-33.2)/34.17] × 100%= 2.84%



4. Replace one of the above masses with an unknown mass. Readjust the positions of the masses until equilibrium is achieved, recording all values. Using the equilibrium condition for rotational motion, calculate the unknown mass. Measure the mass of the unknown on a balance and compare the two masses by finding the percent difference.

mass1: 150g position1: 68.5cm
mass2: 50g position2: 58.5cm
mass3: unknown position2: 4.8cm
mass of clamp: 20g

expected value of mass3(unknown):
W1·Ll+ W2·L2 = W3·L3
[(50+ 20)/1000]kg × 9.8m/s^2 × [(58.5- 48.5)/100]m+ [(150+ 20)/1000]kg × 9.8m/s^2 × [(68.5-48.5)/100]m = [(m3+ 20)/1000]kg× 9.8m/s^2 × [(48.5-4.8)/100]m
m3= 94g

measured value of the unknown mass: 91.5g

percent of error= [(94-91.5)/94] × 100%= 2.66%



5. Place about 200 grams at 90 cm on the meter stick and balance the system by changing the balance point of the meter stick. From this information, calculate the mass of the meter stick. Compare this with the meter stick mass obtained from the balance. Should the clamp holding the meter stick be included as part of the mass of the meter stick? EXPLAIN!

balance point: 77.8cm

the mass of meter stick obtained from the balance: 84.2g

the calculated value of meter stick:
W1·Ll = W2·L2
m × 9.8m/s^2 × [(77.8- 48.5)/100]m = 0.2kg × [(90- 48.5)/100]m
m= 83.28g

percent of error: [(84.2-83.28)/84.2] × 100%= 1.09%

In this case, the mass of clamp is included in 200grams, so it shouldn't be included as part of the mass of the meter stick. Because in part 1, when we got the balance point of the meter stick (48.5cm) , there is no clamp or mass on it. So, here we still use the balance point as 48.5cm, the mass of clamp also shouldn't be included.



6. With the 200 grams still at 90 cm mark, imagine that you now position an additional 100 grams mass at the 30 cm mark on the meter stick. Calculate the position of the center of gravity of this combination (two masses and meter stick). Where should the point of support on the meter stick be to balance this system? Check your result by actually placing the 100 g at the 30 cm mark and balancing this system. Compare the calculated and experimental result.

Let the unknown point equal to x cm.

experimental value: x=65.1cm

calculated value:
W1·Ll+ W2·L2 = W3·L3
[100/1000]kg × 9.8m/s^2 × [(x- 30)/100]m+ [(84.2)/1000]kg × 9.8m/s^2 × [(x-48.5)/100]m
= [200/1000]kg× 9.8m/s^2 × [(90-x)/100]m

x=65.29cm

percent of error: [(65.29-65.1)/65.1] × 100%= 0.29%



Conclusion:

In this lab,we learn how  to determine the center of gravity of a system of masses and  how to investigate the conditions for rotational equilibrium of a rigid bar.
 The magnitude of torque depends on three quantities: the force applied, the length of the lever arm connecting the axis to the point of force application, and the angle between the force vector and the lever arm.

 

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.

5.Working With Spreadsheets

Purpose: To get familiar with electronic spreadsheets by using them in some simple applications.

Equipment: Computer with EXCEL software.

Part 1:

Create a simple spreadsheet that calculates the values of the following function:

f(x)=Asin(Bx+C)

Initially choose value for of A= 5, B= 3 and C= π/3 (1.047).

Create a column for values of x that run from zero to 10 radians in steps of 0.1 radians. Similarly, create in the next column the corresponding values of f(x) by copying the formula shown above down through the same number of rows (100 in roll).

Then copy and paste our data into the graphing program. Put appropriate labels on the horizontal and vertical axes of the graph. Use Curve Fit to find a function that best fit the data.

The best fit function: y= 5Sin(3x +1.05)-1.49×10^(-10)

Part 2:

Repeat the above process for a spreadsheet that calculates the position of a freely falling particle as a function of time. Start off with g= 9.8m/s^2, vo= 50m/s, xo= 1000m and △t= 0.2s.

The formula for free fall: f(t)= ro+vo△t+(1/2)a(△t)^2

We assuming the direction of vo positive.

(i)When g is positive:

f(t)=1000+ 50t+ (1/2)× 9.8t^2

Use Curve Fit to find a function that best fit the data:


The best fit function: f(t)= 4.9t^2+ 50t+ 1000

(ii)When g is positive:

f(t)=1000+ 50t+ (1/2)× (-9.8)t^2

Use Curve Fit to find a function that best fit the data:

 

The best fit function: f(t)= (-4.9)t^2+ 50t+ 1000

Question: How do data from part 1 and part 2 compare to the values we start with in our spreadsheet?

In part 1, we compare the data(A=5, B=3, C=1.05) from Curve Fit to the values(A=5, B=3, C=1.047) that we start with in our spreadsheet. In part 2, we do the same thing as initial values are "g= 9.8m/s^2, vo= 50m/s, xo= 1000m and △t= 0.2s". We find that the data from Curve fit are almost the same to the initial values.

Conclusion:
In this experiment we use excel spreadsheet to solve the problem.

We find that the data from Curve fit are almost the same to the initial values. This experiment helps us to have many data and efficiently do the formulas. 
 

4.Vector Addition of Forces

Purpose: To study vector addition by Graphical means and Using components.

Equipment: circular force table, masses, mass holders, string, protractor, and four pulleys.

Part 1.  Dr. Haag gave us 3 masses in grams which represent the magnitude of three forces and three angles.

	Magnitude	angle
A	100kg	0°
B	100kg	335°
C	100kg	270°

 

We made a vector diagram showing these forces, and find the resultant after adding the three vectors. There is two ways to calculate the resultant:

·         The head to tail vector. This involves lining up the head of one vector with the tail of the other.

·         The parallelogram method to calculate resultant vector. This method involves properties of parallelograms but, in the end boils down to a simple formula.

 

100 cos 0° + 100 cos 335° + 100 cos 270°= 190.631 = X

100 sin 0° + 100 sin 335° + 100 sin 270° = -142.262

Tanθ= 190.631/-142.262

θ= tan^-1 (142.262/190.63)= -36.733 + 180 = 143°

 

 

Conclusion:


When we place a mass on fourth holder equal to the magnitude of the resultant, the ring turns to equilibrium. That means the force of the fourth mass is equal to the resultant force of the first three masses. A vector is a quantity having a magnitude and a direction, and two vectors of the same type can be added.

The sources of error: Some magnitude of vectors are decimals, but we only have the masses with whole numbers.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3.Acceleration of Gravity on an Inclined Plane

The purpose of these lab is to find the acceleration of gravity by studying  the motion of cart on an incline plan.                                                                                                                                         


We set up the track putting the wood friction block under the track support at approximately 50cm raising the end of the track.

To determine the inclination of anlge X we solve a for the triangle.

Since the force of friction acts with the force of gravity when the cart is going up the track and against the force of gravity when the cart is going down the track, we can average the slightly increased acceleration (when going up) with the slightly decreased acceleration (when going down) to obtain an acceleration that depends only on the force of gravity. If we call g the acceleration along the track is g sin θ  where theta is the angle of incline for the track.


g sin θ= (a1+a2)/2

a 1	a 2	g
-0.3316	0.2679	9.83232

There were 3 trials for the first angle 

 

 

For the second angle there were three trials as well.

 

 

 

 





Conclusion:
According to two experiments, we found when θ is larger, our experimental data are closer to actual data(0.5% diff compare to 8.2% diff). The reason is that when θ is larger, the motion of cart is closer to free fall, which is influenced less by the disturbance. The causes of error:  Air resistance also against the motion, Our table is not horizontal, so our θ is not precise enough and the error of the equipment and the error when we read the data.

In this lab, we learned acceleration along the track is gsinθ where θ is the angle of inline for the track. we can use this property to estimate the gravity. We also learned how to control the variable to get another group of data, then try to think abut what cause the difference.



a 1

a 2

g 

-0.3316

0.2679

9.83232

 



2.Falling Body Experiment - Acceleration of Gravity

The purpose of this lab is to determine the acceleration of gravity for a freely falling object and to gain experience using the computer as data collector.

We found the free fall acceleration of a rubber ball tossed into the air by collecting the ball position (x) vs time(t) data. Since the velocity of an object is equal to the slope of the x vs t, the computer also constructs the graph of v vs t because acceleration = ΔV/Δt. For this exercise we conducted 5 trials.

1.        We placed the motion detector in the floor with the basket for protection. Then we tossed the ball into the air until we got a position-time graph of a parabolic shape.

2.       We selected the data in the interval of the parabola we chose Analyze/curve fit and chose Quadratic.

3.       From the velocity vs time graph we found out the acceleration.

4.      Finally we calculate the percent difference for each of the trials.

 

Results from Falling Body Experiment
trial	g exp	% diff	g exp	% diff
1	-9.302		-9.785
2	-9.506		-9.61
3	-10.156		-10.62
4	-9.99		-10.06
5	-9.534		-10.1

 

Conclusion:

In this lab, we determine the gravity is close to 9.8m/s^2.
In order to get the most precise data, we did this experiment for many times and got the average data, which decrease the error accidental error. Our data is 3.18% varying from the actual, that mean this experiment can prove that the gravity for a freely falling object is close to 9.8m/s^2. Our errors are because of : Air resistance, The inevitable experimental error, because the equipment can't be exactly precise. The curve fit is an estimate, so our gravity is also an estimate value.

We also learned how to use Lab Pro interface, Logger Pro Software, motion detector and gained the experience using the computer as a data collector. We also worked as a team to gain and analysis the data. This experience may benefit us in the future.