Lab 11: 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: 58.5cm
      mass2: 50g   position2: 24.1cm

      mass of clamp: 20g

      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 = 0.167 N·m
      torqueright = W2  × L2 = [(150+ 20)/1000]kg × 9.8m/s^2 × [(58.5- 48.5)/100]m = 0.166 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= [(10.047-10)/10.047]× 100%= 0.468%



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 learned how to investigate the conditions for rotational equilibrium of a rigid bar, and how to determine the center of gravity of a system of masses.
     We learned that torque is the tendency of a force to rotate an object about an axis, fulcrum, or pivot. 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. 
     The center of mass is the unique point where the weighted relative position of the distributed mass sums to zero. If the object is first balanced to find its center of mass, then the entire weight of the object can be considered to act at that center of mass. If the object is then shifted a measured distance away from the center of mass and again balanced by hanging a known mass on the other side of the pivot point, the unknown mass of the object can be determined by balancing the torques. 
   
    Causes of error:
    1. Some masses have been rusted, some their masses are not accurate.
    2. Our stick is not exactly horizontal, so the angle between stick and masses is not 90 degrees.

Lab 10: Human Power

Purpose: To determine the power output of a person

Equipment: Two meter metersticks, stopwatch, kilogram bathroom scale

Introduction: Power is defined to be the rate at which work is done or equivalently. the rate at which energy is converted from one form to another. In this experiment you will do some work by climbing from the first floor of the science building to the second floor. By measuring the vertical height climbed and knowing your mass, the change in your gravitational potential energy can be found:

                                                                    △PE = mgh

Where m is the mass, g is the acceleration of gravity, and h is the verticle height gained.
Your power output can be determined by:
     
                            Power = △PE / △t     Where △t is the time to climb the vertical height h


Data:


The height of the stairs is 4.29m

 h: 4.29m
m: 61.02kg
mg: 598N

△t₁ = 8.84s
△t₂ = 9.03s

Power₁= mgh / △t₁= 598N × 4.29m / 8.84s = 290.21 J/s = 290.21 W = 0.39 Hp
Power₂= mgh / △t₂= 598N × 4.29m / 9.03s = 284.1 J/s = 284.1 W = 0.38 Hp

Power_avg =  (Power₁ + Power₂ ) / 2 = (0.39 Hp +0.38 Hp) / 2 = 0.385 Hp

Question:

1. Is it okay to use your hands and arms on the handrailing to assist you in your climb up the stairs? Explain why or why not.

     If we use our hands and arms on the handrailing to assist us, our calculated value of power won't change. Because when you put your hands on the handrailing, the handrailing will give you a normal force upward, but your weight doesn't change, which is still equal to mg. So, according to the formula △PE = mgh, the change of the gravitational potential energy will be the same.


2. Discuss some of the problems with the accuracy of this experiment.
 
    The height of the floor, the mass and the time are all measured values that are not really accurate, so our calculated values of power are also have error.


Followup questions:

1. Two people of the same mass climb the same flight of stairs. Hinrik climbs the stairs in 25 seconds. Valdis takes 35 seconds. Which person does the most work? Which person expands the most power? Explain your answers.

     Because △PE = mgh, their mass and the height are the same, so they both did the same work.
     Hinrik expands more power. Because  Power = △PE / △t , △PE are the same, but Hinrik used less time, so he had a higher power.


2. A box that weights 1000 Newtons is lifted a distance of 20.0 meters straight up by a rope and pulley system. The work is done in 10.0 seconds. What is the power developed in watts and kilowatts.
   
         Power = △PE / △t = (1000N× 20m) / 10s=  2000W= 2KW


3. Brynhildur climbs up a ladder to a height of 5.0 meters. If she is 64 kg:

    a) What work dose she do?
                 
        Work = mgh = 64kg × 9.8m/s^2 × 5m =3136J

    b) What is the increase in the gravitational potential energy of the person at this height?
 
        △PE = mgh = 64kg × 9.8m/s^2 × 5m =3136J
         Increase 3136J

    c) Where does the energy come from to cause this increase in P.E.?

         the force of gravity do the negative work on the person, so the gravitational potential energy will increase by the same amount. The energy the person gets to do the work come from this person's chemical energy. The consumption of chemical energy increases this person's gravitational potential energy.

4. Which requires more work: lifting a 50 kg box vertically for distance of 2m , or lifting a 25kg box vertically for a distance of 4 meters?

  Lifting a 50 kg box:
  Work = mgh = 50kg × 9.8m/s^2 × 2m = 980N
  Lifting a 25 kg box:
  Work = mgh = 25kg × 9.8m/s^2 × 4m = 980N
 
  So, they require the same work.


Conclusion:

     In this lab, we determined the power output when people climb the floors. We also learned the definition of the potential energy and power.
     Potential energy is energy stored in an object. This energy has the potential to do work. Gravity gives potential energy to an object. This potential energy is a result of gravity pulling downwards. The gravitational constant, g, is the acceleration of an object due to gravity. This acceleration is about 9.8 meters per second on earth. The formula for potential energy due to gravity is PE = mgh. As the object gets closer to the ground, its potential energy decreases while its kinetic energy increases. The difference in potential energy is equal to the difference in kinetic energy. After one second, if the potential energy of an object fell ten units than its kinetic energy has risen ten units. Potential energy units are joules.
    Power is the rate at which energy is transferred, used, or transformed. The unit of power is the joule per second (J/s), known as the watt. Energy transfer can be used to do work, so power is also the rate at which this work is performed. The same amount of work is done when carrying a load up a flight of stairs whether the person carrying it walks or runs, but more power is expended during the running because the work is done in a shorter amount of time.

Source of error:
    The height of the floor, the mass and the time are all measured values that are not really accurate, so our calculated values of power are also not accurate.

Lab 8: Motion in One Dimension with Air Drag

Purpose: To analyze how changing force affects motion in one dimension.

Introduction: 
So far we have assumed objects move with constant acceleration. This assumes that a constant force is applied to the object. What if a changing force (which causes a changing acceleration) is applied to an object?

An object under the influence of air drag is one example of this. Calculus provides a way to solve this problem where variables are changing with respect time. A spreadsheet can be used to perform this calculus. Two basic relationships can be used to analyze the object:  

                                                           v_new = v_old + a_avg△t     (1)


Where v_old  is the previous value of velocity and position, and a_avg  is the average value of the acceleration and velocity during the time interval △t.


1. By unit analysis, show that the above equation (1) is valid:

            Left side unit = m/s
            Right side unit = (m/s) + (m/s^2) × s = (m/s) + (m/s) = m/s
          
    So the units of two sides are identical.
 

2. Why do we use a_avg in equation (1).
    
            Because the can be changing all the time, so the average data is more accurate.


3. Come up with an analogous equation relating y_new and y_old:

            y_old = y_o
             
                y_new = y_o + v_old△t + (1/2)a_avg(△t)^2
            or   y_new = y_o + v_avg△t


4. What is the benefit of changing a small △t:

            Because the acceleration is changing all the time, so in a small △t, the a_avg is closer to real value.     So, the v_new will also be more accurate.



Since the acceleration depends on the forces acting on the object, we must specify precisely what these forces are. In our problem the drag force will at first be assumed to depend on the first power of the velocity and can be written as:
                                                                     F_D= -kv
Where k is a proportionality constant.


1. Draw a detailed motion diagram of the object falling down:

a. Now draw a force diagram for the object falling down. Include vectors for all forces, and write a  statement of Newton's Second. Solve for acceleration.

    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 |. 

     When there is no drag, the net force on the ball is gravity force, so the acceleration is equal to -9.8m/s^2.  The motion trajectory is parabola. The velocity is changing with constant slope, decreasing to 0 first then increasing in negative position. 

     When drag is equal to -kv, a= -g (1+ v / v_t ). Because the velocity is changing all the time, so the acceleration is also changing. When the ball goes up, velocity is decreasing, so the acceleration will increase in negative position. The maximum height of the ball will be lower than the ball with no drag. Then the velocity will increase in negative position. The velocity will be constant when drag is equal to the gravity force.

     When drag is equal to |kv^2|, according to the v-t graph, we find its acceleration is lower than D= -kv before it reaches the highest point. So, the its maximum height is higher than the ball when D= -kv and D= 0.For balls with  D= -kv and D= |kv^2|, their terminal velocities are the same, and at the highest point their velocities are all equal to 0. According to the v-t graph, when balls are going down the one with D= |kv^2| has a smaller average acceleration. Because v_t^2- v_o^2= 2ax, so the ball with lower acceleration will go deeper. That's why in this case the ball goes deeper than the ball with D= -kv.

      Causes of error:

      We use average velocities in 0.1s to plot our graphs, so the velocities are approximate values, which are not accurate.