Thursday, December 1, 2011

Lab #9 Torque

Introduction:


What is Torque? Torque is similar to a force, except torque is a rotational force. When we talk about force in general, we are talking about a change in linear motion. We can find torque by following:

\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!

Since r is a cross product of F, we can say:

\tau = rF\sin \theta\,\!

r is the distance from the point of rotation where a force is applied, F is the magnitude of the force, and θ is the angle between the force and a line from the point of rotation.





For the object to be in equilibrium, the net torque must be equal to 0, Στ = 0. So, Lmg = rFsinθ. Where L is the distance from the point of rotation to weight.


Overview:


In this lab, we are using a meter stick balanced at the center of mass of the meter stick and hanging a mass on one side, we used a meter stick, because it is easy to determine r and L. We use a force meter to determine F  as we pull on the force meter to put the meter stick in equilibrium. First, we keep the force meter in a constant r spot, and change θ to see the change in F needed to put the system in equilibrium. Then, we kept θ constant and changed the r of the force meter.

Materials:

-meter stick
-force meter
-fulcrum/balance point
-mass

Data:

Data collected with graphs can be found here

I pulled the graphs so they can be analyzed easier.




Analysis:


From our first graph, we can see as sinθ gets closer to 1, the force needed to put the meter stick in equilibrium decreases. This agrees with the equation \tau = rF\sin \theta\,\!. When sinθ reaches 1, we have reached the minimum force needed to put the meter stick at equilibrium. When sinθ approaches zero, we reach maximum force needed to put the meter stick in equilibrium. This is because when θ = 90, we are pulling with only a vertical force, where as, if we pull when θ = 60, now we have a horizontal and vertical force. The vertical force when θ = 60 is the same as the vertical force when θ = 90, but when we add the horizontal force, we use the Pythagorean Theorem  to get the actual force that we are pulling with.





From our second graph, we can see that as r, or the distance from the rotation point to the force, gets smaller, the force gets larger. This is because as we get closer to the point of rotation its tougher to rotate the object. Imagine if I were to pull right next to the point of rotation. This is practically pulling at the point of rotation, so the meter stick won't be in equilibrium and not even close. 




I work at a furniture store and sometimes we have to build furniture using the allen wrench provided for us. (The spaces are too small to get a drill into). Here is a picture of an allen wrench:

If I were stick the long end into the bolt, as the bolt tightens, if becomes nearly impossible to completely tighten the bolt. This is because the distance from the turning point to the force is really small. It doesn't matter how long I make the allen wrench if my "r" is still the same. When I turn the allen wrench the other way, the short end is now sticking out from the bolt, I can completely tighten the bolt. My torque is much bigger than before because my distance from the turning point is much greater.

Here is a numerical calculation to show this application is true:

Let's say I need a torque of 200N to completely tighten the bolt. If I use the short end to turn the bolt my r is 10cm. The maximum force that I can apply to the short end of the allen wrench is only 1500N. My torque is only 150N and I cannot completely tighten the bolt. However, I use the long end of the allen wrench my r becomes 50cm and my maximum force is still 1500N. Now, my new torque is 750N, which is more than enough to tighten the bolt, in fact, I would probably strip the bolt with this amount of force.

Revision of Lab Report 1 - Air Resistance

Introduction:

In previous labs, we neglected air resistance and found the velocity of falling objects. Now we are going to show how air resistance affects the velocity of a falling object. Air resistance can be found by following:





A is the cross sectional area, C is the drag coefficient, and ρ is the density of air.


We can say that when a falling object is at terminal velocity, then Fair = mg.


Overview:

In this lab, we are dropping coffee filters from a certain height to model air resistance on an object. We chose to use coffee filters because we can stack multiple coffee filters on top of each other so that only the mass will change in our equations. The term ρAC will remain constant throughout the whole experiment. Our goal in this experiment is to show how mass affects the terminal velocity.

Materials:

* 5 of the same coffee filter
* LoggerPro
*Velocity/Position Detector

Data: (Here is a showcase of my awesome HTML skills)

mass of a coffee filter = 1.03g
ρAC is a constant

Here is the data we collected for all filters:

Number
of            Avg.          
Coffee     Terminal       Avg.      Avg.
Filters      Velocity       Height    Time
1 filer .75826m/s 1.2316m 1.674s
2 filters 1.1344m/s 1.2382m 1.23s
3 filters 1.4426m/s 1.218m 1.02s
4 filters 1.8254m/s 1.179m .85s
5 filters 1.959m/s 1.1182m .74s

Uploaded with ImageShack.us
Here is a graph of our data:



If you are unable to see the graph Click Here

The slope of our graph is .768, which is pAC/g. We can use the slope to determine the terminal velocity of any given of inputs or different masses. For instance, if I wanted to know the terminal velocity of 15 coffee filters I would just do, V = sqrt(.768*(15*1.03)) = 3.44m/s

Analysis:


The data is a little messy. We didn't drop from one specific height (even thought we should have). As we can see from our data, as we add more filters the velocity increases. We can compare this to our model:
Fair = .5ρACv^2    and
Fair = mg, so


.5ρACv^2 = mg, and from this model we can calculate velocity


v =  , and from this we can see as the mass increases the velocity will increase and
                        our model agrees with our data

Here is an example:

We drop a 10 kg medicine ball with a diameter of .27m from a tall building and we want to find the terminal velocity. The drag coefficient of a sphere is .47, the density of air is 1.275 kg/m3 , and our cross sectional area is .848 m2


v =  , so V = sqrt((2*10kg*9.8m/s2)/(1.275 kg/m3*.114m2*.47)) = 53 m/s

Now, let's drops a heavier medicine with the same diameter but an increase mass of 5 kg.

              V = sqrt((2*15kg*9.8m/s2)/(1.275 kg/m3*.114m2*.47)) = 65.6 m/s

As you can see, the mass does have an effect on the terminal velocity, if an object has an enormous mass, then it will have an enormous terminal velocity.



Thursday, October 20, 2011

Lab #6 Air Resistance

Introduction:

In previous labs, we neglected air resistance and found the velocity of falling objects. Now we are going to show how air resistance affects the velocity of a falling object. Air resistance can be found by following:





A is the cross sectional area, C is the drag coefficient, and ρ is the density of air.


We can say that when a falling object is at terminal velocity, then Fair = mg.


Overview:

In this lab, we are dropping coffee filters from a certain height to model air resistance on an object. We chose to use coffee filters because we can stack multiple coffee filters on top of each other so that only the mass will change in our equations. The term ρAC will remain constant throughout the whole experiment. Our goal in this experiment is to show how mass affects the terminal velocity.

Materials:

* 5 of the same coffee filter
* LoggerPro
*Velocity/Position Detector

Procedure:

1) Record initial height
2) Drop coffee filter(s)
3) Record the slope near the end of the position vs time graph of the fall (this is our terminal velocity)
4) Record the time of the fall
5) Repeat steps 1-4 about 5 times to obtain a range of data
6) Repeat steps 1-5 for 1,2,3,4,5 coffee filters
7) Record mass of one coffee filter (assume all 5 coffee filters are the same mass)


Data: (Here is a showcase of my awesome HTML skills)

mass of a coffee filter = .00103kg
ρAC is a constant

Here is the data we collected for all filters:

1 Coffee Filter

Terminal
Velocity     Height    Time
.8616m/s 1.298m 1.67s
.6668m/s 1.261m 1.85s
.7242m/s 1.2m 1.65s
.8318m/s 1.208m 1.6s
.7069m/s 1.191m 1.6s
    .084        .046     .105  --  Deviation

2 Coffee Filters

Terminal
Velocity    Height  Time
1.122m/s 1.248m 1.2s
1.036m/s 1.228m 1.2s
1.183m/s 1.218m 1.2s
1.127m/s 1.256m 1.3s
1.204m/s 1.241m 1.25s
    .065        .015     .045  --  Deviation

3 Coffee Filters


Terminal
Velocity    Height  Time
1.49m/s 1.225m .1.05s
1.277m/s 1.216m 1.05s
1.581m/s 1.187m .95s
1.505m/s 1.223m 1s
1.361m/s 1.239m 1.05s
    .122        .019     .045  --  Deviation

4 Coffee Filters


Terminal
Velocity    Height  Time
1.81m/s 1.19m .85s
1.87m/s 1.229m .9s
1.762m/s 1.195m .85s
1.824m/s 1.107m .8s
1.861m/s 1.174m .8s
    .043        .045     .042  --  Deviation

5 Coffee Filters


Terminal
Velocity    Height  Time
1.988m/s 1.144m ..75s
1.88m/s 1.146m ..7s
1.943m/s 1.235m ..9s
2.02m/s 1.068m .7s
1.963m/s .998m .65s
    .052        .09     .096  --  Deviation

Analysis:


The data is a little messy. We didn't drop from one specific height (even thought we should have). As we can see from our data, as we add more filters the velocity increases. We can compare this to our model:
Fair = .5ρACv^2    and
Fair = mg, so


.5ρACv^2 = mg, and from this model we can calculate velocity


v =  , and from this we can see as the mass increases the velocity will increase and
                        our model agrees with our data