Sparker free fall and Experiment Uncertainty Lab

On this lab, we handle determination of g(and learning a bit about Excel) and some statistics for analyzing data

This is a Excel about our experiment's date. the time is increasing 1/60 each time and the distances are what we measure.

And this picture is show what the date classmates found.

As these dates, we get nine different value of g, then we plus all of them and divide nine, then we get the average of g is 956.03. Then we use each value of g to minus it. then we get the deviation from the mean value. After of these, we use this function

Then we use the square of every mean value and plus them all and divide the number how many dates we get, then square root it. Now we get the value of standard deviation of mean, and the correct value of g will be in the area of 956.03+/-20.1234523.

Using Calipers/Micrometer to measure the density ofmetalcylinder;Determintion of an unknow mass

This Lad we try to handle two things, the first one is about the density, the other is about the unknow mass.

In this case, we measure three cylinders diameter and the height. The first one is copper. The second is steel. The last one is aluminum. In my date, i just set the code which is C, S , and O.

According my date, the height are 5cm+/-.01, 5cm+/-0.1, and 4.8cm+/-0.1, The mass are 58.5g, 62.0g, and 21.2g. The diameter are 1.25cm+/-0.1, 1.4cm+/-0.1, and 1.4cm+/-0.1, so we can know the radius are 0.625cm+/-0.05, 0.7cm+/-0.05, and 0.7cm+/-0.05.

As we know, density equal the mass over volume, ρ=m/s. So if we get each mass and volume, we will find the density. Then we get each density are 9.53*10^3kg/m^3, 8.05*10^3kg/m^3, and 2.869*10^3kg/m^3.

As this step, we use equation

we can find what the density will be changed if we change the mass, diameter, and height.

On the second experiment, we try to take some notes about the force and degree with the unknown mass. In the first picture, the F(1)left is 8N; the F(2)right is 6N; the theta(1)left is 48 degree and the theta(2)right is 24 degree.

On the second picture, the F(1) is 5N; the F(2) is 6025N; the theta(1) is 23 degree, and the theta(2) is 47 degree.

At the last, we find what the mass is and what the mass will be changed, when the two forces and angles change.

Modeling the fall of an object falling with air resistance.

Today, we try to handle the problem about fall of an object falling with air resistance, at the first, the equation of resistance is that F(resistance)=kv^n.

This picture my partner and I just text the system. We take a video about the coffee filters, then set the length of the part of falling, and point the coffee filters allow time by time. Then the logger pro with the computer will calculate the function. 

About these five function, we fallow our professor to go to the technology building and make the lab of falling objects. We take some videos and notes about the coffee filters which number are one, two, three, four, and five. Each filter's mass is 0.000926 kg. The acceleration is constant that is g=9.8m/s^2, Then there velocity should be same if there are friction-less about air. However, if we care about the friction about air, the velocity is different. According these picture, we know the velocity is 0.835m/s(one coffee filter), 0.1269m/s(two coffee filters), 0.1379m/s(3), 0.2080m/s(4), and 0.2320m/s(5).

Because the third experiment is a big eror, then we cancel that.

After we make the all experiments, we use logger pro to find the graph about the F(resistance). If we set the function Y=aX^b, Y is the F(resistance);X is the velocity; a is the k; b is the n.

As this graph, we know the k=1.338+/-0.09307, and the n=1.543+/-0.09163. (At this picture, we already calculate the mass to be the force.)

Then we set all case which we know and we find that the velocity and position is increasing but become slow; △v, △x and acceleration are decreasing and become to zero.

Modeling Friction Forces

Today, we do this lab because we want to find the friction and the the relation of friction. We use four steps to find the static friction, the kinetic friction static friction with angle and the kinetic friction from sliding block down and incline.

In this case, my partner and I put water into the cup and pull the block which we let slide the surface of table, If the static friction is bigger than the pulling force, the block will not move, because we all know the maximal static friction is always the constant friction, so there is the maximal static friction when the block suddenly moves. We put the different weight of block on the table: the original and add others, we get different water's weight. If we set the pulley is friction-less, the weight of water is the static friction when the block suddenly move. As the coefficient of static friction equation what I show on the picture, we can get the answer that are o.3697, 0.3358, 0.3139, 0.2997.

On this case,we try to find the kinetic friction, so we pull the block with the constant speed. because the constant speed, the acceleration is zero, it means the pulling force is the kinetic friction because function F-F(k)=ma. By the way, the blocks' mass is same as the blocks of static friction. We get the coefficient of kinetic friction are 0.337, 0.3427, 0.286 and 0.282.

This step show how to find the coefficient of kinetic friction. We find that if the theta is 29 degree, the speed of the slide is constant, then the F=f(k), then the  is tan(theta) is 0.38.

on this case we use a 0.1 kg gram to pull the block on the table, if we set the pulley is friction-less, the acceleration will be a= (Mg-mg )/(M+m). we know the  =0.38, M=0.1 and m=0.1322. Then we get the acceleration is 2.065.

 

24-Feb-2015:Find a relationship between mass and period for an inertialbalance.

In this lab, my partner and I try to find a relationship between mass and period.

These were our tools which used to handle our purpose. We put different mass on the inertial balance. And tested the period about different mass.

We put 0, 100g, 200g, ...800g mass marker and the tape which is 152g, and the black staff which is 176g on the inertail balance. we get some different period.

There are two unknow things.

This is the graph of mass and periof.

We are going to guess that the period is related to mass by some power-law type of equation:

T=A(m+Mtray)^n

If we take the natural logarithm of each side, we will get the other function.

lnT=nln(m+Mtray)+lnA. This function looks like y=mx+b.

On this case, this is the graph of ln which is about the mass and period.

This picture is about that we try to find the mass of tray. we try to focus to find a line which is in the centre between each point.

In this last picture, comparing the lnT=nln(m+Mtray)+lnA and y=mx+b, we know the lnT is period, m is 0.6675 as same as n, and lnA is -5.050 as same as b. Then we get the A is 0.0064 because lnA=-5.05. Then we use this function to find the mass of tray. And we get that is 316.00g.

Finally, we find the whole equation that is lnT=0.6675(m+316.00g)-5.050 which is from T=0.0064(m+316.00g)^0.6675.

Finally, we put the periods which are unknow into the function, we get the two function:

0.38=0.0064(m+316g)^0.6675 or ln0.38=0.6675ln(m+316g)-5.05, and

0.40=0.0064(m+316g)^0.6675 or ln0.40=0.6675ln(m+316g)-5.05.

Then we get the mass of tape is 136g and the black staff is 174g. Maybe the friction, the tape's mass has a big error, but the black staff's is really closed.