#9 relationship between angular speed and angle of the string above the vertical for a particular rotating apparatus.

Relationship between angular speed (ω) and angle of
the string above the vertical (θ) for a particular rotating
apparatus.

introduction:

After we do the activity about the relationship between the angle speed and the centripetal acceleration, we are able to find the relationship between the angle and the angle speed by the function and define. 

Setting:

give the unknown 

Thesis:

the total R we have is r=d+lsinθ

Tension*cosθ=mg

Tension*sinθ=ma(centripetal acceleration)

centripetal acceleration=g tanθ=w^2*r=w^2(d+lsinθ)

we can find that w=[gtanθ/(d+lsinθ)]^1/2

use this formula to find the e=relationship

and compare with the w we find from time

Date and calculate

I use Excel to help me calculate those date and this is what I got 

 I make the graph

Conclusion: 

from my graph, we can say that the w from T is close to the w we found by angle are almost the same.

it means that we prove that  the w^2r = centripetal acceleration

#8 Centripetal acceleration as a function of angular speed

Centripetal acceleration as a function of angular speed 

introduction:

We know that the whirling object have centripetal acceleration, but we can really see it.

the activity we done in class is trying to find the relationship between angular speed and centripetal acceleration.

 Material

Date:

Those are 6 date that we find in class 

we use the timer to measure the time it row 5 rounds that is 5T. By this way we can find more exactly what we want to get

Using angular speed w=2 π/T

then we got w and a

finally we can draw the graph of the relationship between angular speed and centripetal acceleration.

result

Because of the slope is mustily like a straight line, we can say that the relationship between angular speed and centripetal acceleration fit the ac=w^2 *R which is the rule we always use to find the angler speed. 

#6 Determining density of cylinders/mass of suspended object

Determining density of cylinders/mass of suspended object

introduction:

After we learn how to read vernier caliper, we try to measure the cylinders diameter and height. Then measure the mass to find the density. Knowing that every thing we measure have error, we estimates their error and calculate it to find the error of density.

Next, we use two spring scale to determination of an unknown. 

those are the materials we have in this lab
















Thesis:
(1)We estimate the error of mass are + or - 0.1g and the height and diameter are + or -0.01cm.
collect the date and use the method(I will tell in calculate part) to calculate the density. 
(2)find the vertical and horizontal component to find the mass the error estimate of degree and force are + or - 0.1

Calculate:

(1)

these are the date we measure

to find the density(P) error we all of the estimate error then find it by three different way

1.
∂P ∂P ∂P dP = —— dm + —— dr + —— dh ∂m ∂r ∂h
find every single number and fill in

2.

∂m ∂r ∂h dP/p = —— + —— + —— m r h

this on is what we use. First if all, calculate the density without the estimate error
P=mass/volume = 4m/πhd^2      p=8.8126   ,   2.7406    ,   7.7417 g/cm^3
the 1 dp= (0.1/56+2*0.01/1.28+0.01/5) *8.8126= 0.1711
      2 dp=(0.1/27+2*0.01/2.6+0.01/4.9)*2.7406=0.0368
      3 dp=(0.1/58.7+2*0.01/0.99+0.01/9.85)*7.7417=0.1774
3.
using the logarithm method
p= m/πhd^2  using log and take differentials

(2)

the left hand spring got 4N with 51 degree and right hand spring got 3N with 46 degree.

We can find mass= 4*cos51。+3*cos46。= 4.60N

and then 4.6/9.8=0.47(kg)
F= F1*cos51。+F2*cos46。
dF=(0.1/4+0.1/51+0.1/3+0.1/46)*0.47=0.0625*0.47=0.03












result:
(1)the density are 
8.8126±0.1711   ,   2.7406±0.0368    ,   7.7417±0.1774

(2) mass is 0.47±0.03kg