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Hello, So I am doing a physics lab using an Atwood machine to detirmine how vali

ID: 1513003 • Letter: H

Question

Hello,

So I am doing a physics lab using an Atwood machine to detirmine how valid the equation (m1 m2)g=(m1 +m2 + I/R^2 )a is. I created an Atwood pulley system. The system is accelerated due to the difference in weight on both sides of the pulley. After completing all the time trails for various weights I was asked to calculate acceleration using first a = [(m1 – m2)g] / m1 + m2 and then linearize to find slope and y intercept.

Iwas then asked to use the averaging time for each trial and calculate a based on the equation h = y - y0 = 0.5at^2, then graph [m1 – m2)g.

My equation from the linear fit was y=0.002963x +(-0.2186)

The lab then asks " From the slope value, calculate the moment of inertia (I) of the pulley. " and I don't understand how I am supposed to do that. The radius for R is unknown, but I measured the pulley and it was 2.25 cm.

I don't know how to find the inertia but here is the data from the lab, if anyone can help and explain how you did it I would really appreciate it!!

n

m1

m2

t1

t2

t3

t4

t5

t

a

(m1 m2)g

6

128.56

115.35

3.44

2.89

2.72

2.69

3.24

2.99±0.45

0.195

(13.21)(9.8)=129.5

5

130.31

113.6

2.25

2.43

2.69

2.16

2.56

2.42±0.4

0.297

163.76

4

132.06

111.86

2.03

2.44

2.58

2.34

2.10

2.30±0.48

0.3289

197.96

3

133.81

110.1

2.19

1.74

1.94

2.10

1.91

1.98±0.28

0.444

233.34

2

135.56

108.36

1.84

1.79

1.82

1.81

1.68

1.79±0.11

0.544

266.56

1

137.31

106.6

1.72

1.60

1.75

1.66

1.51

1.65±0.14

0.639

300.96

0

139.06

104.85

1.41

1.47

1.44

1.37

1.50

1.44±0.07

0.844

335.26

n

m1

m2

t1

t2

t3

t4

t5

t

a

(m1 m2)g

6

128.56

115.35

3.44

2.89

2.72

2.69

3.24

2.99±0.45

0.195

(13.21)(9.8)=129.5

5

130.31

113.6

2.25

2.43

2.69

2.16

2.56

2.42±0.4

0.297

163.76

4

132.06

111.86

2.03

2.44

2.58

2.34

2.10

2.30±0.48

0.3289

197.96

3

133.81

110.1

2.19

1.74

1.94

2.10

1.91

1.98±0.28

0.444

233.34

2

135.56

108.36

1.84

1.79

1.82

1.81

1.68

1.79±0.11

0.544

266.56

1

137.31

106.6

1.72

1.60

1.75

1.66

1.51

1.65±0.14

0.639

300.96

0

139.06

104.85

1.41

1.47

1.44

1.37

1.50

1.44±0.07

0.844

335.26

Explanation / Answer

y = m x + b

The second term on the right represents the constant b (it has negative sign). The first term on the right represents mx, where m represents the constant g/(m1 + m2 + meq) and x represents (m2 - m1). y represents at, of course. Now, when you make the graph, you'll get a nearly linear relation with slope s and y intercept b. Plot accelerations on the "y" axis, and the mass difference on the "x" axis. Then the slope of the line, s, allows you to calculate meq, since you know all of the other constants. The y intercept (where x = 0; no acceleration) then allows you to calculate mf( mass of pulley). (This value should compare well with the amount of mass unbalance which you experimentally found would just initiate acceleration of the system.

now, you have radius of pulley is there mass you have calculated i.e b

use MR^2 to find moment of pulley

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