Hayley Glassic
Rotational Motion
Open-ended Lab
Purpose:
1. Determine the acceleration of a block when released from a pulley system.
2. Using the relationship between linear and rotational acceleration, determine the rotational inertia of the pulley.
Materials: Pulley with a block as illustrated to the left. Motion detector below the mass to measure linear acceleration by collecting distance vs. time data.
Procedure:
A pulley of unknown mass will be used in the lab experiment, as shown above. A small mass of 51.5 g is attached to a string; the other end is attached to the pulley and wrapped around it several times. The block is released from rest, and distance vs. time data is collected with a motion detector CBL unit.
Data:
The time t is measured for various heights D and the data are recorded in the following table.
Time (s)
Time^2 (s^2)
Distance (m)
0.28
0.0784
0.380947
0.288
0.082944
0.371507
0.296
0.087616
0.360956
0.304
0.092416
0.350127
0.312
0.097344
0.338466
0.32
0.1024
0.326804
0.328
0.107584
0.314032
0.336
0.112896
0.301259
0.344
0.118336
0.287932
0.352
0.123904
0.273494
0.36
0.1296
0.2585
0.368
0.135424
0.243506
0.376
0.141376
0.227958
0.384
0.147456
0.211576
0.392
0.153664
0.195194
0.4
0.16
0.178257
0.408
0.166464
0.163263
0.416
0.173056
0.142716
0.424
0.179776
0.125224
Data Analysis:
1. What quantities should be graphed in order to best determine the acceleration of the block? Explain your reasoning.
In order to best determine the acceleration of the block, one should graph the distance versus time squared data. After taking the slope of the line, one must multiply the slope by two.
2. Plot the quantities determined in (1), title the graph, label the axes, and calculate the linear acceleration of the block. Use this acceleration and Newton’s second law for linear motion to find the tension in the string.
Linear acceleration:
a=2xslope
a=2(-2.5358)
a=-5.0716 m/s2
3. Derive an expression for the relationship between the linear acceleration of the block and the angular acceleration of the pulley and the tension in the string.
aL=α(r). . . α= aL/r
T-mg=m(-a)
T=mg-ma
4. Calculate the rotational inertia of the pulley.
I=mr2
∑t=I α
Txr= I α
I=Tr/ α
T= (.0515kg)(-9.81m/s2)+ (.0515kg)(5.0716 m/s2)
T=.2438N
I=Tr/ α
I=(.2438N)(.0365m)/ (5.0716/.0365m)
I= 6.4043 x 10-5 kg● m2
5. Is your answer reasonable? Why or why not?
Yes the answer is reasonable because the pulley has very little mass and is extremely small. Since the moment of inertia depends on the torque and radius, and the pulley is extremely small, the answer is reasonable. It has very small resistance to rolling when the mass is dropped from the initial height.
Rotational Motion
Open-ended Lab
Purpose:
1. Determine the acceleration of a block when released from a pulley system.
2. Using the relationship between linear and rotational acceleration, determine the rotational inertia of the pulley.
Materials: Pulley with a block as illustrated to the left. Motion detector below the mass to measure linear acceleration by collecting distance vs. time data.
Procedure:
A pulley of unknown mass will be used in the lab experiment, as shown above. A small mass of 51.5 g is attached to a string; the other end is attached to the pulley and wrapped around it several times. The block is released from rest, and distance vs. time data is collected with a motion detector CBL unit.
Data:
The time t is measured for various heights D and the data are recorded in the following table.
Time (s)
Time^2 (s^2)
Distance (m)
0.28
0.0784
0.380947
0.288
0.082944
0.371507
0.296
0.087616
0.360956
0.304
0.092416
0.350127
0.312
0.097344
0.338466
0.32
0.1024
0.326804
0.328
0.107584
0.314032
0.336
0.112896
0.301259
0.344
0.118336
0.287932
0.352
0.123904
0.273494
0.36
0.1296
0.2585
0.368
0.135424
0.243506
0.376
0.141376
0.227958
0.384
0.147456
0.211576
0.392
0.153664
0.195194
0.4
0.16
0.178257
0.408
0.166464
0.163263
0.416
0.173056
0.142716
0.424
0.179776
0.125224
Data Analysis:
1. What quantities should be graphed in order to best determine the acceleration of the block? Explain your reasoning.
In order to best determine the acceleration of the block, one should graph the distance versus time squared data. After taking the slope of the line, one must multiply the slope by two.
2. Plot the quantities determined in (1), title the graph, label the axes, and calculate the linear acceleration of the block. Use this acceleration and Newton’s second law for linear motion to find the tension in the string.
Linear acceleration:
a=2xslope
a=2(-2.5358)
a=-5.0716 m/s2
3. Derive an expression for the relationship between the linear acceleration of the block and the angular acceleration of the pulley and the tension in the string.
aL=α(r). . . α= aL/r
T-mg=m(-a)
T=mg-ma
4. Calculate the rotational inertia of the pulley.
I=mr2
∑t=I α
Txr= I α
I=Tr/ α
T= (.0515kg)(-9.81m/s2)+ (.0515kg)(5.0716 m/s2)
T=.2438N
I=Tr/ α
I=(.2438N)(.0365m)/ (5.0716/.0365m)
I= 6.4043 x 10-5 kg● m2
5. Is your answer reasonable? Why or why not?
Yes the answer is reasonable because the pulley has very little mass and is extremely small. Since the moment of inertia depends on the torque and radius, and the pulley is extremely small, the answer is reasonable. It has very small resistance to rolling when the mass is dropped from the initial height.
| rotation_motion_lab_i_of_a_pulleyhayley.doc |
