Apperetus: A narrow metal ring and a semimcircular plate.
Model: τ=αI
Purpose: Derive expressions for the period of various physical pendulums. Verify your predicted periods by experiment.
Process: Part A:
First, we detemine the outer radius and inner radius, router=0.1465 m, rinner=0.1372 m. Because the ring is thin enough, we assume that the point of pivot is exactly half way between the inner and outer radii and that the notch, which is cut out at the top of the ring, does not impact how uniform the mass distribution is. We hang the ring and set up the sensor. We raise the ring with a small angle, release it, and use the LoggerPro to collect date. By the data, we can get the period.
We get the exprimental period, T=1.06977188 s.
Then, we need to calculate out the theoretical period:
We assume the radius of the ring is the average of the inner radius and outer radius.
R=router+rinner=0.14185 m
We already calculate the equation of the period in the notebook
T=2π(2R/g)^0.5
We get the theoretical period:
T=1.0690 s
We compare them:
Experimental T=1.06977188 s
Theoretical T=1.0690 s
0.7% off
Part B:
Now, we have a semicircular disk of radius R, R=0.1445 m.
We hang the semicirular disk as the picture. We raise the semicirular with a small angle, release it, and use the LoggerPro to collect date. By the data, we can get the period.
We get the exprimental period, T=0.83722056 s.
Then, we need to calculate out the theoretical period:
τ=αI
sinθ*mg*(4/3π)R=α*(1/2)mR^2
α=(8g/3πR)sinθ
When the angle is small enoug, we can assume θ≈sinθ.
α=(8g/3πR)sinθ≈(8g/3πR)θ
ω=(8g/3πR)^0.5
T=2π/ω=2π(3πR/8g)^0.5
We get the theoretical period:
T=0.82812 s
We compare them:
Experimental T=0.83722056 s
Theoretical T=0.82812 s
1.3% off
We hang the semicirular disk as the picture. We raise the semicirular with a small angle, release it, and use the LoggerPro to collect date. By the data, we can get the period.
We get the exprimental period, T=0.81718602 s.
Then, we need to calculate out the theoretical period:
τ=αI
sinθ*mg*(1-4/3π)R=α*(1/2)mR^2
α=[(6π-8)g/(9π-16)R]sinθ
When the angle is small enoug, we can assume θ≈sinθ.
α=[(6π-8)g/(9π-16)R]sinθ≈[(6π-8)g/(9π-16)R]θ
ω=[(6π-8)g/(9π-16)R]^0.5
T=2π/ω=2π[(9π-16)R/(6π-8)g]^0.5
We get the theoretical period:
T=0.811509744 s
We compare them:
Experimental T=0.81718602 s
Theoretical T=0.811509744 s
0.85% off
Apparatus: A mass, several spring, a motion detector.
Model:
Process: We have 5 different spring, so we divided into five group. Each group use one spring to collect data. For Kinetic energy purposes one-third of the spring's mass contributes to the KE of the mass-spring system. This is also true as far as contributing to the oscillating mass in the system when you go to determine the period. The effective mass of the spring is (1/3)mspring.
Therefore, we get:
Oscillating mass=hanging mass + hook mass(3.7 g) + (1/3)mass of spring
First, we are going to find the spring constant of our spring. We hang the spring, and we measure the distance from the end of the spring to the ground, d1=0.37 cm. Then, we hang a 100g weight on the spring and measure the distance from the end of the spring to the ground, d2=0.52 cm. By using:
F=Δxk
Δx=d2-d1
k=6.53 N/m
Now, we want to let the oscillating mass be 115g and find out the period. We measure the mass of the spring with the hook on it. Mspring+hook=11.3g. By calculation:
Mspring=Mspring+hook-Mhook=7.6 g
Mhang=Moscillating-(1/3)Mspring-Mhook=115-2.533-3.7=109g
We set the apparatus by hanging the spring and place the motion detector right below the spring. We hang 109 g mass on the spring, pull down the spring a little bit, release it, and start collect the data.
We get the period from the graph:
T=0.735 s
By sharing the data between the group, we get:
We plug in these number into the LoggerPro. Basing on the equation:
We try to linear fit by equation: T=Ak^-0.5, we get:
From the graph, the constant of the equation A=2.111. In theory,
The theoretical A=2.1307
The experimental A=2.111
1% off
Now, we use our spring and collect different period hanging different mass. We get:
We use Mhang+(1/3)Mspring+Mhook=Moscillating to find out the actual oscillating mass. We plug these data into the LoggerPro. Basing on the equation:
We try to linear fit by equation: T=Am^0.5, we get:
From the graph, the constant of the equation A=2.478. In theory,
The theoretical A=2.4588
The experimental A=2.478
1% off
Apparatus: A ramp, a ball, and a ball-catcher.
Purpose: To investigate the conservation of angular momentum about a point that is external to a rolling ball.
Process: First, we measure the ball's mass and diameter. m=0.0289 kg, D=0.019 m.
In order to determine the horizontal velocity of the ball as it rolls off the end of the ramp, we place the ramp on the edge of a table. We release the ball at a marked point, and we use the carbon paper to mark the landing point of the ball. Then, we measure the distances L and high h. L=0.595 m, h=0.955 m.
We use:
(gt^2)/2=h
Vx*t=L
Vx=1.348 m/s
We set up the rotating disk by using the aluminum top disk, and mount the ball-catcher on top of the small torque pulley using a gray-capped thumbscrew. We hang a mass, m=0.0246 kg, on the pulley, rp=0.02495 m, rotate the disk to raise the mass to the top, and release it. We start the sensor and collect the angular acceleration. |αup|=5.92 rad/s^2, |αdown|=5.27 rad/s^2, |αave|=5.6205 rad/s^2.
We already know the equation for the moment of inertia, and we use it to get the moment:
I=(mgrp)/α-mr^2
I=0.001055 kg*m^2
We poition the ramp perpendicular to the ball-catcher, and the end of the ramp should have same high as the catcher so that the ball can be caught by the catcher. We release the ball in the same mark which is same as the mark in first process.
We place the ramp in two different position so that the ball can be caught by the catcher in different position. We measure the distance from the ball to the center of the rotation, r, and collect the angular velocity of the disk, ω.
① ω=2.269 rad/s r=0.076 m
② ω=1.353 rad/s r=0.042 m
Finally, we want to compare the theoretical angular velocity and the experimental angular velocity. Therefore, we use:
r*m*Vx=Iω+[(3/5)mrb^2+mr^2]
① ω=2.269 rad/s r=0.076 m
Theoretial ω=2.42 rad/s
Experimental ω=2.269 rad/s
6% off
② ω=1.353 rad/s r=0.042 m
Theoretial ω=1.478 rad/s
Experimental ω=1.353 rad/s
8% off
Apparatus: A meter stick and a clay
Purpose: Find out the theoretical maximum high and the experimental high, and compare them.
Model: I=Icm+md^2, Isω1=(Is+Ic)ω2, and ΔKE=ΔGPE
Process: We use an apparatus which can rotate with a small friction as a pivot. We place the pivot at the vary end of the meter stick. We use a support to locate a clay on the flood which can be hit be the meter stick. We wrap some tape around the other end of meter stick and the clay.
We release the meter stick from horizontal. We take a video by phone and use LoggerPro to analyse it. Then, we get:
We can get the experimental high H=0.3541m
Now, we are going to find out the theoretical high. We measure the meter stick and the clay:
Ms=161.2g=0.1612kg, Mc=38.7g=0.0387kg
Then, we calculate the moment of the stick and the clay.
Is=1/12(MsL^2+Msd1^2=0.0513516 kg.m^2
Ic=Mcd2^2=0.0375477 kg.m^2
Isys=Is+Ic=0.0888993 kg.m^2
Then, we use the energy equation:
d1=Δh
MsgΔh=1/2Isω1^2
ω1=5.462663 rad/s
Then, we use the moment equation:
Isω1=(Is+Ic)ω2
ω2=Is/(Is+Ic)ω1=3.155441 rad/s
Then, the energy again:
Hmeterstick=(48.5/98.5)*Hclay
MsgHmeterstick+McgHclay=1/2Isysω2^2
Hclay=H=0.3825 m
Now, we get:
Theoretical H=0.3825 m and Experimental H=0.3541 m
Htheoretical≈Hexperimental
There are 7% off in the experiment. It may cause by the friction on the pivot and by the clay sliding on the flood when it be hit.
Aparatus: A top steel disk, a bottom steel disk, a larger torque pulley, a hanging mass, a right triangular thin plate, and the Pasco rotational sensor.
Purpose: To determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle.
Model: Iparallel axis=Icm + Mdparallel axis displacement^2 and I=(mgr)/α-mr^2
m is the mass of hanging mass, α is the angular acceleration of the disk, and r is the radius of the pulley.
Process: We will collect three group of data in different set up. First, we do not mount the triangle on a holder and disk. Second, we mount the triangle on a holder and disk vertically. Third, we mount the triangle on a holder and disk horizontally. A string is wrapped around a pulley on top of and attached to the disk and goes over a freely-rotation "frictionless" pulley to a hanging mass. The tension in the string exerts a torque on the pulley-disk combination. By measuring the angular acceleration of the system, we can determine the moment of inertia of the system.
Then, you can mount the triangle onto the disk-pulley-holder system and measure α again and determine the I of the new system. The moment of inertia of the triangle is the difference between the moment of inertia from these two measurements.
We determine the mass of the hanging mass and the radius of the pulley, and then we use the equation that we get from last experiment: I=(mgr)/α-mr^2 to find the moment of inertia.
Iempty=0.0027021 kg*m^2
Ivertical=0.0029899 kg*m^2
Ihorizontal=0.0033051 kg*m^2
We already know the theory moment of inertia:
I1=(MB^2)/18=0.000246275 kg*m^2
I2=(MH^2)/18=0.000567324 kg*m^2
According to the parallel axis theorem:
Ivertical-
Iempty=0.0002878 kg*m^2 ≈ 0.000246275 kg*m^2 =I1
Ihorizontal-Iempty=0.000603 kg*m^2 ≈ 0.000567324 kg*m^2 =I2
The the theoty moment of inertia approximately equal to the experiment moment of inertia.
Apparatus: A disk and a frictionless cart.
Purpose: Find out the moment ot inertia of the disk and frictional torque. Calculate the time that the cart should take to slipe down a distance. Compare the theory time and experiment time.
Model: τc-τf=Iα
Process:
The "disk" is make up of two cylinder and a disk; therefore, in order to find out the moment of inertia of the entry "disk", we need to calculate their moment of inertia separately. First, we measure the diameter and the heigh of two cylinder and the the disk. We get:
Then, we use the equation V=πh(d/2)^2 to find out the volume.
V1=39.3381 cm^3
V2=479.9128 cm^3
V3=40.1093 cm^3
Vtotal=V1+V2+V3=559.3602 cm^3
In order to find out the mass of each part, we use V/Vtotal=M/Mtotal. We can get the total mass from the mark on the disk. Mtotal=4.724 kg.
M1=(V1/Vtotal)*Mtotal=0.3322 kg
M2=(V2/Vtotal)*Mtotal=4.053 kg
M3=(V3/Vtotal)*Mtotal=0.3387 kg
Now, we can use calculate out the moment of inertia.
cylinder: I1=1/2M1R1^2=4.0942*10^-5
I2=1/2M2R2^2=2.0366*10^-2
disk: I3=1/2M3R3^2=4.0949*10^-5
Itotal=I1+I2+I3=0.020448 kg*m^2
Then, we need to find out the frictional torque. Before that, we need the angular deceleration. We spin the disk, and we measure the time it take to stop and the number of revolution. We get:
Δθ=9π, Δt=5.35 s, ωf=0 m/s
We use: ω0-αt=ωf
ω0t-(αΔt^2)/2=Δθ
By these two equation: (αΔt^2)/2=Δθ
α=(2Δθ)/(t^2)=1.9757 rad/s
We can get the frictional torque: τf=Itotal*α=0.040399 N*m
Then TR3-τf=Iα
sinθMg-T=Ma
θ=49°
By them, we get a=(sinθMgR3^2-τfR3)/(MR+I)=0.012931 m/s
By (at^2)/2=D, D=1 m
We get the theory time t=12.35 s
Then, we set up the inclind, tie one end of a string on a cylinder and the other end of the string on the frictionless cart which is 0.502 kg. We adjust teh disk and the ramp in order to let the string be parallel to the ramp.
The cart rolls down the inclined track for a distance of 1 meter, and we calculate how long it take for the cart to travel 1 meter from rest. We do it several time and take the average.
t1=12.27 s
t2=12.92 s
t3=12.14 s
taverage=12.44 s
We compare the theory time and experiment time:
theory time=12.35 s≈ 12.44 s =experiment time
There are 0.7% off.
The frictionless cart is not actually frictionless. Therefore, the actual time may be longer than the theory time. It makes sense for our data.
Apparatus: a top steel disk, a bottom steel disk, a top aluminum disk, a smaller torque pulley, a larger torque pulley, some hanging mass, and the Pasco rotational sensor.
Purpose: We want to apply a known torque to an object that can rotate, and measure the angular acceleration. Eventually with this torque and angular acceleration data. We can find a measured value for the moment of inertia.
Model: τ=Iα, τ=TL, I=1/2MR^2
τ is the torque on the disk, I is the moment of inertia, T is the tension, M is the mass of the rotating disk, R is the radius of the rotation disk.
Process: First, we measure the diameter and mass of the top steel disk, the bottom steel disk, the top aluminum disk, the smaller torque pulley, and the larger torque pulley. We measure the mass of the hanging mass.
Then, we plug the the pasco rotational sensor into computer and set up the computer. We only neet to discern which is measuring the top disk. Because there is no defined sensor for this rotational apparatus, we need to create something that work with this equipment. We choose the Rotary Motion, and we set the equation in the sensor setting to 200 counts per rotation because there are 200 marks on the top disk.
We turn on the compressed air so that the disk can rotate separately. We can use the pin to let the top disk rotate only or the top disk and the bottem disk rotate together.
In the experiment 1, 2, and 3: Effect of changing the hanging mass.
In the experiment 1 and 4: Effect of changing the radius and which the hanging mass exerts a torque.
In the experiment 4, 5, and 6: Effect of changing the rotating mass.
For each experiment, we get the graph velocity vs. time as the photo showing:
By looking for the slope of the line, we can get the angular acceleration.
We collect the data which is the mass of hanging mass, the kind of torque pulley, the number and kind of rotating disk, and the angular acceleration of going up and down for each experiment.
After getting all data that we need, we start comparing them by using the model:
τ=Iα
In the comparison group 1, 2, and 3:
We change the hanging mass and keep other factors constant.
We get: m1/m2≈α1/α2
m1/m3≈α1/α3
m2/m3≈α2/α3
In the comparison group 1 and 4:
We change the kind of torque pulley which have different radius and keep other factors constant.
We get: r1/r4≈α1/α4
In the comparison group 4, 5, and 6:
We change the mass of the rotation disk and keep other factors constant.
We get: I4/I5≈α5/α4
I4/I6≈α6/α4
I5/I6≈α6/α5
In comparison group 2 and 4:
We change the mass of the hanging mass and the radius of the torque pulley in the same time and keep other factors constant.
We get: (m2/m4)*(r2/r4)≈α2/α4
The number in the comparison group are not the same because the tension is not equal to the weigh of the hanging mass and there are some friction existing between the disk.
Part 2:
We use our data to determine the moment of inertia of the disks or disk combinations from previous experiments.
Because there is frictional torque in the system. The rotating disk is not trully frictionless and there is some mass in the frictionless pulley which we can approximate as a frictional torque. The angular acceleration of the system when the mass is desecending is not the same as when it is ascending.
We make some assumption in this experiments. The frictional torque in the system is independent of direction of angular velocity. The disk are solid whick without a hole in the center. The moments of inertia of the pulley is 1000 times smaller than the disk, so we choose to ingore it.
Then, we use the model and do some calculation:
Tr=Iα
mg-T=ma
a=αr
By these model:
(Iα)/r=mg-m(αr)
I=(mgr)/α-mr^2
We use I=(mgr)/α-mr^2 to find out the experiment moments of inertia of the disk.
We use I=(M*R^2)/2 to find the theory moments of inertia of the disk.
We compare the theory moments of inertia and experiment moments of inertia:
I1≈Itop steel
I2≈Itop steel
I3≈Itop steel
I4≈Itop steel
I5≈Itop aluminum
I6≈Itop steel+Idown steel
Bass on the assumption that we make, we can say the theory and the experiment are close enough.
