Rotational Motion and Moments of Inertia
March 12 & 19, 2002
Objectives:
To measure an object’s moment of inertia in regards to the distribution of its mass, to then compare this measured moment of inertia with theoretical values, to study the parallel axis theorem and the equations of motion for rotation.
6.5.2 Compensation for Friction
Procedure:
A horizontal wheel is attached to a photogate pulley system by a lightweight string. A weight holder is attached to the other end of the string and allowed to fall freely and rotate the wheel. The motion of the wheel is recorded by the movement of the photogate pulley system using SensorNet. The acceleration of the weight holder without any weight added.
Data & Analysis:
Mass of Weight holder = .005+/-.001kg Acceleration of Weight holder = .056m/s2
Due to low friction in the wheel bearings, the acceleration of the weight holder could not be slowed any further than .056m/s2.
Red – Distance vs. Time Blue – Velocity vs. Time Black – Acceleration vs. Time
6.6.1 Determining Io for the Wheel
Procedure:
A horizontal wheel is attached to a photogate pulley system by a lightweight string. A weight holder is attached to the other end of the string and allowed to fall freely and rotate the wheel. The motion of the wheel is recorded by the movement of the photogate pulley system using SensorNet. The acceleration of the weight holder with a 100g weight added was measured in four trials and moment of inertia for the wheel was calculated.
Data & Analysis:
Total mass minus mass of weight holder = .100kg Radius of Wheel = .199m
Trial Number / Acceleration / Io of Wheel#1 / 1.219+/-.001m/s2 / .576+/-.001kg*m2
#2 / 1.232+/-.001m/s2 / .576+/-.001kg*m2
#3 / 1.183+/-.001m/s2 / .578+/-.001kg*m2
#4 / 1.220+/-.001m/s2 / .576+/-.001kg*m2
Mean Acceleration: a = 1.214m/s2 Standard Deviation of Acceleration: σ = .021 m/s2
Mean Moment of Inertia: Io = .576 kg*m2 Standard Deviation of Moment of Inertia: σ = .001kg*m2
The Moment of Inertia for the wheel was calculated using the formula Io = [mr2(g-a1)]/a2
The acceleration fluctuated very slightly due to slight variation in the release mechanism of the weight holder at the beginning of freefall. The calculated moments of inertia were almost identical though with a very small standard deviation.
6.6.2 “Point” Masses at Various Radii and Symmetry
Procedure:
A horizontal wheel is attached to a photogate pulley system by a lightweight string. A weight holder is attached to the other end of the string and allowed to fall freely and rotate the wheel. The motion of the wheel is recorded by the movement of the photogate pulley system using SensorNet. Cyndrical weights are added at various positions on the wheel. The wheel is spun by the freefall acceleration of the weighthold supporting a 100g weight.
Data & Analysis:
Total mass minus mass of weight holder = .100+/-.001kg
Mass of each cylinder = .225+/-.001kg
Distance from center of wheel to masses in Figures 6.5,6.6, and 6.7 =.129+/-.001m
6.5 / .828+/-.001m/s2 / .602+/-.001kg*m2 / .026+/-.001kg*m2
6.6 / .822+/-.001m/s2 / .603+/-.001kg*m2 / .027+/-.001kg*m2
6.7 / .812+/-.001m/s2 / .603+/-.001kg*m2 / .027+/-.001kg*m2
6.8 / 1.280+/-.001m/s2 / .527+/-.001kg*m2 / -.004+/-.001kg*m2
For all setups, the actual moment of inertia for the objects was calculated by taking the total moment of inertia and subtracting the moment of inertia of the wheel.
I = Io + Im
Figure Type / The theoretical moment of inertia for the object (Im) / The difference between Im and actual moment of inertia6.5 / .015 kg*m2 / .011 kg*m2
6.6 / .015 kg*m2 / .012 kg*m2
6.7 / .015 kg*m2 / .012 kg*m2
6.8 / .000 kg*m2 / .004 kg*m2
The theoretical moments of inertia were calculated using the formula Im = Σmiri2
This formula assumes that the masses are point masses and not cylinders like they are in reality. Therefore, differences arise between the theoretical and actual moments of inertia due to this assumption. If the geometry of the masses was taken into account when calculating moments of inertia, the theoretical and actual results would be extremely similar.
6.6.3 PVC Ring – Centered
Procedure:
A horizontal wheel is attached to a photogate pulley system by a lightweight string. A weight holder is attached to the other end of the string and allowed to fall freely and rotate the wheel. The motion of the wheel is recorded by the movement of the photogate pulley system using SensorNet. A PVC ring is centered on the wheel. The wheel is spun by the freefall acceleration of the weighthold supporting a 100g weight.
Data & Analysis:
Mass of PVC ring = .889+/-.001kg
Distance from center or wheel to outside of ring = .129+/-.001m
Acceleration of the system: a = .860m/s2
The theoretical moment of inertia for the ring (Im = Σmiri2) was .015 kg*m2.
The actual moment of inertia for the ring was:
I = Io + Im
Im = .600-.576 = .024 kg*m2.
Although the radius of the PVC ring and the cylinders from 6.6.2 was the same, the mass of the PVC was only 89% as large as the four cylinders combined. However, the difference in the actual and theoretical moments of inertia for the PVC ring and the cylinders was very similar. Therefore, there was larger error in the trial using the PVC ring than in the trials using cylinders. The error may have been due to the cylinder having a width of several millimeters and not being extremely thin as is assumed in the moment of inertia calculations.
6.6.4 Rectangular Plastic Plate
Procedure:
A horizontal wheel is attached to a photogate pulley system by a lightweight string. A weight holder is attached to the other end of the string and allowed to fall freely and rotate the wheel. The motion of the wheel is recorded by the movement of the photogate pulley system using SensorNet. A plastic rectangular plate is centered on the wheel. The wheel is spun by the freefall acceleration of the weight holder supporting a 100g weight.
Data & Analysis:
Mass of the rectangular plastic plate = .830+/-.001kg
The plate is square with a side length of .304+/-.001m
Acceleration of the system: a = .858m/s2
The theoretical moment of inertia for the plate was .013 kg*m2.
The actual moment of inertia for the plate was:
I = Io + Im
Im = .600-.576 = .024 kg*m2.
There was a larger difference between the actual and theoretical yield than in any of the other previous experiments. Like before, error may have occurred due to the plate being slightly off center and changing the distribution of mass on the wheel.
6.6.5 The Parallel Axis Theorem
Procedure:
A horizontal wheel is attached to a photogate pulley system by a lightweight string. A weight holder is attached to the other end of the string and allowed to fall freely and rotate the wheel. The motion of the wheel is recorded by the movement of the photogate pulley system using SensorNet. A PVC ring is placed on the wheel so that it is off center. The system is accelerated by the free falling weight holder supporting a 100g weight
Data & Analysis:
Mass of the PVC ring was = .889+/-.001kg
The distance between the center of the ring and the center of the wheel = .072+/-.001m
Acceleration of the system: a = .794 m/s2
The theoretical moment of inertia for the ring was .029 kg*m2.
The actual moment of inertia for the ring was:
Id = Ic + Md2
Id = .604-.576 = .028 kg*m2.
The theoretical and actual moments of inertia were almost identical.
Questions:
1. (b) The car, if experiencing a centripetal force, must be traveling in a circular motion. Therefore, it is going around a curve. If the car is also experiencing a tangential acceleration, it is either slowing down or speeding up its linear motion. Therefore, choice “b” is the best of all five.
2. An angular acceleration of α is produced. If the force is of the same magnitude, the angular acceleration will be the same at all points on the disk regardless of the distance from the center of the disk. It is the linear acceleration that changes with distance from the center.
3. The bicycle wheel has the greatest moment of inertia and the solid sphere has the least moment of inertia. All three shapes have the same equation for moment of inertia (mr2), but the coefficient in front of the variables is different. The solid sphere has the smallest coefficient (2/5), the hollow sphere has the next greatest (2/3), and the bicycle wheel has the largest coefficient (1).
6. The angular velocity at the end of the time equals ωf = 2.857 rad/s.
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