As shown by the equation of rotational motion: θ = θ0 + ω0t + 0.5αt², the greater the initial angular velocity (ω0), the greater the angle/angular displacement. But does it affect the time? Well, if we use the second equation:
ω = ω0 + αt
ω = 0 + αt
t = ω/α
This equation shows that the initial angular velocity is proportional to the time the fidget spinner stays in motion.
We can double check this by using my code: I can modify the initial angular velocity and see how long it stays in motion.
By default, it uses the initial angular velocity of the video: 10.272 rad/sec.
This value resulted in it staying in motion for: 34.717 seconds
However, after testing out two other values, I got these three results.
w0 = 5.136 -> t = 17.367 seconds
w0= 10.272 -> t = 34.717 seconds
w0 = 20.544 -> t = 69.417 seconds
From these results, we can see that if we half the initial angular velocity, the time also halves. If we double the initial angular velocity, the time also doubles. Thus the initial angular velocity is proportional to the time the fidget spinner stays in motion.
For my specific fidget spinner, we found the inertia to be 0.000579 kg*m². If the fidget spinner was instead a disc, we can find the inertia using 0.5mr². The fidget spinner has a radius of 0.0453 m and a mass of 0.054 kg. Thus if the fidget spinner was in the shape of a disc, its inertia would be 0.5*(0.054)*(0.0453)² = 0.00005540643 kg*m². This value is smaller than the actual inertia, which means that the inertia of a fidget spinner is greater than the inertia of a disc. This is most likely because most of the mass of the fidget spinner is on the outside (the three rings). Whereas a disc is spread evenly throughout. When more mass is on the outside, it is more difficult to rotate, and thus its inertia is also higher. This concept can be seen when comparing the inertia of a disc (0.5mr²) to the inertia of a hoop (mr²). The hoop has all of its mass on the very outside rim, whereas the disc does not, causing the disc to have a significantly lower inertia.
I chose to combine my answer to these two questions, because my answers are closely related.
In the video analysis of video 1, I calculated the angular momentum using a calculated column in logger pro. From the analysis, I found that the angular momentum of the fidget spinner decreases over time. Initially, it was around 0.004 kg*m²/s, however by the time it stopped rotating, the angular momentum was 0. This makes sense as L = 𝐼 ω. When it stops rotating, the angular velocity is 0, and thus the angular momentum is also 0.
This is also shown by the impulse being equal to the starting momentum, meaning that the angular momentum must have ended with a value of 0.
In most of the problems we do in class, angular momentum is conserved in the system, however in this case, it isn't. The reason angular momentum is not conserved in this case is because of the torque created by the friction between the fidget spinner and its inner bearings. As the spinner rotates, it continues to stay in contact with the bearings. Although these bearings are designed to minimize friction, the friction is still there, creating a negative torque that slows down the fidget spinner. This torque of friction is what causes the fidget spinner to slow down and not spin forever.