Conservation of Angular Momentum Experiment: Moment of Inertia

AI Thread Summary
The experiment involves a rotating solid disk with a hollow cylinder dropped at its center to test angular momentum conservation. The moment of inertia for both the disk and the hollow cylinder is calculated under the assumption that the cylinder is centered. When the cylinder is placed off-center, the parallel axis theorem should be applied to adjust the moment of inertia. This theorem allows for the calculation of the new moment of inertia by adding the product of the cylinder's mass and the square of the distance from the center of the disk to the new axis of rotation. Understanding this adjustment is crucial for accurately determining the final moment of inertia and calculating percent error.
chrismoon
Messages
1
Reaction score
0

Homework Statement


I did a lab where there was a rotating solid disk with mass= 0.915kg and diameter=0.253m.
This was rotating horizontally with an initial angular velocity of 3 different values ω radians/second. After recording the initial angular velocity, I dropped a thin-walled hollow cylinder with mass=0.708kg and diameter=0.125m in the center and measured the final angular velocity, testing the conservation of angular momentum.

Issue: If I placed the ring off center of the disk by say, 1cm (0.01m), how will that affect my moment of inertia?


Homework Equations


Idisk=(1/2)(Mass)(radius)2
Ihoop/hollow cylinder=(Mass)(radius)2
Li=Idiskωdisk initial
Lf=(Idisk+Ihoop/hollow cylindercombined final

The Attempt at a Solution


First, I calculated the moments of inertia-
Idisk=(1/2)(Mass)(radius)2=(1/2)(0.915kg)(0.253m/2)2=0.00732kgm2

Ihoop/hollow cylinder=(Mass)(radius)2=(0.708kg)(0.125m/2)2=0.00277kgm2

Icombined=(0.00732kgm2)+(0.00277kgm2)=0.01009kgm2

The Icombined is for the ideal situation of the ring being completely centered, but I have no idea what I would do to get the experimentally flawed moment of inertia. Would I just change the radius of the hoop/cylinder by 1cm? If so, would I add or subtract? I'm really not sure how I'd calculate it. I understand this all generally pretty well, but executing this has me a little stumped. I need a way to get the new final moment of inertia instead of the ideal (Idisk+Ihoop/hollow cylinder) to calculate a percent error.
 
Physics news on Phys.org
Hi, chrismoon. Have you studied the "parallel axis theorem"? You can use it to calculate the moment of inertia of the hollow cylinder when it is placed off-center on the disk.
 
have you used the parallel axis theorem?

ok then...

late again, sorry. Its getting strange how I am right behind Tnsy.

I will now bow out again.
 
Last edited:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top