So should i do something like:
AM(student-initial) + AM(wheel-initial)=AM(student-final)+AM(wheel-final)
and then solve it for the AM of the student-final.
so it would be:
3.6*0 +.37*14 = 3.6*x+.37*-14
and solve for the final angular velocity.
x = 2.89 rad/s
Is this correct...
The Angular Momentum of the stool and student must change to make the equation still valid, since Angular Momentum is conserved.
So to calculate the Angular Momentum of the wheel:
AM(wheel) = .37*(-14) = -5.18
so then:
AM(wheel) + AM(Student) = constant.
-5.18 + 3.6*ω(student) =...
Sorry, I was a bit confused while writing my previous post. I mean does the moment of inertia for the student and the stool matter?
So if I flip over the bicycle wheel, its angular velocity should become negative, right? (relative to the system, its now spinning in the other direction)...
If I treat this system as two parts, then the angular momentum is just that of the wheel in the beginning, .37 kg*m^2. So does the angular momentum become negative if we flip the wheel on its z axis, becoming -.37kg*m^2?
this would mean that the angular velocity of the system would flip...
Homework Statement
In a demonstration, a bicycle wheel with moment of inertia = .37 kg*m^2 is spun up to 14 rad/s, rotating about a vertical axis. A student hold the wheel while sitting on a rotatable stool. The student and the stool are initially stationary and have a moment of inertia equal...
Hello everyone!
I am new here, so I am not sure how things normally work.
anyways, on the the problem.
I have a ramp made out of channeling, approximately 8 foot long, that i am going to roll spherical objects down. i want to find out the theoretical time it will take. friction doesn't...