Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
<span>
Of course. Wind is air in motion, and the gases in air are composed of
all the usual familiar stuff ... atoms, molecules, mass, etc. That's how
the wind moves things ... it has momentum and kinetic energy, which
get transferred to the things that move in the wind.</span>
I don't completely understand your drawing, although I can see that you certainly
did put a lot of effort into making it. But calculating the moment is easy, and we
can get along without the drawing.
Each separate weight has a 'moment'.
The moment of each weight is:
(the weight of it) x (its distance from the pivot/fulcrum) .
That's all there is to a 'moment'.
The lever (or the see-saw) is balanced when (the sum of all the moments
on one side) is equal to (the sum of the moments on the other side).
That's why when you're on the see-saw with a little kid, the little kid has to sit
farther away from the pivot than you do. The kid has less weight than you do,
so he needs more distance in order for his moment to be equal to yours.
Answer:
the moment of inertia with the arms extended is Io and when the arms are lowered the moment
I₀/I > 1 ⇒ w > w₀
Explanation:
The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,
L₀ = L_f
I₀ w₀ = I w
w =
w₀
where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms
I₀ = I_cylinder + 2 m r²
where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.
If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be
I <I₀
I₀/I > 1 ⇒ w > w₀
therefore the angular velocity (rotations) must increase
in this way the skater can adjust his spin speed to the musician.
