A person stands on a platform, initially at rest, that can rotate freely without friction. The moment of inertia of the person plus the platform is IP. The person holds a spinning bicycle wheel with its axis horizontal. The wheel has moment of inertia IW and angular velocity ωW. Take the ωW direction counterclockwise when viewed from above.Part A

What will be the angular velocity ωP of the platform if the person moves the axis of the wheel so that it points vertically upward?Part B

What will be the angular velocity ωP of the platform if the person moves the axis of the wheel so that it points at a 60 ∘ angle to the vertical?Part C

What will be the angular velocity ωP of the platform if the person moves the axis of the wheel so that it points vertically downward?Part D

What will ωP be if the person reaches up and stops the wheel in part A?

Question

contestada

Respuesta :

DeniceSandidge DeniceSandidge · Answer 1 · 2020-04-18T07:35:11+02:00

Answer:

answer is given below

Explanation:

first we use here conservation of angular momentum in the vertical direction that is express as

angular velocity platform = [tex]\frac{\omega W\times IW}{IP}[/tex] clock-wise ..........................1

so the net angular momentum = 0

and

here when vertical angular momentum will be 0 than

here angular velocity of the platform will be

ωP = [tex]\frac{1}{2} \times \frac{\omega W\times IW}{IP}[/tex] clock-wise ..........................2

and

the angular velocity of the platform

ωP = [tex]\frac{\omega W\times IW}{IP}[/tex] anti-clock-wise ............................3

and

when here Platform will stop than

ωP = 0 ............................4

so as that conserve the angular momentum