Question

. Conservation along the horizontal using a bicycle wheel: Stand on the platform holding a bicycle wheel with its axis horizontal with your arms fully extended. While holding the platform stationary, let a colleague spin the wheel. Release the platform then turn the wheel’s spin axis "up." Observe the platform turn with a ‘down’ spin vector, ω. Return the wheel to the starting position and the platform should stop. Turn the wheel’s spin axis "down." Return the wheel to the starting position. Describe the platform’s reaction in terms of conservation of vertical angular momentum: Use L1 as the angular momentum of the person and the platform, and L2 as the wheel’s angular momentum. Use vector diagrams to show how the vertical angular momentum of the system, Ltotal = L1 + L2, starts at zero and remains zero throughout the exercise. That is, show that L1i + L2i = L1f + L2f.

Answer 1

Answer:

w = I₂ / (I₁ -I₂) w₀ ,    L₂ = 2 L₁

Explanation:

This is an angular momentum exercise,

      L = I w

where bold indicates vectors

We must define the system as formed by the bicycle wheel, the platform, we create a reference system with the positive sign up

At the initial moment the wheel is turning and the platform is without rotation

The initial angle moment is

                 Lo = L₂ = I₂ w₀

L₁ is the angular momentum of the platform and L₂ is the angular momentum of the wheel.

In the Final moment, when the wheel was turned,

                $L_{f}$ = L₁ - L₂

                L_{f} = (I₁ - I₂) w

the negative sign of the angular momentum of the wheel is because it is going downwards since the two go with the same angular velocity

as all the force are internal, and there is no friction the angular momentum is conserved,

             L₀ =L_{f}

             I₂ w₀ = (I₁ -I₂) w

             w = I₂ / (I₁ -I₂) w₀

we can see that the system will complete more slowly

 we can also equalize the angular cognition equations

                 L₀ = Lf

                 L₁ = L₂-L₁

                 L₂ = 2 L₁

In this part we can see that the change in the angular momentum of the platform is twice the change in the angular momentum of the wheel.