Question

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?

Answer 1

In Part A, the angular velocity of the platform is equal to the angular velocity of the wheel. In Part B, the angular velocity of the platform is calculated using the moment of inertia of the wheel and the platform. In Part C, the angular velocity of the platform is calculated using the moment of inertia of the wheel and the platform. In Part D, the angular velocity of the platform becomes zero when the wheel is stopped.

Step-by-step explanation:

Part A: To find the angular velocity ωP of the platform when the person moves the axis of the wheel vertically upward, we need to consider the conservation of angular momentum. The angular momentum before the movement is given by the product of the moment of inertia of the person plus the platform (IP) and the angular velocity of the platform (ωP), and after the movement, it is given by the product of the moment of inertia of the person plus the platform (IP) and the new angular velocity (ωP). Since the angular momentum is conserved, we can set the initial and final angular momenta equal to each other and solve for ωP:

IP * ωP = (IP + IW) * ωP
ωP = IW / IP

Part B: To find the angular velocity ωP of the platform when the person moves the axis of the wheel at a 60° angle to the vertical, we can use the same conservation of angular momentum principle as in Part A. Since the axis of the wheel is inclined at a 60° angle, the moment of inertia of the wheel (IW) can be decomposed into two components: one perpendicular to the platform and one parallel to the platform. The perpendicular component does not contribute to the angular momentum of the system, so we only need to consider the parallel component. We can then use the same equation as in Part A:

ωP = (IW * sin^2(60°)) / IP

Part C: To find the angular velocity ωP of the platform when the person moves the axis of the wheel vertically downward, we can again use the conservation of angular momentum principle. Since the axis of the wheel is now vertically downward, the moment of inertia of the wheel (IW) can be decomposed into two components as in Part B. However, this time we need to consider the perpendicular component, as it will contribute to the angular momentum of the system. We can then use the same equation as in Part A:

ωP = (IW * cos^2(60°)) / IP

Part D: To find the angular velocity ωP of the platform when the person stops the wheel, we need to consider the conservation of angular momentum again. Initially, the angular momentum of the system is given by IP * ωP. When the person stops the wheel, the angular momentum of the wheel becomes zero. Therefore, the final angular momentum of the system is simply given by the product of the moment of inertia of the person plus the platform (IP) and the new angular velocity (ωP):

IP * ωP = 0
ωP = 0