A professor sits on a rotatable stool, and with feet on the ground, spins a 2.5

Question

A professor sits on a rotatable stool, and with feet on the ground, spins a 2.5 kg bicycle wheel, holding the axle horizontal so the wheel is straight in front of him. The wheel has a radius of 0.7 m, and has very light weight spokes. Once he gets the wheel spinning at 3 radians per second he lifts his feet and lifts the wheel over his head, so the axis now points upward and the wheel is parallel to the floor. Now he and the stool spin slowly, at 0.05 radians per second.

(A) What is the moment of inertia of the wheel?

(B) What is the angular momentum of the wheel?

(C) Relative to the rotation of the wheel, explain the direction of rotation of the prof.

(D) What is the moment of inertia of the prof plus stool (about their rotation axis)?

Explanation / Answer

The moment of inertia a wheel that rotates with respect to its axis

Icm = ½ m R2         Icm = ½ 2.5 0.72 = 0.6125 Kg m2

The angular momentum of the wheel   

L = I w

L= 0.6125 3 = 1.8375 Kg m2/s

We use conservation of the quantity of motion Lt0= Lsp + Lpr = Lsp 0

Ltf = Lsp f + Lpr

Lt0=Ltf

Since torque is used to change the movement of the wheel is internal system, therefore the teacher should begin to turn opposite to the wheel

remember that the moment of inertia is a vector quantity

Lspf = Lsp0         Lsp0 = -Lsp0 +Lpr                      Lpr= Lsp0 – Lspf

Lpr= Ipr wpr              wpr= 0.05 rad/s

The scalar value Lpr= Lsp

Ipr = Lpr/w        Ipr = 1.8375 / 0.05 = 36.75 Kg m2

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