A student sits on a pivoted stool while holding a pair of weights. The stool is free to rotate about a vertical axis with negligible friction. The moment of inertia of student, weights, and stool is 5.25 kg. m². The student is set in rotation with arms outstretched, making one complete turn every 2.2 s, arms outstretched. Consider opposite of turning direction as positive. Quiz 6 Question4
What is the initial angular velocity of the system?

Answer 1
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As he rotates, he pulls the weights inward so that a new moment of inertia of the system (student, objects, and stool) becomes 4 kg. m2. What is the new angular velocity of the system?

Answer 2
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a) 4.99 rad/sec b) 6.24 rad/sec c) 7.03 J

Explanation:

a)  If the student completes one turn in 1.26 sec, this is called the period of the movement.

If we take into account that the angle rotated during one turn is 2π rads, by definition of angular velocity, we can get this value as follows:

ω = Δθ / Δt = 2*π rad / 1.26 seg = 4.99 rad/sec.

b) As no external torques are acting on the system, the total angular momentum must be conserved, so we can write the following equation:

Li = Lf   ⇒  I₁ * ω₁  = I₂* ω₂

So, we can solve for ω₂, as follows:

ω₂ = (I₁ * ω₁) / I₂ = 6.24 rad/sec

c) Appying the work-energy theorem, we know that the work done by the student, must be equal to the change in the kinetic energy, which in this case is only rotational, so we can write:

W = 1/2 I₂* ω₂² - 1/2 I₁ ω₁²

W =1/2 ((2.25 kg.m² * (6.24)²) (rad/sec)² - (1.8 kg.m²* (4.99)²) (rad/sec)²)  

W = 7.03 J

need to see circuit to get time constant