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, weight and stool is 2.25 kgm^2. The student is set in rotation with arms outstretched, making one complete turn every 1.26 second, arm outstretched. a) What is the initial angular speed of the system? b) As he rotates, he pulls the weights inward so that the new moment of inertia of the system becomes 1.80 kg m^2. What is the new angular speed of the system? c) Find the work done by the student on the system while pulling in the weights.

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