Answer:
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
