Question

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a mass of 117.0 kg, a radius of 2.00 m, and a rotational inertia of 4.68×102 kgm2 about the axis of rotation. A student of mass 50.0 kg walks slowly from the rim of the platform toward the center. If the angular speed of the system is 1.75 rad/s when the student starts at the rim, what is the angular speed when she is 0.650 m from the center? 2.98 rad/s

Answer 1

Answer: 2.39 rad/sec

Explanation:

Under the assumption that no external torques are applied, total angular momentum must be conserved:

L₁ = L₂

The initial angular momentum, is given by the following expression:

L₁ = I₁ ω₁ , where I₁, is the total rotational inertia  of the system, and is equal to the sum of the rotational inertia of the disk, plus the one due to the student standing just in the rim of the disk, as follows:

I₁ = Id + Is = 4.68.10² Kg m² + 50.0 (2.00) ² kg.m²= 668 Kg.m²

So, L₁ = 668 kg. m² . 1.75 rad/s = 1,169 Kg.m². rad/s

Final angular momentum, is obtained as follows:

L₂ = I₂ . ω₂.

Now, I₂, is just the same term as in I₁, with the difference, that Is now given by the following expression:

Is₂ = 50. 0 . (0.65)² kg.m² = 21.1 kg.m²

So, total I₂ is as follows:

I₂ = 468 kg. m² + 21.1 kg.m² = 489.1 Kg.m²

So, equating L₁ and L₂, and solving for ω₂, we have:

489.1 Kg. m² . ω₂ = 1,169 kg. m² rad/sec ⇒ ω₂ = 2.39 rad/sec