A system consists of a 2.0 kg uniform disk with radius 0.40 m, and ring of negligible thickness and mass 1.0 kg with a radius of 0.20 m, as shown. If the system has a rotational speed of 50.0 rad/s, what is its kinetic energy in Joules?

Question

A system consists of a 2.0 kg uniform disk with radius 0.40 m, and ring of negligible thickness and mass 1.0 kg with a radius of 0.20 m, as shown.  If the system has a rotational speed of 50.0 rad/s, what is its kinetic energy in Joules?

BlackTom AI · Accepted Answer

To determine the kinetic energy of a rotating system, we sum the moments of inertia of all components and then use KE = (1/2) I_total ω^2.

First, compute the moment of inertia for each piece:
- Uniform disk: I_disk = (1/2) M R^2 = (1/2)(2.0 kg)(0.40 m)^2 = 1.0 × 0.16 = 0.16 kg·m^2.
- Ring (hoop) with negligible thickness: I_ring = M R^2 = (1.0 kg)(0.20 m)^2 = 0.04 kg·m^2.

Total moment of inertia: I_total = I_disk + I_ring = 0.16 + 0.04 = 0.20 kg·m^2.

Next, apply the rotational kinetic energy formula with ω = 50.0 rad/s:
KE = (1/2) I_total ω^2 = (1/2)(0.20 kg·m^2)(50.0 rad/s)^2.
Compute ω^2: (50.0)^2 = 2500 (rad^2/s^2).
So KE = 0.10 × 2500 = 250 J.

If your multiple-choice options included 250, that would be the value corresponding to the given configuration. Note that any discrepancy in units or in the radii/ masses would change the result, so careful substitution of the provided numbers is essential.

01:750:124 ANALYTICAL PHYSICS IB Mini-Exam 2 (Modules 3,4,5) - 9 questions, 50 minutes

A system consists of a 2.0 kg uniform disk with radius 0.40 m, and ring of negligible thickness and mass 1.0 kg with a radius of 0.20 m, as shown. If the system has a rotational speed of 50.0 rad/s, what is its kinetic energy in Joules?

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