Question

A boy stands at the center of a platform that is rotating without friction at 1.0 rev/s. The boy holds weights as far from his body as possible. At this position the total moment of inertia of the boy, platform, and weights is 5.0kg·m2. The boy draws the weights in close to his body, thereby decreasing the total moment of inertia to 1.5kg·m2. (a) What is the final angular velocity of the platform? (b) By how much does the rotational kinetic energy increase?

a) (a) 3.0 rev/s, (b) 12.5 J
b) (a) 2.0 rev/s, (b) 7.5 J
c) (a) 2.5 rev/s, (b) 10.0 J
d) (a) 4.0 rev/s, (b) 15.0 J

Answer 1

The final angular velocity of the platform is 3.0 rev/s and the increase in rotational kinetic energy is 11.0 J.

Step-by-step explanation:

To calculate the final angular velocity of the platform, we can use the conservation of angular momentum. The total initial angular momentum of the system is equal to the total final angular momentum of the system. The initial moment of inertia of the system (including the boy, platform, and weights) is 5.0 kg·m² and the final moment of inertia is 1.5 kg·m². Let's assume the final angular velocity of the platform is ωf.

Using the equation:

Iiωi = Ifωf

5.0 kg·m² × (1 rev/s) = 1.5 kg·m² × ωf

Simplifying the equation, we find:

ωf = (5.0 kg·m² × (1 rev/s)) / 1.5 kg·m²

ωf = 3.0 rev/s

The final angular velocity of the platform is 3.0 rev/s.

To calculate the increase in rotational kinetic energy, we can use the equation:

ΔKE = KEf - KEi

where KEf is the final rotational kinetic energy and KEi is the initial rotational kinetic energy. The initial rotational kinetic energy is given by:

KEi = (1/2)Ii(ωi)²

KEi = (1/2)(5.0 kg·m²)(1 rev/s)²

KEi = 2.5 J

The final rotational kinetic energy is given by:

KEf = (1/2)If(ωf)²

KEf = (1/2)(1.5 kg·m²)(3 rev/s)²

KEf = 13.5 J

Finally, calculating the increase in rotational kinetic energy:

ΔKE = 13.5 J - 2.5 J

ΔKE = 11.0 J

The rotational kinetic energy increases by 11.0 J.