A wheel with a moment of inertia of 0.10 kg-m^2 rotates about a shaft at an angular speed of 160 rev/minute . A second wheel is set into rotation at 300 rev/minute and is then coupled to the same shaft, causing both

Correct answer: B. 0.04 kg-m^2 By conservation of angular momentum: I_1 ω_1 + I_2 ω_2 = (I_1 + I_2) ω Given I_1 = 0.10 kg-m^2 , ω_1 = 160 rev/minute , ω_2 = 300 rev/minute , and ω = 200 rev/minute . Substituting the values: 0.10 × 160 + I_2 × 300 = (0.10 + I_2) × 200 16 + 300 I_2 = 20 + 200 I_2 100 I_2 = 4 I_2 = 0.04 kg-m^2

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Rotational Mechanics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Rotational Mechanics

A wheel with a moment of inertia of 0.10 kg-m^2 rotates about a...

A wheel with a moment of inertia of

0.10 \text{ kg-m}^2

rotates about a shaft at an angular speed of

160 \text{ rev/minute}

. A second wheel is set into rotation at

300 \text{ rev/minute}

and is then coupled to the same shaft, causing both wheels to finally rotate at a common angular speed of

200 \text{ rev/minute}

. Determine the moment of inertia of the second wheel.

Options

A $0.05 \text{ kg-m}^2$
B $0.04 \text{ kg-m}^2$
C $0.36 \text{ kg-m}^2$
D $0.08 \text{ kg-m}^2$

Correct answer

B. $0.04 \text{ kg-m}^2$

Step-by-step solution

By conservation of angular momentum:

I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega

Given

I_1 = 0.10 \text{ kg-m}^2

\omega_1 = 160 \text{ rev/minute}

\omega_2 = 300 \text{ rev/minute}

, and

\omega = 200 \text{ rev/minute}

.

Substituting the values:

0.10 \times 160 + I_2 \times 300 = (0.10 + I_2) \times 200

16 + 300 I_2 = 20 + 200 I_2

100 I_2 = 4

I_2 = 0.04 \text{ kg-m}^2

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