## Physics: Principles with Applications (7th Edition)

Published by Pearson

# Chapter 8 - Rotational Motion - Problems: 43

#### Answer

a. $1.90 \times 10^3 kg \cdot m^2$ b. $8900 N \cdot m$

#### Work Step by Step

a. The rotational inertia of one rod/blade around an end is $\frac{1}{3}ML^2$. There are three rods. $$I=3\frac{1}{3}MR^2=(135 kg)(3.75m)^2 \approx 1.90 \times 10^3 kg \cdot m^2$$ b. The angular acceleration is $\alpha=\frac{\Delta \omega}{\Delta t}=4.71 \frac{rad}{s^2}$. Calculate the torque using $\tau = I \alpha \approx 8900 N \cdot m$.

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