Two identical masses of 4.00 kg each are attached to the opposite ends of a thin

Question

Two identical masses of 4.00 kg each are attached to the opposite ends of a thin, 0.420 m rod with a mass of 2.75 kg. The system is rotating with an angular velocity of omega = 5.60 rad/s about a vertical axel at the center of the rod. Use this system to answer questions 12-13. What is the moment of inertia of the entire system, including the two masses and the rod? 0.0404 kgm^2 0176 kgm^2 0.353 kgm^2 0.393 kgm^2 Someone forgot to attach the masses properly, and they go flying off the ends of the rod. If the masses do not carry any angular momentum away with them, what is the new angular velocity of the unencumbered rod? 5.60 rad/s 54.5 rad/s 48.9 rad/s 24.3 rad/3

Explanation / Answer

12.moment of inertia of rod=ML^2/12=0.040425 kgm^2

moment of inertia of both mass=2*4*(0.21^2)=0.3528 kgm^2

total moment of inertia=0.3932 kgm^2

(d) is correct

13.by consrvation of angular momentum

I1*w1=I2*w2

w2=I1*w1/I2=0.393*5.6/0.040425=54.47 rad/sec

(b) is correct

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