Question

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 12.00 m long, weighs 20.00 N, and rotates at 350 rev/min clockwise when seen from above. Calculate its rotational inertia about the axis of rotation.

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Calculate the angular momentum of the rod about that axis.

A man stands at the center of a platform that rotates without friction with an angular speed of 1.2 rev/s. His arms are outstretched, and he holds a heavy weight in each hand. The moment of inertia of the man, the extended weights, and the platform is 13.3 kg*m^2. When the man pulls the weights inward toward his body, the moment of inertia decreases to 2.4 kg*m^2.

What is the resulting angular speed of the platform?

41.78 rad/s

You are correct. Your receipt no. is 158-3841

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What is the kinetic energy of the system before the man pulls the weights to his body?

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What is the kinetic energy of the system after the man pulls the weights to his body??

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What is the change in kinetic energy of the system?

Answer 1

Find the mass of the rod using the weight. Use the angular velocity and moment of inertia to find the angular momentum as shown below

- 1) a) Weight of nod is, - W = mg => 20N = mg > M= 20N - 20N 9.8m/s2 = 2:04 Kg Izmete mazme = 197.9614 me I = M22 197.96/kg.m2 b) L: Tu = 1/350 X2+ raals) = 3590.4 19.meys 60

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This concludes the answers. If there is any mistake or omission, let me know immediately and I will fix it....

- 1) a) Weight of nod is, - W = mg => 20N = mg > M= 20N - 20N 9.8m/s2 = 2:04 Kg Izmete mazme = 197.9614 me I = M22 197.96/kg.m2 b) L: Tu 1 50 *2raols) - 3590.4 19.me 60