Rotational Energy of 3 Point Masses along a Rigid Rod

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The discussion focuses on calculating the kinetic energy of a system consisting of three point masses on a rigid rod rotating about the y-axis. Two masses of 2.1 kg are positioned at the ends of the rod, while a 7.29 kg mass is located at the center. The rotation occurs about a point 1.28 m from one end, which affects the distance of each mass from the axis of rotation. To determine the kinetic energy, the linear speed of each mass is calculated using their distances from the axis and the constant angular speed of 5.38 rad/s. The distance of 1.28 m is crucial as it influences the linear velocities and, consequently, the total kinetic energy of the system.
shenwei1988
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Three point masses lie along a rigid, massless rod of length L = 5.61 m :

- Two particles, both of mass m = 2.1 kg, lie on opposite ends of the rod.
- Mass M = 7.29 kg is in the center of the rod.

Assume the rod lies along the x-axis, and rotates about the y-axis. about a point 1.28 m from one end at constant angular speed ω = 5.38 rad/s.

Find the kinetic energy of this system



could someone explain the ( about a point 1.28m from one end) . dose the distance 1.28 contribute any Kinetic energy to the system?
thank you so much
 
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Use the distance from the axis of rotation and the angular speed to find the linear speed of each point mass. Use that to find their kinetic energy.
 
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