A uniform thin rod of mass M = 3.55 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m = 0.291 kg, are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is I = 0.845 kg-m² ? L = M m

Principles of Physics: A Calculus-Based Text
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Chapter10: Rotational Motion
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Problem 23P: Find the net torque on the wheel in Figure P10.23 about the axle through O, taking a = 10.0 cm and b...
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A uniform thin rod of mass M = 3.55 kg pivots about an axis through its center and perpendicular to its length. Two small
bodies, each of mass m = 0.291 kg, are attached to the ends of the rod. What must the length L of the rod be so that the
moment of inertia of the three-body system with respect to the described axis is I = 0.845 kg-m² ?
L =
m
L
Transcribed Image Text:A uniform thin rod of mass M = 3.55 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m = 0.291 kg, are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is I = 0.845 kg-m² ? L = m L
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