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mahaypatto
09.11.2019 •
Physics
Auniform thin rod of mass m = 4.17 kg pivots about an axis through its center and perpendicular to its length. two small bodies, each of mass m = 0.519 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 = 0.913 kgm2?
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Ответ:
The length of the rod must be 1.226 meter
Explanation:
Step 1: Data given
mass of rod = 4.17 kg
mass of the bodies = each 0.519 kg
length of the rod = L
I = 0.913 kg*m²
Step 2: Calculate moment of inertia of the rod
Consider L = length of the rod.
⇒ Moment of inertia of the rod = (1/12)*4.17*L²
Step 3: Moment of inertia of the 2 bodies
Moment of inertia of each of the two bodies = 0.519*(L/2)²
Step 4: Total moment of inertia
Total moment of inertia of the three-body system =
(1/12)*4.17*L² + 2*0.519*(L/2)² = 0.913 kg*m²
0.3475* L² +0.2595 *L²= 0.913
0.607 L² = 0.913
L² = 1.504
L = 1.226 m
The length of the rod must be 1.226 meter
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