A horizontal spring with spring constant 85 N/m extends outward from a wall just above floor level. A 5.5 kg box sliding across
a frictionless floor hits the end of the spring and compresses it 6.5 cm before the spring expands and shoots the box back out. How fast was the box going when it hit the spring
1 answer:
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Answer:
Moment of inertia of the system is 289.088 kg.m^2
Explanation:
Given:
Mass of the platform which is a uniform disk = 129 kg
Radius of the disk rotating about vertical axis = 1.61 m
Mass of the person standing on platform = 65.7 kg
Distance from the center of platform = 1.07 m
Mass of the dog on the platform = 27.3 kg
Distance from center of platform = 1.31 m
We have to calculate the moment of inertia.
Formula:
MOI of disk =
Moment of inertia of the person and the dog will be mr^2.
Where m and r are different for both the bodies.
So,
Moment of inertia of the system with respect to the axis yy.
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⇒
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The moment of inertia of the system is 289.088 kg.m^2