A force of 32.5N stretches a spring of 0.500cm. What force will stretch a similar spring 1.60 cm?
1 answer:
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Answer:
≈ 2.1 R
Explanation:
The moment of inertia of the bodies can be calculated by the equation
I = ∫ r² dm
For bodies with symmetry this tabulated, the moment of inertia of the center of mass
Sphere = 2/5 M R²
Spherical shell = 2/3 M R²
The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass
I = + M D²
Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation
Let's start with the spherical shell, axis is along a diameter
D = 2R
Ic = + M D²
Ic = 2/3 MR² + M (2R)²
Ic = M R² (2/3 + 4)
Ic = 14/3 M R²
The sphere
Is = + M [
²
Is = Ic
2/5 MR² + M ² = 14/3 MR²
² = R² (14/3 - 2/5)
= √ (R² (64/15)
= 2,066 R