A travelling spring

Classical Mechanics Level 2

Suppose the Earth is a perfect sphere of radius R R and its density function is a constant. If the angular velocity of the Earth is Ω , \Omega, what is the difference in displacement of a spring of angular frequency ω \omega when mesured on one of the poles and on the equator if we deem the spring massless?

( ω Ω ) 2 R \left( \frac{\omega}{\Omega} \right)^{2} R ( Ω ω ) 2 R \left( \frac{\Omega}{\omega} \right)^{2} R ( ω + Ω ω ) 2 R \left( \frac{\omega + \Omega}{\omega} \right)^{2} R ( Ω − ω ω ) R \left( \frac{\Omega - \omega}{\omega} \right) R

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1 solution

The force while the spring is on the poles is

F 1 = m g = k x 1 , F_1 = mg = k x_1,

where m m is the mass attached to the spring. When on the equator:

F 2 = m g − m Ω 2 R = k x 2 . F_2 = mg - m \Omega^2 R = k x_2.

Taking the difference between the two forces we have

F 1 − F 2 = m Ω 2 R = k ( x 1 − x 2 ) , F_1 - F_2 = m \Omega^2 R = k (x_1 - x_2),

and therefore

x 1 − x 2 = ( Ω ω ) 2 R , x_1 - x_2 = \left( \frac{\Omega}{\omega} \right)^{2} R,

since ω 2 = k / m . \omega^2 = k/m.

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We also know that the limit of the expression must be zero as omega goes to infinity. All but one of the possible answer choices are obviously wrong in this light.

Steven Chase - 3 years, 5 months ago

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Also, if the Earth stopped, the answer should be zero.

A Former Brilliant Member - 3 years, 5 months ago

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