A toy globe rotates freely without friction with an initial angular velocity $ω_{0}$ . A bug starting at one pole N travel to the other pole S along a meridian with constant velocity $υ$ . The axis of rotation of the globe is held fixed. Let M and R denote the mass and radius of the globe (a solid sphere, moment of inertia $I_{0} = 2 M R^{2} / 5$ ), m the mass of the bug, and T the duration of the bug’s journey.

Show that, during the time the bug is travelling, the globe rotates through an angle.

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$Δ θ = \frac{2 π ω_{0} R}{υ} \sqrt{\frac{2 M}{2 M + 7 m}}$

$Δ θ = \frac{π ω_{0} R}{υ} \sqrt{\frac{2 M}{(2 M + 5 m)}}$

$Δ θ = \frac{2 π ω_{0} R}{υ} \sqrt{\frac{2 M}{2 M + 5 m}}$

$Δ θ = \frac{π ω_{0} R}{υ} \sqrt{\frac{2 M}{2 M + 7 m}}$

answer is A.

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Detailed Solution

$\frac{2}{5} M R^{2} ω = (\frac{2}{5} M R^{2} + m R^{2} \sin^{2} θ) \int ω_{1} d t = \int \frac{M ω_{0}}{M + \frac{5}{2} m \sin^{2} θ} d t Δ θ = ω_{0} \int \frac{\frac{M \cos e c^{2} θ R d θ}{V}}{M \cot^{2} θ + (\frac{5 m}{2} + M)} Δ θ = \frac{R ω_{0}}{V} \int \frac{\cos e c^{2} θ d θ}{\cot^{2} θ + \frac{M + \frac{5 m}{2}}{M}} = \frac{R ω_{0}}{V} \int \frac{d t}{t^{2} + \frac{M + \frac{5 m}{2}}{M}} = \frac{π R ω_{0}}{V} \sqrt{\frac{2 M}{2 M + 5 m}}$