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A body of mass mm moves in an elliptical path with a constant angular speed ω\omega (see the figure). It can be shown that the force acting on the body is always directed toward the origin and has magnitude given by

F=mω2a2cos2ωt+b2sin2ωtt0F=m \omega^2 \sqrt{a^2 \cos ^2 \omega t+b^2 \sin ^2 \omega t} \qquad t \geq 0

where aa and bb are constants with a>ba>b. Find the points on the path where the force is greatest and where it is smallest. Does your result agree with your intuition?


Answered 2 years ago
Answered 2 years ago
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We begin by taking FF':

F=(mω22)(2a2cosωt(sinωt))+2b2sinωtcosωt)a2cos2ωt+b2sinωt=(mω22)(sin2ωt)(b2a2)a2cos2ωt+b2sinωt\begin{align*} F' &= \left(\frac{m\omega^2}{2}\right)\frac{(2a^2\cos{\omega t}(-\sin{\omega t})) + 2b^2\sin{\omega t}\cos{\omega t})}{\sqrt{a^2\cos^2{\omega t} + b^2\sin{\omega t}}}\\ &= \left(\frac{m\omega^2}{2}\right)\frac{(\sin{2\omega t})(b^2 - a^2)}{\sqrt{a^2\cos^2{\omega t} + b^2\sin{\omega t}}} \end{align*}

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