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Find the magnitude and direction of the force acting on the particle of mass m during its motion in the plane xy according to the law x=a sin omega t, y=b cos omega t, where a, b, and omega are constants. |
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Answer» Solution :OBVIOUSLY the RADIUS vector DESCRIBING the position of the PARTICLE relative to the origin of coordinate is `vecr=xveci+yvecj=a sin omega tveci+b cos omega t vecj` DIFFERENTIATING twice with respect the time: `vecw=(d^2vecr)/(dt^2)=-omega^2(a sin omega t veci+b cos omega t vecj)=-omega^2vecr` (1) Thus `vecF=mvecw=-momega^2vecr` |
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