An object of mass m is circulating in a circular path of radius r with help of the centripetal force $- \frac{k}{r^{2}}$ $-k/r^2$ . Total kinetic energy of the object is???(rotation of rigid body)

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Answer 1 · 2017-08-27T08:53:20

See below.

The object is rotating in a circular path then

$- \frac{k}{r^{2}} + m ω^{2} r = 0$ $-k/r^2+m omega^2r = 0$

The kinetic energy is given by

$E = \frac{1}{2} J ω^{2} = \frac{1}{2} m r^{2} \frac{k}{m r^{3}} = \frac{1}{2} \frac{k}{r}$ $E = 1/2J omega^2 = 1/2 m r^2 k/(m r^3) = 1/2k/r$

and

$ω = \sqrt{\frac{k}{m r^{3}}}$ $omega=sqrt(k/(m r^3))$

An object of mass m is circulating in a circular path of radius r with help of the centripetal force −kr2-k/r^2. Total kinetic energy of the object is???(rotation of rigid body)