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  • A particle is moving in a circle of radius r under the action of a force which is direct

A particle is moving in a circle of radius r under the action of a force F=\alpha r^{2} which is directed towards centre of the circle. Total mechanical energy (kinetic energy+potential energy) of the particle is (take potential energy=0 for r=0) :

 

Option: 1

\alpha r^{3}


Option: 2

\frac{1}{2}\alpha r^{3}


Option: 3

\frac{4}{3}\alpha r^{3}


Option: 4

\frac{5}{6}\alpha r^{3}


Answers (1)

best_answer

 

 

 F = \alpha r^2

\frac{mv^2}{r} = \alpha r^2 \: \: or \: \: mv^2 = \alpha r^3

Kinetic energy  = 1/2 mv^2=\alpha r^3/2

Potential energy = U=\int_{0}^{r} \alpha r^{2} d r=\frac{a r^{3}}{3} \ \ \ (\because dU = \vec{F}.d\vec{r})

 

Total energy = \frac{\alpha r^3}{2}+ \alpha \frac{r^3 }{3}

E = \frac{5 \alpha r^3}{6}

 

 

 

 

 

 

 

 

Posted by

manish

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