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  • Please help! A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then :

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then :

  • Option 1)

    T∝Rn/2

  • Option 2)

    T∝R3/2 for any n.

  • Option 3)

    T\alpha R^{\frac{n}{2}+1}

  • Option 4)

    T∝R(n+1)/2

 

Answers (1)

As we learnt that

 

Relation of Angular Velocity frequency and Time period -

\omega=\frac{\theta}{t}=\frac{2\pi}{T}=2\pi n

\omega=\frac{2\pi}{T}=2\pi n

- wherein

\omega= Angular Velocity

n = frequency

T = time period

 

 F=\frac{K}{R_{n}}=m\omega ^{2}R

\therefore \omega ^{2}\propto \frac{1}{R^{n+1}}

\therefore T=\frac{2\pi }{\omega }=2\pi .R^{(\frac{n+1}{2})}


Option 1)

T∝Rn/2

Option 2)

T∝R3/2 for any n.

Option 3)

T\alpha R^{\frac{n}{2}+1}

Option 4)

T∝R(n+1)/2

Posted by

Vakul

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