A thin circular plate of mass M and radius R has its density varying as \rho (r)=\rho\, _{0}r with \rho _{0} as constant and r is the distance from its center. The moment of Inertia of the circular plate and passing through its edge is I=a\; MR^{2}. The value of the coefficient a is :
                                     

  • Option 1)

    3/5

  • Option 2)

    1/2

  • Option 3)

    8/5

     

  • Option 4)

    3/2

 

Answers (1)

take a small circular ring at a distance x from centre of thickness dx.

  dI=dmx^{2}+dm\; R^{2}

                                          dI=dm (x^{2}+R^{2})

& dm=\rho _{o}.2\pi x.dx

            =\rho _{o} x.2\pi x.dx

dm=\rho _{o} .2\pi x^{2}.dx

m=\int dm=\rho _{o}2\pi \int x^{2}dx=\rho _{o}\times2\pi \left [ \frac{R^{3}}{3} \right ]

I = \int dI = \varrho _{o}\times2\pi\int (x^{2}+R^{2})x^{2}dx

=\rho _{o}\times2\pi\left [ \int x^{4}dx +R^{2}\int x^{2}dx\right ]

I=\rho _{o}\times2\pi\left \{ \left [ \frac{x^{5}}{5} \right ]_{o}^{R}+R^{2} \left [ \frac{x^{3}}{3} \right ]_{o}^{R}\right \}

I=\rho _{o}\times2\pi\left [ \frac{R^{5}}{5}+\frac{R^{5}}{3} \right ]

I=\frac{16\pi}{15}\rho _{o}\pi R^{5}, M=2\pi \rho _{o}(\frac{R^{3}}{3})

So I = \frac{8}{5}mR^{2}\Rightarrow a=\frac{8}{5}


Option 1)

3/5

Option 2)

1/2

Option 3)

8/5

 

Option 4)

3/2

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