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Calculate the gravitational field intensity and potential at the centre of the base of a solid hemisphere of mass m, radius R |
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Answer» Solution :We consider the shaded elemetnal disc of radius `Rsintheta`and thickness `Rd theta` its mass `dM=M/(2/3piR^(3))pi(Rsintheta)^(2)(Rd thetasintheta)` or `dM=(3M)/2SIN^(3)theta d theta` FIELD due to this plate at `O`, `dE=(2GdM(1-costheta))/((Rsintheta)^(2))` (see field due to a uniform disc) or `dE=(3GMsintheta(1-costheta)d theta)/(R^(2))` Therefore, `E=int_(0)^((pi)/2) dE=int_(0)^((pi)/2) =(3GMsintheta(1=costheta))/(R^(2))d theta` `=(3GM)/(R^(2))[-costheta+(cos^(2)theta)/2]_(0)^((p)i/2)` or `E=(3GM)/(2R^(2))`  Now potential due to the element under consider at the cente of the baase of the HEMISPHERE `DV=-2GdM//r (cosectheta-cottheta)` (See potential due to a circular plate) or `dV=(-3GMsin^(3)theta(cosectheta-cottheta)d theta)/((Rsintheta)` Therefore, `V=-(3GM)/Rint_(0)^((pi)/2) (sintheta-costhetasintheta)d theta` `=-(3GM)/R[-costheta+(cos^2theta)/2]_0^(pi/2)` or `V=(-3GM)/(2R)` Alternative method: Consider a HEMISPHERICAL shell of radius `r` and thicknes `dr` It mass `dm=M/(2/3piR^(3))(2pi^(2)dr)` or `dm=(3Mr^(2)dr)/(R^(3))` Since all points of this hemispherical shell are at the same distnce `r` from `O` hence potential at `O` is `dV=(-Gdm)/r=(-3GMrdr)/(R^(3))` `:. V=int_(0)^(R)dV=(-3GM)/(2R)` |
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