A hemispherical bowl of radius $$R=0.1m$$ is rotating about its own axis (which$$ is $$vertical) with an angular velocity ω. A particle of mass $$10-2$$ kg on the $$friction-less$$ inner surface of the bowl is also rotating with the same w. The particle is at height h from the bottom of the bowl. It is desired to measure ‘g’ using the $$set-up$$ by measuring h accurately. Assuming that R and ω are known precisely and that the least count in the measurement of h is $$10-4$$ m. The minimum possible error (Dg) in the measured value of g is n×$$10$$−$$2m/s2$$. Then n is (h$$

Question

A hemispherical bowl of radius $$R=0.1m$$ is rotating about its own axis (which$$ is $$vertical) with an angular velocity ω. A particle of mass $$10-2$$ kg on the $$friction-less$$ inner surface of the bowl is also rotating with the same w. The particle is at height h from the bottom of the bowl. It is desired to measure ‘g’ using the $$set-up$$ by measuring h accurately. Assuming that R and ω are known precisely and that the least count in the measurement of h is $$10-4$$ m. The minimum possible error (Dg) in the measured value of g is n×$$10$$−$$2m/s2$$. Then n is (h$$ << $$R)

Infinity Learn · Accepted Answer

(a) R is the radius of the bowl​$$sin$$θ$$=rR$$⇒$$r=Rsin$$θ−−−$$1$$​Ncosθ$$=mg$$−−−$$2$$​and​Nsinθ$$=mr$$ω$$2$$−−−−$$3$$​⇒Nsinθ$$=m$$ω$$2Rsin$$θ​⇒$$N=mR$$ω$$2$$−−−$$4$$​from $$2$$ & $$4$$​∴mRω$$2cos$$θ$$=mg$$⇒$$g=$$ω$$2R$$−$$h=$$ω$$2R1$$−hR​ taking 'log' both side ​⇒$$logg=log$$ω$$2R+log1$$−hR​⇒$$logg=log$$ω$$2R+$$−hR−$$h22R2$$−$$h33R3+$$…Differentiating with respect to ‘h’​⇒$$1gdgdh=0$$−$$1R$$−… higher order of hR neglected​⇒Δ$$g=gR$$×Δ$$h=100.1$$×$$10$$−$$4=10$$×$$10$$−$$3=1$$×$$10$$−$$2$$

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