Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 27326 by NECx last updated on 05/Jan/18

$$Whatistherelationshipbetween$$$$thecentreofgravityandthecentreof$$$$mass?$$

Answered by prakash jain last updated on 05/Jan/18

$$\mathrm{Center}\mathrm{of}\mathrm{gravity}\mathrm{and}\mathrm{center}\mathrm{of}\mathrm{mass}$$$$\mathrm{is}\mathrm{at}\mathrm{the}\mathrm{same}\mathrm{point}\mathrm{if}\mathrm{gravitational}$$$$\mathrm{field}\mathrm{is}\mathrm{same}\mathrm{at}\mathrm{all}\mathrm{points}\mathrm{of}\mathrm{the}\mathrm{body}.$$$$$$$$\mathrm{It}\mathrm{is}\mathrm{different}\mathrm{if}\mathrm{the}\mathrm{gravitation}\mathrm{field}$$$$\mathrm{varies}\mathrm{for}\mathrm{example}\mathrm{a}\mathrm{rod}\mathrm{placed}$$$$\mathrm{vertifically}\mathrm{on}\mathrm{earth}\mathrm{with}\mathrm{height}=\frac{{\mathrm{R}}_e}{2}$$$$\mathrm{then}\mathrm{we}\mathrm{cannot}\mathrm{assume}\mathrm{gavitation}$$$$\mathrm{field}\mathrm{is}\mathrm{same}\mathrm{at}\mathrm{all}\mathrm{points}.$$

Commented by mrW1 last updated on 06/Jan/18

$$Itrytocalculate.$$$$M=massofearth$$$$R=radiusofearth$$$$m=massofrod\left(verticallyplaced\right)$$$$L=lengthofrod$$$$x_G=positionofCoG$$$$dm=\frac{m}{L}dx$$$$F=\int \frac{GMdm}{{(R+x)}^2}=\frac{GMm}{L}{\int }_0^L\frac{dx}{{(R+x)}^2}$$$$=\frac{GMm}{L}{[-\frac{1}{R+x}]}_0^L$$$$=\frac{GMm}{L}[\frac{1}{R}-\frac{1}{R+L}]$$$$=\frac{GMm}{R(R+L)}$$$$F\times x_G=\int \frac{xGMdm}{{(R+x)}^2}=\frac{GMm}{L}{\int }_0^L\frac{xdx}{{(R+x)}^2}$$$$=\frac{GMm}{L}{[\frac{R}{R+x}+\ln (R+x)]}_0^L$$$$=\frac{GMm}{L}[\frac{R}{R+L}-1+\ln \frac{R+L}{R}]$$$$=\frac{GMm}{L}[-\frac{L}{R+L}+\ln \frac{R+L}{R}]$$$$\frac{GMm}{R(R+L)}\times x_G=\frac{GMm}{L}[-\frac{L}{R+L}+\ln \frac{R+L}{R}]$$$$\Rightarrow x_G=\frac{R(R+L)}{L}[-\frac{L}{R+L}+\ln \frac{R+L}{R}]$$$$\Rightarrow \frac{x_G}{L}=\frac{R(R+L)}{L^2}[-\frac{L}{R+L}+\ln \frac{R+L}{R}]$$$$let\lambda =\frac{L}{R}orL=\lambda R$$$$\Rightarrow \frac{x_G}{L}=\frac{R(R+\lambda R)}{{\lambda }^2{R}^2}[-\frac{\lambda R}{R+\lambda R}+\ln \frac{R+\lambda R}{R}]$$$$\Rightarrow \frac{x_G}{L}=\frac{(1+\lambda )}{{\lambda }^2}[-\frac{\lambda }{1+\lambda }+\ln (1+\lambda )]$$$$\Rightarrow \frac{x_G}{L}=\frac{(1+\lambda )\ln (1+\lambda )-\lambda }{{\lambda }^2}=\alpha$$$$\Rightarrow x_G=\alpha L$$$$\alpha isnotaconstantand\alpha <\frac{1}{2}$$$$\Rightarrow x_G<\frac{L}{2}$$$$but{x}_M=\frac{L}{2}$$$$itmeanstheCoGlieslowerthan$$$$theCoM.$$

Commented by NECx last updated on 05/Jan/18

$$thanksboss$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com