What-is-the-relationship-between-the-centre-of-gravity-and-the-centre-of-mass-

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

W h a t i s t h e r e l a t i o n s h i p b e t w e e n

t h e c e n t r e o f g r a v i t y a n d t h e c e n t r e o f

m a s s ?

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

Center of gravity and center of mass

is at the same point if gravitational

field is same at all points of the body .

It is different if the gravitation field

varies for example a rod placed

vertifically on earth with height = \frac{R_{e}}{2}

then we cannot assume gavitation

field is same at all points .

Commented by mrW1 last updated on 06/Jan/18

I t r y t o c a l c u l a t e .

M = m a s s o f e a r t h

R = r a d i u s o f e a r t h

m = m a s s o f r o d (v e r t i c a l l y p l a c e d)

L = l e n g t h o f r o d

x_{G} = p o s i t i o n o f C o G

d m = \frac{m}{L} d x

F = \int \frac{G M d m}{{(R + x)}^{2}} = \frac{G M m}{L} \int_{0}^{L} \frac{d x}{{(R + x)}^{2}}

= \frac{G M m}{L} {[- \frac{1}{R + x}]}_{0}^{L}

= \frac{G M m}{L} [\frac{1}{R} - \frac{1}{R + L}]

= \frac{G M m}{R (R + L)}

F \times x_{G} = \int \frac{x G M d m}{{(R + x)}^{2}} = \frac{G M m}{L} \int_{0}^{L} \frac{x d x}{{(R + x)}^{2}}

= \frac{G M m}{L} {[\frac{R}{R + x} + \ln (R + x)]}_{0}^{L}

= \frac{G M m}{L} [\frac{R}{R + L} - 1 + \ln \frac{R + L}{R}]

= \frac{G M m}{L} [- \frac{L}{R + L} + \ln \frac{R + L}{R}]

\frac{G M m}{R (R + L)} \times x_{G} = \frac{G M m}{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}]

l e t λ = \frac{L}{R} o r L = λ R

\Rightarrow \frac{x_{G}}{L} = \frac{R (R + λ R)}{λ^{2} R^{2}} [- \frac{λ R}{R + λ R} + \ln \frac{R + λ R}{R}]

\Rightarrow \frac{x_{G}}{L} = \frac{(1 + λ)}{λ^{2}} [- \frac{λ}{1 + λ} + \ln (1 + λ)]

\Rightarrow \frac{x_{G}}{L} = \frac{(1 + λ) \ln (1 + λ) - λ}{λ^{2}} = α

\Rightarrow x_{G} = α L

α i s n o t a c o n s t a n t a n d α < \frac{1}{2}

\Rightarrow x_{G} < \frac{L}{2}

b u t x_{M} = \frac{L}{2}

i t m e a n s t h e C o G l i e s l o w e r t h a n

t h e C o M .

Commented by NECx last updated on 05/Jan/18

t h a n k s b o s s

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