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Question Number 24186 by Tinkutara last updated on 14/Nov/17

(n − 1) equal point masses each of mass  m are placed at the vertices of a regular  n-polygon. The vacant vertex has a  position vector a with respect to the  centre of the polygon. Find the position  vector of centre of mass.

$$\left({n}\:−\:\mathrm{1}\right)\:\mathrm{equal}\:\mathrm{point}\:\mathrm{masses}\:\mathrm{each}\:\mathrm{of}\:\mathrm{mass} \\ $$$${m}\:\mathrm{are}\:\mathrm{placed}\:\mathrm{at}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{regular} \\ $$$${n}-\mathrm{polygon}.\:\mathrm{The}\:\mathrm{vacant}\:\mathrm{vertex}\:\mathrm{has}\:\mathrm{a} \\ $$$$\mathrm{position}\:\mathrm{vector}\:{a}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{polygon}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{position} \\ $$$$\mathrm{vector}\:\mathrm{of}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{mass}. \\ $$

Commented by mrW1 last updated on 14/Nov/17

−(1/(n−1))a

$$−\frac{\mathrm{1}}{{n}−\mathrm{1}}\boldsymbol{{a}} \\ $$

Commented by Tinkutara last updated on 14/Nov/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Commented by mrW1 last updated on 14/Nov/17

let′s say the center of (n−1)m is x  from the center of polygon. we know  if we also put a mass m at the vacant vertex,  then we have n mass m, and the  center of nm is the center of polygon.  (n−1)mg∣x∣=mg∣a∣  ⇒∣x∣=((∣a∣)/(n−1))  x is opposite to a  ⇒x=−(1/(n−1))a

$${let}'{s}\:{say}\:{the}\:{center}\:{of}\:\left({n}−\mathrm{1}\right){m}\:{is}\:{x} \\ $$$${from}\:{the}\:{center}\:{of}\:{polygon}.\:{we}\:{know} \\ $$$${if}\:{we}\:{also}\:{put}\:{a}\:{mass}\:{m}\:{at}\:{the}\:{vacant}\:{vertex}, \\ $$$${then}\:{we}\:{have}\:{n}\:{mass}\:{m},\:{and}\:{the} \\ $$$${center}\:{of}\:{nm}\:{is}\:{the}\:{center}\:{of}\:{polygon}. \\ $$$$\left({n}−\mathrm{1}\right){mg}\mid{x}\mid={mg}\mid{a}\mid \\ $$$$\Rightarrow\mid{x}\mid=\frac{\mid{a}\mid}{{n}−\mathrm{1}} \\ $$$${x}\:{is}\:{opposite}\:{to}\:{a} \\ $$$$\Rightarrow\boldsymbol{{x}}=−\frac{\mathrm{1}}{{n}−\mathrm{1}}\boldsymbol{{a}} \\ $$

Commented by mrW1 last updated on 14/Nov/17

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