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Question
(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.
Solution
The centre of mass of a regular polygon with n sides lies on its geometric centre. If mass m is placed at all the n vertices, then the C.O.M Is again at the geometric centre. Let `vecr` be the position vector of the COM and `vecA` of the vacant vertex. Then
`r_cm = ((n - 1)mr + ma)/((n - 1)m + m)` = 0 .....(when mass is placed at nth vertex also)
`(n - 1)mr + ma` = 0
`r = - (ma)/((n - 1)m)`
`vecr = - veca/((n - 1))`
The negative sign depicts that the C.O.M lies on the opposite side of nth vertex.
Hence the center of mass of n particles is a weighted average of the position vectors of n particles making up the system.
The centre of mass of a regular n-polygon lies at its geometrical centre.
Let the position vector of each centre of mass or regular n polygon is r.
(n – 1) equal point masses each of mass m are placed at (n – 1) vertices of the regular n-polygon, therefore, for its centre of mass.
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Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:
- Show pi = p’i + miV Where pi is the momentum of the ith particle (of mass mi) and p′ i = mi v′ i. Note v′ i is the velocity of the ith particle relative to the centre of mass. Also, prove using the definition of the centre of mass `sum"p""'"_"t" = 0`
-
Show K = K′ + 1/2MV2
where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of the
system when the particle velocities are taken with respect to the centre of mass and MV2/2 is the
kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the
system). The result has been used in Sec. 7.14. - Show where `"L""'" = sum"r""'"_"t" xx "p""'"_"t"` is the angular momentum of the system about the centre of mass with
velocities taken relative to the centre of mass. Remember `"r"_"t" = "r"_"t" - "R"`; rest of the notation is the standard notation used in the chapter. Note L′ and MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles. - Show `"dL"^"'"/"dt" = ∑"r"_"i"^"'" xx "dP"^"'"/"dt"`
Further show that `"dL"^'/"dt" = τ_"ext"^"'"`
Where `"τ"_"ext"^"'"` is the sum of all external torques acting on the system about the centre of mass.
(Hint: Use the definition of centre of mass and third law of motion. Assume the internal forces between any two particles act along the line joining the particles.)
