
Six identical particles each of mass $'m'$ are arranged at the corners of a regular hexagon of side length . If the mass of one of the particle is doubled, the shift in the centre of mass is:
A) $L$
B) $\dfrac{{6L}}{7}$
C) $\dfrac{L}{7}$
D) $\dfrac{L}{{\sqrt 3 }}$
Answer
232.8k+ views
Hint: We can find the shift within the center of mass because the product of mass removed and therefore the distance between original center of mass and away from each other when divided by the mass of the remaining portion is also known as centre of mass.
Complete step by step answer:
The center of mass of the primary arrangement is six equal masses on the six corners of a regular hexagon side length and is at the center of the hexagon. Then the new arrangement if the mass of one of the particles is doubled, and find the value of the shift in the center of mass.
The total masses will be $6m + m = 7m$.
When, the six identical particles are present at six different positions of the hexagon. So, the moment will be $2m \times L$ for that shifted particle. So, the total will be
$\dfrac{2mL+\dfrac{-mL}{2}+\dfrac{-mL}{2} +mL-mL}{7m}$
$ \Rightarrow \dfrac{{2mL - mL}}{{7m}}$
Which one is solving we will get the value of position of the center is,
$ \Rightarrow \dfrac{L}{7}$
Since, the masses are the same for all positions except the shifted mass of $2m$ hence the moment can be added.
Hence, the correct answer is option (C).
Additional Information:
A scientific method for locating the middle of mass is to suspend the thing from two locations and to drop plumb lines from the suspension points. The intersection of the 2 lines is that the centre of mass. The shape of an object might already be mathematically determined, but it's getting to be too complex to use a known formula.
Note: Shift in Centre of Mass, when the certain mass is removed. If a particular mass from the system is removed, the centre of mass obviously changes from its existing position. When some mass of the system is removed, a new centre of mass shifts towards the heavier a neighbourhood of the system.
Complete step by step answer:
The center of mass of the primary arrangement is six equal masses on the six corners of a regular hexagon side length and is at the center of the hexagon. Then the new arrangement if the mass of one of the particles is doubled, and find the value of the shift in the center of mass.
The total masses will be $6m + m = 7m$.
When, the six identical particles are present at six different positions of the hexagon. So, the moment will be $2m \times L$ for that shifted particle. So, the total will be
$\dfrac{2mL+\dfrac{-mL}{2}+\dfrac{-mL}{2} +mL-mL}{7m}$
$ \Rightarrow \dfrac{{2mL - mL}}{{7m}}$
Which one is solving we will get the value of position of the center is,
$ \Rightarrow \dfrac{L}{7}$
Since, the masses are the same for all positions except the shifted mass of $2m$ hence the moment can be added.
Hence, the correct answer is option (C).
Additional Information:
A scientific method for locating the middle of mass is to suspend the thing from two locations and to drop plumb lines from the suspension points. The intersection of the 2 lines is that the centre of mass. The shape of an object might already be mathematically determined, but it's getting to be too complex to use a known formula.
Note: Shift in Centre of Mass, when the certain mass is removed. If a particular mass from the system is removed, the centre of mass obviously changes from its existing position. When some mass of the system is removed, a new centre of mass shifts towards the heavier a neighbourhood of the system.
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