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A system of four particles is x - y plane. Of these, two particles each of mass m are located at (-1, 1) and (1, -1). The remaining two particles each of mass 2m are located at (1, 1) and (-1, 1). The xy component of the moment of inertia tensor of this system of particles is -
- a)10 m
- b)2 m
- c)-10 m
- d)-2 m
Correct answer is option 'B'. Can you explain this answer?
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A system of four particles is x - y plane. Of these, two particles eac...
The xy component of the moment of inertia tensor
Two particles of mass m located at (-1,1) and (1, -1) other particles of mass 2m are located at (1, 1) and (-1, 1)
W = - [m1x1y1 + m2x2y2 + m3X3y3 + m4x4y4]
= - [(mx -1 x 1 ) + (m x 1 x -1 ) + (2m x 1x1) + (2m x 1x-1)]
= - [-m -m + 2m - 2m]
= [2m] ⇒ Ixy = 2m
Two particles of mass m located at (-1,1) and (1, -1) other particles of mass 2m are located at (1, 1) and (-1, 1)
W = - [m1x1y1 + m2x2y2 + m3X3y3 + m4x4y4]
= - [(mx -1 x 1 ) + (m x 1 x -1 ) + (2m x 1x1) + (2m x 1x-1)]
= - [-m -m + 2m - 2m]
= [2m] ⇒ Ixy = 2m
Most Upvoted Answer
A system of four particles is x - y plane. Of these, two particles eac...
Solution:
The moment of inertia tensor is given by:
I = ∑m(i)(r(i)^2) - m(i)r(i) * r(j)
where m(i) is the mass of the ith particle, r(i) is the position vector of the ith particle, and r(j) is the position vector of the jth particle.
Given that two particles of mass m are located at (-1, 1) and (1, -1) and two particles of mass 2m are located at (1, 1) and (-1, 1), we can write the position vectors as:
r(1) = (-1, 1, 0)
r(2) = (1, -1, 0)
r(3) = (1, 1, 0)
r(4) = (-1, -1, 0)
The xy component of the moment of inertia tensor is given by:
I(xy) = ∑mi(xi^2 + yi^2) - mi(xi*yj + yi*yj)
where xi and yi are the x and y components of the ith particle's position vector.
Substituting the position vectors and masses, we get:
I(xy) = m[(2 + 2) - (-1*1 + 1*(-1))] + 2m[(2 + 2) - (1*1 + 1*1)]
Simplifying the expression, we get:
I(xy) = 2m + 4m - 2m = 2m
Therefore, the correct answer is option B, 2m.
The moment of inertia tensor is given by:
I = ∑m(i)(r(i)^2) - m(i)r(i) * r(j)
where m(i) is the mass of the ith particle, r(i) is the position vector of the ith particle, and r(j) is the position vector of the jth particle.
Given that two particles of mass m are located at (-1, 1) and (1, -1) and two particles of mass 2m are located at (1, 1) and (-1, 1), we can write the position vectors as:
r(1) = (-1, 1, 0)
r(2) = (1, -1, 0)
r(3) = (1, 1, 0)
r(4) = (-1, -1, 0)
The xy component of the moment of inertia tensor is given by:
I(xy) = ∑mi(xi^2 + yi^2) - mi(xi*yj + yi*yj)
where xi and yi are the x and y components of the ith particle's position vector.
Substituting the position vectors and masses, we get:
I(xy) = m[(2 + 2) - (-1*1 + 1*(-1))] + 2m[(2 + 2) - (1*1 + 1*1)]
Simplifying the expression, we get:
I(xy) = 2m + 4m - 2m = 2m
Therefore, the correct answer is option B, 2m.
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A system of four particles is x - y plane. Of these, two particles each of mass m are located at (-1, 1) and (1, -1). The remaining two particles each of mass 2m are located at (1, 1) and (-1, 1). The xy component of the moment of inertiatensor of this system of particles is -a)10 mb)2 mc)-10 md)-2 mCorrect answer is option 'B'. Can you explain this answer?
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