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A system of four particles is in x-y plane of these two particle of masses 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 moment of inertia tensor of the system of particle is
- a)10m
- b)-10m
- c)0m
- d)-2m
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A system of four particles is in x-y plane of these two particle of ma...
The xy component of inertia tensor is
Ixy = -[(m x1x1) + (m x-1x1) + (2m x1x-1) + 2m(-1x-1)]
= - [ m - m - 2m + 2m ]
= 0
Ixy = -[(m x1x1) + (m x-1x1) + (2m x1x-1) + 2m(-1x-1)]
= - [ m - m - 2m + 2m ]
= 0
Most Upvoted Answer
A system of four particles is in x-y plane of these two particle of ma...
To find the xy component of the moment of inertia tensor of the system of particles, we need to calculate the moment of inertia of each particle about the xy plane and then sum them up.
1. Moment of Inertia of Particle 1 (m1):
The moment of inertia of Particle 1 about the xy plane is given by: I1 = m1 * r1^2, where r1 is the perpendicular distance of the particle from the xy plane. In this case, r1 = 1 (since the particle is located at (1, 1)).
Therefore, I1 = m * 1^2 = m.
2. Moment of Inertia of Particle 2 (m2):
The moment of inertia of Particle 2 about the xy plane is given by: I2 = m2 * r2^2, where r2 is the perpendicular distance of the particle from the xy plane. In this case, r2 = 1 (since the particle is located at (-1, 1)).
Therefore, I2 = m * 1^2 = m.
3. Moment of Inertia of Particle 3 (m3):
The moment of inertia of Particle 3 about the xy plane is given by: I3 = m3 * r3^2, where r3 is the perpendicular distance of the particle from the xy plane. In this case, r3 = 1 (since the particle is located at (1, -1)).
Therefore, I3 = 2m * 1^2 = 2m.
4. Moment of Inertia of Particle 4 (m4):
The moment of inertia of Particle 4 about the xy plane is given by: I4 = m4 * r4^2, where r4 is the perpendicular distance of the particle from the xy plane. In this case, r4 = 1 (since the particle is located at (-1, -1)).
Therefore, I4 = 2m * 1^2 = 2m.
5. Summing up the Moments of Inertia:
Now, we can sum up the moments of inertia of all the particles to get the moment of inertia of the system of particles about the xy plane.
Ixy = I1 + I2 + I3 + I4
= m + m + 2m + 2m
= 6m.
Therefore, the xy component of the moment of inertia tensor of the system of particles is 6m. However, the given options do not have 6m as a choice. So, it seems that there might be an error in the options provided.
In conclusion, the correct answer cannot be determined from the given options.
1. Moment of Inertia of Particle 1 (m1):
The moment of inertia of Particle 1 about the xy plane is given by: I1 = m1 * r1^2, where r1 is the perpendicular distance of the particle from the xy plane. In this case, r1 = 1 (since the particle is located at (1, 1)).
Therefore, I1 = m * 1^2 = m.
2. Moment of Inertia of Particle 2 (m2):
The moment of inertia of Particle 2 about the xy plane is given by: I2 = m2 * r2^2, where r2 is the perpendicular distance of the particle from the xy plane. In this case, r2 = 1 (since the particle is located at (-1, 1)).
Therefore, I2 = m * 1^2 = m.
3. Moment of Inertia of Particle 3 (m3):
The moment of inertia of Particle 3 about the xy plane is given by: I3 = m3 * r3^2, where r3 is the perpendicular distance of the particle from the xy plane. In this case, r3 = 1 (since the particle is located at (1, -1)).
Therefore, I3 = 2m * 1^2 = 2m.
4. Moment of Inertia of Particle 4 (m4):
The moment of inertia of Particle 4 about the xy plane is given by: I4 = m4 * r4^2, where r4 is the perpendicular distance of the particle from the xy plane. In this case, r4 = 1 (since the particle is located at (-1, -1)).
Therefore, I4 = 2m * 1^2 = 2m.
5. Summing up the Moments of Inertia:
Now, we can sum up the moments of inertia of all the particles to get the moment of inertia of the system of particles about the xy plane.
Ixy = I1 + I2 + I3 + I4
= m + m + 2m + 2m
= 6m.
Therefore, the xy component of the moment of inertia tensor of the system of particles is 6m. However, the given options do not have 6m as a choice. So, it seems that there might be an error in the options provided.
In conclusion, the correct answer cannot be determined from the given options.
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A system of four particles is in x-y plane of these two particle of masses 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 moment of inertia tensor of the system of particle isa)10mb)-10mc)0md)-2mCorrect answer is option 'C'. Can you explain this answer?
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