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Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD is
- a)3 ml2
- b)ml2
- c)2 ml2
- d)√3 ml2
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Four point masses, each of value m, are placed at the corners of a squ...
IA=0 (on the Axis .)
IB=ID=mr2
=m(l/√2)2
=ml2/2
IC=m(√2l)2
IC=2ml2
I=IA+IB+IC+ID
I=0+ml2/2+ml2/2+2ml2
I=3ml2
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Four point masses, each of value m, are placed at the corners of a squ...
Calculation of Moment of Inertia of Four Point Masses
Given:
- Four point masses of value m are placed at the corners of a square ABCD of side l.
- The axis passes through A and is parallel to BD.
To find: The moment of inertia of this system.
Solution:
- Let's assume the square ABCD is in the XY plane.
- The position vectors of the four masses with respect to the origin O are:
- r1 = (0, 0, 0)
- r2 = (l, 0, 0)
- r3 = (l, l, 0)
- r4 = (0, l, 0)
- The moment of inertia of each mass about the axis passing through A and parallel to BD is:
- I = m * (d^2 + h^2)
- Here, d is the distance of the mass from the axis and h is the perpendicular distance of the mass from the plane containing the axis and the center of mass of the system.
- The center of mass of the system is at the point O' = (l/2, l/2, 0).
- The distances of the masses from the axis are:
- d1 = l/2
- d2 = l/2 * sqrt(2)
- d3 = l/2
- d4 = l/2 * sqrt(2)
- The perpendicular distances of the masses from the plane containing the axis and the center of mass of the system are:
- h1 = l/2
- h2 = l/2
- h3 = l/2 * sqrt(2)
- h4 = l/2 * sqrt(2)
- Therefore, the moment of inertia of the system about the axis passing through A and parallel to BD is:
- I = m * (d1^2 + h1^2 + d2^2 + h2^2 + d3^2 + h3^2 + d4^2 + h4^2)
- I = m * (l^2 + l^2/2 + l^2/2 + l^2/2 + l^2/2 + l^2 + l^2/2 + l^2/2)
- I = 3 * m * l^2
Answer: The moment of inertia of the system about the axis passing through A and parallel to BD is 3ml^2.
Given:
- Four point masses of value m are placed at the corners of a square ABCD of side l.
- The axis passes through A and is parallel to BD.
To find: The moment of inertia of this system.
Solution:
- Let's assume the square ABCD is in the XY plane.
- The position vectors of the four masses with respect to the origin O are:
- r1 = (0, 0, 0)
- r2 = (l, 0, 0)
- r3 = (l, l, 0)
- r4 = (0, l, 0)
- The moment of inertia of each mass about the axis passing through A and parallel to BD is:
- I = m * (d^2 + h^2)
- Here, d is the distance of the mass from the axis and h is the perpendicular distance of the mass from the plane containing the axis and the center of mass of the system.
- The center of mass of the system is at the point O' = (l/2, l/2, 0).
- The distances of the masses from the axis are:
- d1 = l/2
- d2 = l/2 * sqrt(2)
- d3 = l/2
- d4 = l/2 * sqrt(2)
- The perpendicular distances of the masses from the plane containing the axis and the center of mass of the system are:
- h1 = l/2
- h2 = l/2
- h3 = l/2 * sqrt(2)
- h4 = l/2 * sqrt(2)
- Therefore, the moment of inertia of the system about the axis passing through A and parallel to BD is:
- I = m * (d1^2 + h1^2 + d2^2 + h2^2 + d3^2 + h3^2 + d4^2 + h4^2)
- I = m * (l^2 + l^2/2 + l^2/2 + l^2/2 + l^2/2 + l^2 + l^2/2 + l^2/2)
- I = 3 * m * l^2
Answer: The moment of inertia of the system about the axis passing through A and parallel to BD is 3ml^2.
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