3 masses m1, m2, and m3 are kept at vertices of an equilateral triangle a, b,c respectively. find the moment of inertia when triangle is rotated about an altitude passing through m1
To find the moment of inertia when the equilateral triangle is rotated about an altitude passing through m1, we can use the parallel-axis theorem. The parallel-axis theorem states that the moment of inertia of a body about any axis parallel to and a distance 'd' away from an axis through the center of mass is equal to the sum of the moment of inertia about the center of mass and the product of the mass and the square of the distance 'd' between the two axes.
Here's how we can apply the parallel-axis theorem in this case:
1. Determine the moment of inertia of each individual mass about the axis passing through its center of mass:
The moment of inertia of a point mass about an axis perpendicular to the plane containing the mass and passing through the mass is given by the formula I = mr^2, where 'm' is the mass of the object and 'r' is the distance of the object from the axis. In this case, since the masses are located at the vertices of the equilateral triangle, the distance of each mass from the axis is equal to the length of the side of the equilateral triangle.
So, the moment of inertia of each mass about the axis passing through its center of mass is:
I1 = m1a^2 (moment of inertia of mass m1 about its center of mass)
I2 = m2a^2 (moment of inertia of mass m2 about its center of mass)
I3 = m3a^2 (moment of inertia of mass m3 about its center of mass)
2. Find the distance between the two axes:
The distance between the axis passing through m1 and the axis passing through the center of mass of the equilateral triangle is equal to the altitude of the equilateral triangle. Let's denote this distance as 'h'.
3. Apply the parallel-axis theorem:
The moment of inertia of each mass about the axis passing through m1 can be given by:
I'1 = I1 + m1h^2
I'2 = I2 + m2h^2
I'3 = I3 + m3h^2
4. Calculate the total moment of inertia:
The moment of inertia of the entire system when the triangle is rotated about an altitude passing through m1 is given by the sum of the individual moments of inertia:
Moment of inertia = I'1 + I'2 + I'3
Substituting the expressions:
Moment of inertia = (m1a^2 + m1h^2) + (m2a^2 + m2h^2) + (m3a^2 + m3h^2)
Simplifying further, we can factor out 'a^2' from the first term and 'h^2' from the second term:
Moment of inertia = a^2(m1 + m2 + m3) + h^2(m1 + m2 + m3)
Therefore, the moment of inertia when the equilateral triangle is rotated about an altitude passing through m1 is given by:
Moment of inertia = a^2(m1 + m2 + m3) + h^2(m1 + m2 + m3)