Three identical point masses of mass M are fixed at the corners of an equilateral triangle of sides l as shown. Axis A runs through a point equidistant from all three masses, perpendicular to the plane of the triangle. Axis B runs through M1 and is perpendicular to the plane of the triangle. Axes C, D, and E, lie in the plane of the triangle and are as shown.1. Determine an expression for the moment of inertia of the masses about Axis E in terms of M and l. In the image, there is a triangle with vertices labeled \( M_1 \), \( M_2 \), and \( M_3 \). The sides of the triangle are intersected by several axes, labeled as Axis A, Axis B, Axis C, Axis D, and Axis E.- **Vertices and Axis B:** - Vertex \( M_1 \) is at the top of the triangle, with an arrow pointing to the right labeled Axis B. - **Axis A:** - Axis A is a line starting from vertical of \( M_1 \) and intersecting the triangle. It is marked with a red arrow pointing towards the center.- **Axis C:** - Axis C is a diagonal line intersecting the triangle and extending to the right.- **Axis D:** - Axis D is a horizontal line passing through the triangle below \( M_1 \) and above \( M_2 \) and \( M_3 \).- **Axis E and Points \( M_2 \) and \( M_3 \):** - Axis E is at the base of the triangle, running horizontally along the points \( M_2 \) and \( M_3 \). - The distance between \( M_2 \) and \( M_3 \) is labeled \( l \).The diagram likely represents a geometric or vector analysis, with axes depicting different directions or forces.

Solution: To Determine an expression for the moment of inertia of the masses about Axis E in terms of M and l is given as: If we draw a Triangle to represent the Axis E is shown below: From the above Figure, By Pythagoras Theorem, We have l2=R2+l22R2=l2-l24 =3 l24R=3 l2 The points that lie on the axis-E are M2 and M3. Therefore, The moment of inertia about axis E is equal to zero. Thus, The moment of inertia along axis E is only with M1 is given as: IE=M1R2We have,R=3l2So,The above equation becomes,IE=M 3l22IE =3 M l2 4 Thus, The moment of Inertia of the masses about Axis E in terms of M and l axis is IE =3 M l2 4

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Three identical point masses of mass M are fixed at the corners of an equilateral triangle of sides l as shown. Axis A runs through a point equidistant from all three masses, perpendicular to the plane of the triangle. Axis B runs through M1 and is perpendicular to the plane of the triangle. Axes C, D, and E, lie in the plane of the triangle and are as shown. 1. Determine an expression for the moment of inertia of the masses about Axis E in terms of M and l.

Three identical point masses of mass M are fixed at the corners of an equilateral triangle of sides l as shown. Axis A runs through a point equidistant from all three masses, perpendicular to the plane of the triangle. Axis B runs through M1 and is perpendicular to the plane of the triangle. Axes C, D, and E, lie in the plane of the triangle and are as shown. 1. Determine an expression for the moment of inertia of the masses about Axis E in terms of M and l.

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