Question 225.
What will be moment of inertia about an axis perpendicular to the plane of star made by using 5 rods and
passing through one of the 5 vertices?
Answer 225.
M.I. of all 5 rods, each having mass M and length L, with respect to the axis through their centers and perpendicular to their lengths
= 5 * (ML^2) / 12
M.I. of these rods about an axis through the center of the star perpendicular to the plane of the star (using parallel axes theorem)
= (5/12) ML^2 + 5M * [(L/2)tan18°]^2
M..I. of these rods about an axis through one of the 5 vertices
perpendicular to the plane of the star (using parallel axes theorem again)
= (5/12) ML^2 + 5M * [(L/2)tan18°]^2 + 5M * [(L/2)sec18°]^2
= 5ML^2 [1/12 + (1/4)tan^2 18° + (1/4)sec^2 18°]
= 5ML^2 [0.0833 + 0.0264 + 0.2764]
= 1.9305 ML^2.
What will be moment of inertia about an axis perpendicular to the plane of star made by using 5 rods and
passing through one of the 5 vertices?
Answer 225.
M.I. of all 5 rods, each having mass M and length L, with respect to the axis through their centers and perpendicular to their lengths
= 5 * (ML^2) / 12
M.I. of these rods about an axis through the center of the star perpendicular to the plane of the star (using parallel axes theorem)
= (5/12) ML^2 + 5M * [(L/2)tan18°]^2
M..I. of these rods about an axis through one of the 5 vertices
perpendicular to the plane of the star (using parallel axes theorem again)
= (5/12) ML^2 + 5M * [(L/2)tan18°]^2 + 5M * [(L/2)sec18°]^2
= 5ML^2 [1/12 + (1/4)tan^2 18° + (1/4)sec^2 18°]
= 5ML^2 [0.0833 + 0.0264 + 0.2764]
= 1.9305 ML^2.
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