A rigid rod has a mass M and a length L. This rod is formed from two equal-lengt

Question

A rigid rod has a mass M and a length L. This rod is formed from two equal-length uniform rods made of different materials that are fastened together at their ends. One of the halves of the rod has four times the mass of the other. Show that center of mass of the rod is 13/20 of its total length from its lighter end. Find the rotational inertia of the rod about its center of mass in terms of M and L Strings are attached to both ends of the rod, and it is hung from the ceiling, attached to it at points that are by a distance greater than L (so the strings do not hang vertically). If the string holding up the heavier end of the makes and angle of 12 degree with the vertical, find the angle the other string makes with the vertical.

Explanation / Answer

Let the masses be m1 and m2

m1 + m2 = M

m1 = 4m2

Hence, m1 = 4/5M and m2 = M/5

Lets write the COM from lighter end

X = M/5 * L/4 + 4M/5 * 3L/4/M

X = 13/20 L

THe moment of inertia will be given as

I = m1+ m2 (L/4)^2 / 3

I = ML^2/48

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