Two equal masses m are fastened to the ends of a straight rod of length 21 and o

Question

Two equal masses m are fastened to the ends of a straight rod of length 21 and of negligible weight. the rod is attached to the center of a vertical shaft of length L at an angle alpha as shown. If the vertical shaft rotates at a constant angular velocity w, find the dynamic reactions at the bearings. The system is to be dynamically balanced by the addition of two concentrated masses m_1. These masses are to be located in planes at a distance a from the bearings (see picture). Show where these masses should be attached and find the radius at which they should be located.

Explanation / Answer

The mass ‘m’ at angle to the shaft of length L produces a moment about the axis = m2LSin LCos. = m2L2Sin Cos

There are 2 such masses. Hence total moment or couple that produced = 2m2L2sin Cos = m2L2Sin2

This couple has to be balanced by equal and opposite forces at the bearings. The force on bearings = m2L2Sin2/L = m2LSin2

If the couple has to be balance the added mass ‘m1’ at both sides, should produce the same magnitude in opposite direction. Let ‘e be the eccentricity of the mass m1, then the couple produced by m1 =2 m1e 2(L/2-a) and this must be equal to the unbalanced couple acting is m2L2sin2.

2 m1e 2(L/2-a) = m2L2sin2

After simplification, we get ‘e’ = eccentricity ( or radius) of mass m1 = mLSin2/{m1(L-2a)}

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