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How to solve this? A smooth hemispherical bowl, of internal radius r, is fixed with its rim horizontal, a thin uniform rod of length l(l < 4r) rests with one end inside the bowl and the other projecting beyond the rim. If the rod is inclined at an angle θ to the horizontal, show that 4r cos2θ = lcosθ. Find also the reactions between the rod and the bowl in terms of θ and w (the weigh of the rod) only
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