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This ques. was askd by biki... karthik solvd it in an unusual method... i cudn't get it. read the ques. and then go through karthik's sol^n to understand what my problem is A rod of mass m and length l , of youngs modulus = Y and area of c.s. A is rotated with angular velocity omega(w) about an axis perpendicular to the plane of the rod and passing through one end of the rod as shown. Neglect gravity and find strain at the middle point of the rod. Here is karthik's sol^n consider the 2nd half of the rod. This can be concieved of as a single mass at a distance of 3L/4 from the origin. In other words, the second half, ie, the one with the mass m/2, rotates in a radius of 3L/4. Hence, we get
F = m/2 x w2 x 3L/4.
Hence, stress = 3mw2L/8AY
here, we have calculated the force exerted by the second half, on the first half.That is, this force is uniform throughout the first half. Note that we have taken the radius to be 3L/4. My Confusion:- But how is that possible, is centrifugal force on the 2nd half = force at the middle...are the second part and the first part the same bodies ??? Howcome the force acting on the second part become equal to force acting on the middle ??? force on second part = m/2 . w^2. (3L/4)
howcome this is the force on the middle ???
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Moderation Team
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