Question

A uniform rod of mass m and length l is rotating with constant angular velocity w about its axis which passes through its one end and perpendicular to the length of rod. the area of cross section of rod is A and its Youngs Modulus is Y. The strain at the midpoint of the rod (neglect gravity)

Answer 1

@hartnn ???? @ganeshie8

Answer 2

@Callisto ??

Answer 3

@jhonyy9 ??@thomaster ??

Answer 4

i drew FBD of right half and got the force as \[F=\frac{ m \omega^2(l^2-x^2) }{ 2l }\]

Answer 5

as we know that \[\Delta l=\frac{ FL }{ AY }\]

Answer 6

\[\Delta L=\int\limits_{0}^{\frac{ L }{ 2 }}\frac{ m(l^2-x^2)\omega^2.dx }{ 2LAY }\]

Answer 7

iam getting answer as \[\frac{ 11m \omega^2 }{ 48AY }\] but it is not correct cant find my mistake

Answer 8

@hartnn ??

Answer 9

@Kainui ??

Answer 10

@thomaster ?

Answer 11

@calculusxy

Answer 12

We are missing some of the question posed. The contribution to centripetal force dF due to the acceleration for an element of mass dm at distance r from the axis of rotation will be dm r w^2 and one would need to integrate this. Since "midpoint" is mentioned, perhaps using the center of mass approximation will suffice, giving you a force of approximately m (l/2) w^2 with which to estimate deflection by including A and the modulus of elasticity..