Saved Bookmarks
| 1. |
A thin uniform rod of length L and radius r rotates with an angular velocity in a horizontal plane about a vertical axis passing through one of its ends. The density of the material is rho and Youngs modulus is Y. 1) Find tension in the rod at distance Y from the axis. 2) Find elongation of the rod |
|
Answer» Solution :1) Consider and element dx at a DISTANCE from the axis of rotation. 2) The centripetal force on this portion is `dt= dmr OMEGA^(2) dT=(rho A dr) R omega^(2)` This force is provided by the tension in the rod `int dT= overset(L) underset(r) int rho A dr omega^(2), T rho A omega^(2)overset(L) underset(r) intrdr, T= rho A omega^(2)[ (r^(2))/(2)]_(r)^(L)` `rArr T=(rhoA omega^(2))/(2)[ L^(2)-r^(2)] T=(1)/(2) rho A omega^(2)(L^(2)-r^(2))` (3) strain `=("stress")/(Y)` If dy is the elongation in element of length dx `(dy)/(dx)=(T)/(AY) rArr dy=(Tdx)/(AY) rArr dy=((1)/(2) rho A omega^(2)(L^(2)-r^(2)) dr)/(AY)` `int dy=(rho omega^(2))/(2y) overset(L) underset(0) int (L^(2)-r^(2))dr rArr DeltaL=(rho omega^(2))/(2y)[ L^(2)r-(r^(3))/(3)]_(0)^(L)` `DeltaL=(rho omega^(2))/(2Y)[L^(2)-(L^(3))/(3)] rArr DeltaL=( rho omega^(2))/(2y)[(2L)/(3)], DeltaL-=( rho omega^(2)L^(3))/(3Y)` |
|