A pendulum bob of mass .m. is held out in the horizontal position and then released from rest. If the string is of length l(I) what is the velocity of the bob and tension in the string when the bob reaches the lowest position (2) If the breaking of the string.

A pendulum bob of mass .m. is held out in the horizontal position and

1.	A pendulum bob of mass .m. is held out in the horizontal position and then released from rest. If the string is of length l(I) what is the velocity of the bob and tension in the string when the bob reaches the lowest position (2) If the breaking of the string.
Answer» Solution :From law of consevation of mechanical energy. `0 + MGL = (1)/(2) mv^(2) + 0 "" v=sqrt(2GL)` The tension in the string at the lowest position `=(mv^(2))/(r)+mg=(m(2gl))/(L)+mg(thereforer=l)=3MG` (2) `mgl(1-costheta)=(1)/(2)mv^(2)orv^(2)=2gl(1-costheta)` Tension in the string at the lowest position `T=(mv^(2))/(l)+mg` `thereforenmg=(m)/(l)[2gl(1-costheta)]+mg` on simplification `cos theta = (3-n)/(2)`