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 (1) what is the velocity of the bob and tension in the string when the bob reached the lowest position.(2) If the breaking strength of the string is .nmg.. Find the maximum angle .theta. with respect to the vertical at which the bob is to be released to avoid 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 (1) what is the velocity of the bob and tension in the string when the bob reached the lowest position.(2) If the breaking strength of the string is .nmg.. Find the maximum angle .theta. with respect to the vertical at which the bob is to be released to avoid breaking of the string.
Answer» Solution :(1) From law of conservation 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 (because r = l)=3MG` (2)`mgl (1-cos theta)=(1)/(2) mv^(2)` or `v^(2)=2gl (1-cos theta)` Tension in the string at the lowest position `T = (mv^(2))/(l)+mg therefore nmg.=(m)/(l)[2gl(1-cos theta)]+mg` on simpliication `cos theta = (3-n)/(2)`