- #1
Satvik Pandey
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- 12
Homework Statement
A thin rigid rod of length ##l## is placed perfectly vertical on a smooth ground. A slight disturbance on the upper end of rod causes lower end of rod to slip along the ground and rod starts falling down. Find velocity of center of mass of rod when the rod makes ##30## with horizontal.
2.Relevant equations
The Attempt at a Solution
I first tried to find the instantaneous axis of rotation--
So distance of IAOR from Com of rod is ## l cos\theta /2 ##
Now by conservation of energy
##\frac { mgl(1-sin\theta ) }{ 2 } =\frac { 1 }{ 2 } \left( \frac { m{ l }^{ 2 } }{ 12 } +\frac { m{ l }^{ 2 }{ cos }^{ 2 }\theta }{ 4 } \right) { \omega }^{ 2 }##
Also ##{ V }_{ CM }=\omega \frac { lcos\theta }{ 2 } ##
From these two equations I got ##{ V }_{ CM }=\sqrt { \frac { 9gl }{ 26 } } ##
I have confusion that why the lower end of the rod started moving backward.
I know that as only vertical force mg acts on the rod so the CoM of the rod falls vertically.
But why the lower end of the rod sliped backward?
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