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DC Pandey Mechanics Part 2 Q. A uniform bar of length l stands vertically touching a wall OA, when slightly displaced, its lower end begins to slide along the floor. obtain an expression for the angular velocity of the bar as a function of . neglect friction everywhere. [ D C PANDEY page 27, introductory exercise 9.5, problem 2 ]

DC Pandey Mechanics Part 2


Q. A uniform bar of length l stands vertically touching a wall OA, when slightly displaced, its lower end begins to slide along the floor. obtain an expression for the angular velocity of the bar as a function of  . neglect friction everywhere.

[ D C PANDEY page 27, introductory exercise 9.5, problem 2 ]

Grade:

1 Answers

noel thekkekara
31 Points
9 years ago

the point of zero velocity  lies at a distance of l/2 from the centre of mass of the rod.

moment of inertia of the rod about that point i=(ml^2/12)+m(l/2)^2=(ml^2)/3

we know that

decrease inpotential energy=increase in kinetic energy

(mgl/2)(1-cos )=I^2

(mgl/2)(1-cos )=(ml^2)/3 * ^2

(3g/2l)(1-cos)=^2

hence =sqrt{3g(1-cos)/2l}

please approve my answer!!!!!!!!

 

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