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End `A` of a rod `AB` is being pulled on the floor with a constant velocity `v_(0)` as shown. Taking the length of the rod as `l`, at an instant when the rod makes an angle `37^@` with the horizontal, calculate the velocity of the `CM` of the rodA. `5/7v_(0)` at `tan^(-1)4/3` below horizontalB. `5/7v_(0)` at `tan^(-1)3/4` below horizontalC. `5/6v_(0)` at `tan^(-1)3/4` below horizontalD. `5/6v_(0)` at `tan^(-1) 4/3` below horizontal |
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Answer» Correct Answer - D `x_(CM)=x/2` `impliesV_(CM)=1/2(dx)/(dt)0=(v_(0))/2(rarr)` `implies y_(CM)=y/2` `y_(CM)=1/2 (dy)/(dt)=(v_(B))/2=2/3v_(0)darr` `V_(CM)=sqrt((v_(0)^(2))/4+(4v_(0)^(2))/9)=5/6v_(0)` at `tan^(-1)(4/3)` below horizontal |
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