Constrained Motion (JEE)

One end A A of rod A B AB (length 4 m 4 \ \text{m} ) is pivoted to the ground, while the other end B B is pivoted to the rod B C BC (also length 4 m 4 \ \text{m} ). The end C C of rod B C BC is such that it can only slide on the ground in a straight line. It cannot be lifted.

If the rod A B AB rotates about A A with ω = 1 rad/s \omega=1 \ \text{rad/s} as shown, then find the speed, in m/s , of the end C C when A B C = 6 0 o \angle ABC=60^{o} .


The answer is 6.928.

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2 solutions

Mvs Saketh
Nov 3, 2014

Let the length of AC be 'h'

h ˙ = v c = ( 2 l ( c o s θ ) ˙ ) = 2 l s i n θ ( w ) = 8 s i n θ \dot { h } ={ v }_{ c }=-\left( \dot { 2l(cos\theta ) } \right) =2lsin\theta (w)=8sin\theta

when angle is 60 degree,, it becomes equillateral triangle,, thus substituting, we get answer as

4 3 4\sqrt { 3 }

+1 For it saketh

Karan Shekhawat - 6 years, 7 months ago

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