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The rod of length l=1 m rotates with an ...

The rod of length `l=1 m` rotates with an angular velocity `omega=2 rads^(-1)` an the point `P` moves with velocity `v=1ms^(-1)` and acceleration `a=1ms^(-2)`. Find the velocity and acceleration of `Q`.

Text Solution

Verified by Experts

`v_(Q)=|vecv_(QP)+vecv_(P)|=sqrt(l^(2)omega^(2)+v^(2)+2vlomega120^@)`

`=sqrt((1xx2)^(2)+1^(2)+2.2(1)(2)xx(1/2))=1ms^(-1)`
`a_(Q)=|veca_(QP)+veca_(P)|=sqrt(a^(2)_(QP)+a^(2)_(P)+2a_(QP)a_(P)cos30^@)`
where `a_(QP)=lomega^(2)=2(2)^(2)=4ms^(-2)` and `a_(P)=1`
`sqrt(4^(2)+1^(2)+2.4xx1xx(sqrt(3))/2)=sqrt(17+4sqrt(3))ms^(-2)`
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