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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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Knowledge Check

  • A rod of length 1 m rotates in the xy plane about the fixed point O in the anticlockwise sense, as shown in figure with velocity omega=a+bt where a=10rads^(-1) and b=5rads^(-2) . The velocity and acceleration of the point A at t=0 is

    A
    `+10hatims^(-1)` and `+5hatims^(-2)`
    B
    `+10hatjms^(-1)` and `(-100hati+5hatj)ms^(-2)`
    C
    `-10hatjms^(-1)` and `(100hati+5hatj)ms^(-2)`
    D
    `-10hatjms^(-1)` and `-5hatjms^(-1)`

  • A uniform rod AB of length l rotating with an angular velocity omega while its centre moves with a linear velocity v=(omegal)/(6) If the end A of the rod is suddenly fixed, the angular velocity of

    A
    `(3)/(4)omega`
    B
    `(omega)/(3)`
    C
    `(omega)/(2)`
    D
    `(2)/(3)omega`

  • A body executing SHM has a maximum velocity of 1ms^(-1) and a maximum acceleration of 4ms^(-1) . Its time period of oscillation is

    A
    `3.14s`
    B
    `1.57s`
    C
    `6.28s`
    D
    `0.25s`