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In the instant shown in the diagram the ...

In the instant shown in the diagram the board is moving up (vertically) with velocity `v`. The drum winds up at a constant rate `omega`. If the radius of the drum is `R` and the board always remains horizontal, find the value of velocity in terms of `R, theta, omega`.

Text Solution

Verified by Experts

The correct Answer is:
`(omegaR)/(1+costheta)`
`I=y+z` (total length at any time) `=y+sqrt(y^(2)+l_(1)^(2))`
`(dl)/(dt)=(dy)/(dt)+y/(sqrt(y^(2)+l_(1)^(2)))(dy)/(dt)`
`omega R=v(1+costheta), v=(omegaR)/(1+costheta)`
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Knowledge Check

  • In the diagram 100kg block is moving up with constant velocity, then find out the tension at point P :

    A
    `1330N`
    B
    `490N`
    C
    `1470N`
    D
    `980N`

  • In the block moves up with constant velocity v m//s. Find F

    A
    `F=(mg)/(2)`
    B
    `F=(2mg)/(3)`
    C
    `F=(mg)/(3)`
    D
    `F=(m)/(3)`

  • If block A is moving horizontally with velocity v_(A) , then find the velocity of block B at the instant as shown in fig.

    A
    `(hv_(A))/(2sqrt(x^(2)+h^(2)))`
    B
    `(xv_(A))/(sqrt(x^(2)+h^(2)))`
    C
    `(xv_(A))/(2sqrt(x^(2)+h^(2)))`
    D
    `(hv_(A))/(sqrt(x^(2)+h^(2)))`