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A rod of length l is pivoted about a hor...

A rod of length `l` is pivoted about a horizontal , frictionless pin through one end. The rod is released from ret in a vertical position. Find the velocity of the `CM` of the rod when the rod is inclined at an angle `theta` from the vertical.

Text Solution

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The fall in postion of the `CM` of the rod, `h=1/2(1-costheta)`
In the process, decrease in `E` is equal to increase in rotational
`KE` of the rod, so `mgh=1/2Iomega^(2)`
or `mg1/2((3g)/l(1-costheta))`
The velocity of the CM of the rod `V_(CM)=omegar` is
`sqrt((3g)/l(1-costheta))xxl/2=sqrt((3gl(1-costheta))/4`
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Knowledge Check

  • A slender rod of mass m and length L is pivoted about a horizontal axis through one end and released from rest at an angle of 30^@ above the horizontal. The force exerted by the pivot on the rod at the instant when the rod passes through a horizontal position is

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    `(sqrt(10))/4mg` alonng a line making an angle of `tan^-1(1/3)` with the horizontal
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    `(sqrt(10))/4`mg along a line making an angle of `tan^(-1)(3)` with the horizontal

  • A rod of length L whose lower end is fixed along the horizontal plane starts to topple from the vertical position. The velocity of the upper end of the rod when it hits the ground is

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  • In the above problem, if the rod is released from horizontal position, the angular velocity of the rod as it passes the vertical position is (l=length fo rod)

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