1.

A uniform rod of length l is placed with one end in contact with the horizontal and is then inclined at an angle a to the horizontal and allowed to fall, without slipping at contact point. When it becomes horizontal, its angular velocity will be

Answer»

`omega=sqrt((3gsinalpha)/l)`
`omega=sqrt((2l)/(3gsinalpha))`
`omega=sqrt((6gsinalpha)/l)`
`omega=sqrt(l/(gsinalpha))`

Solution :By principle of conservation of energy, potential energy of the ROD is converted into rotational energy.
P.E. of rod= MG`.l/2sinalpha`
Rotational K.E. =`1/2Iomega^(2)=1/2(ML^(2))/3omega^(2)`
`mgl/2sinalpha=1/2.(ml^(2))/3.omega^(2)rArromega=sqrt((3gsinalpha)/l)`


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