InterviewSolution
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InterviewSolution
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A solid cylinder rolls down an inclined plane of height `3m` and reaches the bottom of plane with angular velocity of `2sqrt2 rad//s`. The radius of cylinder must be [take `g=10m//s^(2)`]A. 5 cmB. 0.5 cmC. `sqrt(10)` cmD. `sqrt(5)` m |
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Answer» Correct Answer - D The relation between linear velocity `(v)` and angular velocity `(omega)` is `v=r omegarArr r=(v)/(omega)` Total kinetic energy `= (1)/(2)mv^(2)+(1)/(2)Iomega^(2)` `=(1)/(2)mv^(2)+(1)/(2)xx(1)/(2)mr^(2)omega^(2)=(1)/(2)mv^(2)+(1)/(4)mv^(2)=(3)/(4)mv^(2)` `(3)/(4)mv^(2)=mg.3rArrv=2 sqrt(10)` `because omegar=vrArrr=(v)/(omega)=(2sqrt(10))/(2 sqrt(2))=sqrt(5)` m. |
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