1.

A ring rolls down an inclined plane. At the bottom its kinetic energy is E. Theratio of its rotational K.E. to the translational K.E. is :

Answer»

`1:4`
`1:2`
`1:1`
`2:1`

SOLUTION :Rotational K.E. `=(1)/(2)IOMEGA^(2)=(1)/(2)MR^(2)OMEGA^(2)=(1)/(2)mv^(2)`
Translational K.E. = `(1)/(2)mv^(2)`
`therefore (R.K.E.)/(T.K.E.)=(1)/(1)`


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