1.

A solid sphere of mass m rolls down an inclined plane without slipping from rest at the top of an inclined plane.The linear speed of the sphere at the bottom of the inclined plane is v .The kinetic energy of the sphere at the bottom is

Answer»

`(1)/(2)MV^(2)`
`(5)/(3)mv^(2)`
`(2)/(5)mv^(2)`
`(7)/(10)mv^(2)`

Solution :When the SPHERE rolls down the inclined PLANE without slipping then total K.E. = Linear K.E. + Rot. K.E.
`THEREFORE` Total K.E. `=(1)/(2)mv^(2)+(1)/(2)Iomega^(2)`
`=(1)/(2)mv^(2)+(1)/(2)MK^(2).(v^(2))/(r^(2))=(1)/(2)mv^(2)1+(K^(2))/(r^(2))`
For sphere `(K^(2))/(r^(2))=(2)/(5)`
`therefore` Total `K.E=(1)/(2)mv^(2)1+(2)/(5)=(7)/(10)mv^(2)`


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