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

Options

A. `(1)/(2)mv^(2)`
B. `(5)/(3)mv^(2)`
C. `(2)/(5)mv^(2)`
D. `(7)/(10)mv^(2)`

Answer

D

Explanation

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)`