If a solid sphere of mass 1 kg and radius 0.1 m rolls without slipping at a uniform velocity of 1 m/s along a straight line on a horizontal floor, the kinetic energy is ……….

Options

A. `(7)/(5)J`
B. `(2)/(5)J`
C. `(7)/(10)J`
D. `1J`

Answer

C

Explanation

Total kinetic energy of ball rolling on a horizontal surface
= (translational kinetic energy) + (rotational kinetic energy)
`=(1)/(2)mv^(2)+(1)/(2)I.omega^(2)`
Moment of inertia `I=(2)/(5)MR^(2)` (for sphere) and `omega=(v)/(R )`
`therefore` Total kinetic energy
`=(1)/(2)mv^(2)+(1)/(2).(2)/(5)mR^(2).(v^(2))/(R^(2))`
`=(1)/(2)mv^(2)+(1)/(5)mv^(2)`
`=(7)/(10)mv^(2)`
Putting `m=1kg, v=1m//s`
Total energy `=(7)/(10)xx1xx(1)^(2)`
`=(7)/(10)J`