A ring of mass m is rolling without slipping with linear velocity v as shown is figure. A rod of identical mass is fixed alone one of its diameter. The total kinetic energy of the system is

Options

A. `7/5 mv^(2)`
B. `2/3 mv^(2)`
C. `4/3 mv^(2)`
D. `5/3 mv^(2)`

Answer

D

Explanation

`K=K_(T) + K_(R) = 1/2(2m)(v^(2)) + 1/2[mR^(2) + (m(2R)^(2))/12][v/R]^(2) = 5/3 mv^(2)`