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A ring of mass m is rolling without slip...

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

A

`7/5 mv^(2)`

B

`2/3 mv^(2)`

C

`4/3 mv^(2)`

D

`5/3 mv^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`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)`
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