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A ring mass m and radius R has three par...

A ring mass `m` and radius `R` has three particle attached to the ring as shown in the figure. The centre of the centre `v_(0)`. Find the kinetic energy of the system. (Slipping is absent).
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Text Solution

Verified by Experts

The correct Answer is:
`6mv_(0)^(2)`
About instantaneous axis of rotation rolling system can be considered as pure rotation
`KE = (1)/(2) I omega^(2)` …(1)
here `I` = moment of inertia about instantaneous axis of rotation
`I = 2m (sqrt(2) R)^(2) + m(2R)^(2) + m(sqrt(2) R)^(2) + I_(ring)`
`I = 2m (sqrt(2) R)^(2) + m(2R)^(2) + m(sqrt(2) R)^(2) + mR^(2)`
`I = 12 mR^(2)`
putting the value of `I` in equation (1)
`KE = 6mR^(2) omega^(2)`
`KE = 6m(R omega)^(2)`
`KE = 6mv_(o)^(2)`.
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Knowledge Check

  • A uniform ring of mass m and radius R is in uniform pure rolling motion on a horizontal surface. The velocity of the centre of ring is V_(0) . The kinetic energy of the segment ABC is:

    A
    `(mV_(0)^(2))/(2)-(mV_(0)^(2))/(pi)`
    B
    `(mv_(0)^(2))/(2)+(mv_(0)^(2))/(pi)`
    C
    `(mv_(0)^(2))/(2)`
    D
    `mv_(0)^(2)`

  • Three rings each of mass M and radius R are arranged as shown in the figure. The moment of inertia of the system about YY¢ will be

    A
    `3MR^(2)`
    B
    `(3)/(2)MR^(2)`
    C
    `5MR^(2)`
    D
    `(7)/(2)MR^(2)`

  • For a uniform ring of mass M and radius R at its centre

    A
    field and potential both are zero
    B
    field is zero but potential is `(GM)/(R)`
    C
    field is zero but potential is `-GM//R`
    D
    magnitude of field is `(GM)/(R^(2))` and potential `-(GM)/(R)`

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