A ring of mass m and radius R has three parti | vedclass

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(A) The system is rolling without slipping on a horizontal surface. The point of contact with the ground is the instantaneous axis of rotation $(IAR)$

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(A) The system is rolling without slipping on a horizontal surface. The point of contact with the ground is the instantaneous axis of rotation $(IAR)$.For pure rolling,the velocity of the centre of the ring is $v_0 = R\omega$,where $\omega$ is the angular velocity.The kinetic energy of the system is given by $KE = \frac{1}{2} I_{IAR} \omega^2$.First,we calculate the moment of inertia $I_{IAR}$ about the point of contact.The ring has mass $m$ and radius $R$. The particles have masses $m$,$2m$,and $m$ at positions $(0, R)$,$(-R, 0)$,and $(R, 0)$ relative to the centre,respectively.The point of contact is at $(0, -R)$.The squared distances of the particles from the point of contact are:For particle of mass $m$ at $(0, R)$: $r_1^2 = 0^2 + (R - (-R))^2 = (2R)^2 = 4R^2$.For particle of mass $2m$ at $(-R, 0)$: $r_2^2 = (-R - 0)^2 + (0 - (-R))^2 = R^2 + R^2 = 2R^2$.For particle of mass $m$ at $(R, 0)$: $r_3^2 = (R - 0)^2 + (0 - (-R))^2 = R^2 + R^2 = 2R^2$.For the ring of mass $m$,$I_{ring, IAR} = I_{cm} + mR^2 = mR^2 + mR^2 = 2mR^2$.Total $I_{IAR} = m(4R^2) + 2m(2R^2) + m(2R^2) + 2mR^2 = 4mR^2 + 4mR^2 + 2mR^2 + 2mR^2 = 12mR^2$.Substituting $I_{IAR}$ and $\omega = v_0/R$ into the kinetic energy formula:$KE = \frac{1}{2} (12mR^2) (v_0/R)^2 = 6mR^2 (v_0^2/R^2) = 6mv_0^2$.

$A$ ring of mass $m$ and radius $R$ has three particles attached to the ring as shown in the figure. The centre of the ring has a speed $v_0$. The kinetic energy of the system is: (Slipping is absent) (in $, mv_0^2$)

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