$A$ uniform bar of length $6l$ and mass $8m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ moving in the same horizontal plane with speed $2v$ and $v$ respectively,strike the bar (as shown in the figure) and stick to the bar after collision. The total rotational kinetic energy about the centre of mass $c$ will be:

  • A
    $\frac{2mv^2}{5}$
  • B
    $\frac{mv^2}{5}$
  • C
    $\frac{3mv^2}{5}$
  • D
    $mv^2$

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$A$ ring starting from rest rotates under a constant angular acceleration of $8 \ rad \ s^{-2}$ due to an applied torque. How many revolutions will the ring complete in $5 \ s$? How many revolutions will it complete in the $6^{th}$ second? If the torque becomes zero after $6 \ s$,how many revolutions will the ring complete in the $7^{th}$ second?

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$A$ uniform circular disc of radius $R$,lying on a frictionless horizontal plane,is rotating with an angular velocity $\omega$ about its own axis. Another identical circular disc is gently placed on top of the first disc coaxially. The loss in rotational kinetic energy due to friction between the two discs,as they acquire a common angular velocity,is ($I$ is the moment of inertia of the disc).

$A$ uniform disc is acted upon by two equal forces of magnitude $F$. One of them acts tangentially to the disc,while the other acts at the central point of the disc. The friction between the disc surface and the ground surface is $nF$. If $r$ is the radius of the disc,then the value of $n$ would be:

$A$ uniform rod of mass $M$ is hinged at its upper end. $A$ particle of mass $m$ moving horizontally strikes the rod at its mid-point elastically. If the particle comes to rest after the collision,find the value of $M/m$.

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$STATEMENT-1$ If there is no external torque on a body about its center of mass,then the velocity of the center of mass remains constant. because
$STATEMENT-2$ The linear momentum of an isolated system remains constant.

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