A man of mass `m` stands on a horizontal platform in the shape of a disc of mass `m` and radius `R`, pivoted on a vertical axis through its centre about which it can freely rotate. The man starts to move around the centre of the disc in a circle of radius `r` with a velocity `v` relative to the disc. Calculate the angular velocity of the disc.
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Since there is no torque acting about the axis of rotation of the disc, so the angular momentum of the system (disc`+`man) remains constant. Initially it is zero. Suppose `omega` is the angular velocity of the disc (take positive in the sense of motion of the man). The velocity of the man with respect to the ground observer will be: `vecv_("man, disc")=vecv_("man")-vecv("disc")` `impliesvecv_("man")=vecv("man,disc")-vecv_("disc")` or `vecv_("man")=v-omegar` Thus, angular momentum of the man `=m(v-omegar)r` And angular momentum of the disc is `I_("disc") omega=(MR^(2))/2omega` By conservation of angular momentum, we have `m(v-omegar)=(MR^(2))/2omega` After solving, we get `omega=(mvr)/((mvr^(2)+(MR^(2))/2)`
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