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A uniform solid cone of mass m, base radius ‘R’ and height 2R, has a smooth groove along its slant height as shown in figure. The cone is rotating with angular speed `omega`, about the axis of symmetry. If a particle of mass ‘m’ is released from apex of cone, to slide along the groove, then angular speed of cone when particle reaches to the base of cone is A. `(3 omega)/13`B. `(4 omega)/13`C. `(5 omega)/13`D. `( 9 omega)/13` |
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Answer» Initially `I_(1)=3/10 mR^(2) & omega_(1)=omega` Finally `I_(2) =13/10 mR^(2)& omega_(2)=omega_("new")` Using conservation of anular momentum `I_(1)omega_(1)=I_(2)omega_(2)` `omega_(2)=omega_("new")=(3omega)/13` |
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