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A circular table with smooth horizontal surface is rotating at an angular speed `omega` about its axis. A groove is made on the surface along a radius and a small particle is gently placed inside the groove at a distance l from the centre. Find the speed of the particle with respect to the table as its distance from the centre becomes L.A. `v=omegal`B. `v=omega(l-a)`C. `v=(omega(l+a))/(2)`D. `v=omegasqrt(l^(2)-a^(2))` |
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Answer» `a=omega^(2)xor(vdv)/(dx)=omega^(2)x` `int_(0)^(v) vdv =int_(a)^(t)omega^(2)xdx` `v=omegasqrt(l^(2)-a^(2))` |
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