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Q. A coin is placed on a rotating turn table slips if it is placed at a minimum distance of 9cm from the centre. If the angular velocity of the turn table is tripled, it wilt just slip if the distance from the centre is:

EAMCETEAMCET 1998

Solution:

Centripetal force on coin placed on a rotating table is $ =mr{{\omega }^{2}} $ In both case the force required is same. Hence, $ m{{r}_{1}}\omega _{1}^{2}=m{{r}_{2}}\omega _{2}^{2} $ Given: $ {{r}_{1}}=9cm $ $ {{\omega }_{2}}=3{{\omega }_{1}} $ $ 9\times \omega _{1}^{2},={{r}_{2}}{{(3{{\omega }_{1}})}^{2}} $ $ 9\omega _{1}^{2}={{r}_{2}}9\omega _{1}^{2} $ $ {{r}_{2}}=\frac{9}{9}=1cm $