InterviewSolution
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InterviewSolution
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A body moves on a horizontal circular road of radius `r`, with a tangential acceleration `a_(t)`. The coefficient of friction between the body and the road surface Is `mu`. It begins to slip when its speed is `v`. (i) `v^(2)=murg` (ii) `mug=(v^(4)/(r^92))+a_(t))` (iii) `mu^(2)g^(2)=(v^(4)/(r^(2)+a_(t)^(2))` (iv) The force of friction makes an angle `tan^(-1)(v^(2)//a_(t)r)` with the direction of motion at the point of slipping.A. `v^(2)=murg`B. `mug=v^(2)/r+a_(T)`C. `mu^(2)g^(2)=v^(4)/r^(2)+a_(T)^(2)`D. The force of friction makes an angle `tan^(-1)((v^(2))/(a_(T)xxr))` with direction of motion of point of slipping. |
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Answer» Correct Answer - C::D At time of slipping `f=mumg` `f cos theta=ma_(T), f^(2)=(ma_(T))+((mv^(2))/r)` `(mumg)^(2)=(ma_(T))^(2)+((mv^(2))/r)^(2)` `rArr mu^(2)g^(2)=a_(T)^(2)+((v^(4))/r^(2))`,Also `tan theta=v^(2)/(a_(T)r)` |
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