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) = mu rg`B. `mu g=(v^(2))/(r) +a_(t)`C. `mu^(2)g^(2)=(u^(4))/(r^(2))+a_(t)^(2)`D. The force of friction makes an angle `tan^(-1)(v^(2)//a_(t)r)` with the direction of motion at the point of slipping. |
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Answer» Correct Answer - c,d |
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