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Sahay Sir > Question Answers > Cengage NEET by C.P Singh > Part 1 > Chapter 9 - Circular Motion > A body moves on a horizontal circular road of radius r, with a tangential acceleration at. The coefficient of friction between the body and the road surface Is μ. It begins to slip when its speed is v. (i) v^2=μrg (ii) μg=(v^4/r^2)+at) (iii) μ^2g^2=(v^4/r^2+a^2t (iv) The force of friction makes an angle tan^−1(v^2/atr) with the direction of motion at the point of slipping.

A body moves on a horizontal circular road of radius r, with a tangential acceleration at. The coefficient of friction between the body and the road surface Is μ. It begins to slip when its speed is v. (i) v^2=μrg (ii) μg=(v^4/r^2)+at) (iii) μ^2g^2=(v^4/r^2+a^2t (iv) The force of friction makes an angle tan^−1(v^2/atr) with the direction of motion at the point of slipping.

A body moves on a horizontal circular road of radius r A string of length L is fixed at one end and carries a mass M at the other end. The string makes 2/π revolution per second around the vertical axis through the fixed end as shown in the figure then tension in the string is.