Question

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.

(a) v² = μrg

(b) μg = v²/r + a_t

(c) μ²g² = v⁴/r² +a_t²

(d) The force of friction makes on angle tan ^-1 (v²/a_tr) with the direction of motion at the point of slipping

Answer 1

Correct answer is: (a, d)

The tangential acceleration a_t and the radial acceleration ar = v²/r being mutually perpendicular, have a resultant acceleration a = [a_t² +(v²/r)²] ^1/2 .The only horizontal force on the body is the force of friction, F = μmg =ma,acting in the direction of a.