The maximum permissible speed of a car on a banked rough road can be obtained as under.Ncosθ = mg + fsinθ

Nsinθ + fcosθ = mv²/R

If v = v_{max} then, f = f_{max} = μ_{s}N

On substituting then simplifying we get

\({{v}_{max}}={{\left( \left( \frac{\left( {{\mu }_{s}}+tan\theta \right)}{\left( 1-{{\mu }_{s}}tan\theta \right)} \right)Rg \right)}^{1/2}}\).

**EXAMPLE 1:- **A small block of mass m is moving over a smooth horizontal surface on a circular path at some moment. Length of string is L and velocity of block at some moment behind the tension in the string.

**Solution:** Here Tension provides, the required centripetal force

T = mv²/r = mvₒ²/L

**EXAMPLE 2:- **Find the tangential acceleration of the bob of mass m at a moment when string makes angle θ with vertical as shown.

**Solution:** Let us make F.B.D. of bob, tension is radial, only force along tangent is

F_{T} = mgsinθ

a_{T} = F/m = gsinθ