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Question: A 2000 kg car has to go over a turn whose radius is 750 m and the angle of slope is \[5^\circ \]. Th...

A 2000 kg car has to go over a turn whose radius is 750 m and the angle of slope is 55^\circ . The coefficient of friction between the wheels and the road is 0.50.5. What should be the maximum speed of the car so that it may go over the turn without slipping?

Explanation

Solution

The maximum velocity is directly proportional to the square root of the product of the radius of turn, the acceleration due to gravity and the effective coefficient of friction. The effective coefficient of friction is affected by the coefficient of friction between the surfaces and the banking angle of the road.
Formula used: In this solution we will be using the following formulae;
v=rgμsv = \sqrt {rg{\mu _s}} where vv is the maximum velocity of an object rounding a bend, rr is the radius of the turn, gg is the acceleration due to gravity and μs{\mu _s} is the effective coefficient of friction.
μs=μ+tanθ1μtanθ{\mu _s} = \dfrac{{\mu + \tan \theta }}{{1 - \mu \tan \theta }} where μ\mu is the coefficient of friction between the object and the surface at which it bends, θ\theta is the slant angle of the surface from the horizontal (called banking angle).

Complete Step-by-Step solution:
To find the velocity we shall recall the velocity
v=rgμsv = \sqrt {rg{\mu _s}} where, rr is the radius of the turn, gg is the acceleration due to gravity and μs{\mu _s} is the effective coefficient of friction.
But . μs=μ+tanθ1μtanθ{\mu _s} = \dfrac{{\mu + \tan \theta }}{{1 - \mu \tan \theta }} where μ\mu is the coefficient of friction between the object and the surface at which it bends, θ\theta is the slant angle of the surface from the horizontal (called banking angle).
Hence by inserting all known values we have
μs=0.5+tan51(0.5)tan5{\mu _s} = \dfrac{{0.5 + \tan 5^\circ }}{{1 - \left( {0.5} \right)\tan 5^\circ }}
μs=0.5+0.087510.5(0.0875)\Rightarrow {\mu _s} = \dfrac{{0.5 + 0.0875}}{{1 - 0.5\left( {0.0875} \right)}}
By computing, we get
μs=0.6144{\mu _s} = 0.6144
Hence,
v=rgμs=750(9.8)(0.6144)v = \sqrt {rg{\mu _s}} = \sqrt {750\left( {9.8} \right)\left( {0.6144} \right)}
Hence by computation, we get
v=67.2m/sv = 67.2m/s

Note: For clarity, observe that the banking presence of the banking angle and the coefficient of friction increases the maximum velocity than if only one was present. As an application, this is why many highways are slightly banked at their roundabouts.