Question

What is the maximum speed a car can take in a 20m-radius curve along a road if the coefficient of static friction between the tyres and the road is 0.60?

Answer 1

$v=10.84\ m.s^{-1}$ is the maximum speed that car can take at the turn.

Explanation:

Given:

radius of the turn, $r=20\ m$

coefficient of friction between the road and tyres, $\mu=0.6$

Now we know that during the turn of a car its centripetal force balances the friction to avoid skidding of car:

$F_c=f$

$m.\frac{v^2}{r} =\mu.N$

where:

m = mass of the car

v = speed of car at the turn

r = radius of the turn

N = normal reaction

so,

$m.\frac{v^2}{r} =\mu.m.g$

$\frac{v^2}{r} =\mu.g$

$v=\sqrt{\mu.g.r}$

$v=\sqrt{0.6\times 9.8\times 20}$

$v=10.84\ m.s^{-1}$

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TOPIC: centripetal force

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Answer 2

Answer:

a car moves in a 20m-radius curve along a road if the coefficient of static friction between the tyres and the road is 0.60, the maximum speed which this car can take  will be $10.95\ m/s$

Explanation:

If 'r' is the radius of the curve, 'g' is the acceleration due to gravity,' $v$' is the maximum speed of the car, 'm' be the mass of the car and 'μ' be the coefficient of static friction.

The centripetal force , $F=\frac{mv^{2}}{r}$ ...(1)

The centripetal force will equivalise frictional force (f) which is given by,

$f =\mu mg$  ...(2)

Using (1) and (2),

$\frac{mv^{2}}{r} = \mu mg$   ...(3)

By rearranging the above equation, we can obtain the expression for v.

Then,

  $v^{2} = \mu rg\\ v \ = \sqrt{ \mu rg}$ ...(4)

In the question, it is given that,

r =  $20$ m

g = $10\ m/s^{2}$

$\mu = 0.60$

Substitute all these values into equation (1).

Then,

$v \ = \sqrt{ 0.60\times20\times10}$

   $= \sqrt{120}$

   $= 10.95\ m/s$

Maximum speed  =  $10.95\ m/s$