Question

pls answer this question correctly and explanation plss...

Q11 only.....
Obtain the formula for the maximum safe speed of a vehicle on a level curved road......​

Answer 1

Answer:

It doesn't matter if you are driving a Hummer or a Hugo (well, at least as far as the physics of turning, anyway...), the maximum speed on this curve will be 12 m/s for any vehicle with the same coefficient of friction.

Answer 2

Question:

Obtain the formula for maximum safe speed of a vehicle on a level curved road.

Answer:

$\bigstar{\bold{V_{max}=\sqrt{\mu_s\:r\:g}} }$

Explanation:

➤ Consider a car of weight mg moving along a circular track of radius r with constant velocity v.

➤ Here the force of friction f₁ act inward the inner tyre and force f₂ act on the outer tyre.

➤ Consider R₁ and R₂ as normal reactions exerted by the tyres on the ground.

➤ Hence,

    $\sf{f_1=\mu_s\:R_1}$

    $\sf{f_2=\mu_s\:R_2}$

   where $\mu_s$ is the coefficient of static friction.

➤Hence the total frictional force F is given by

   F = f₁ + f₂

   $\sf{F=\mu_s\:R_1+\mu_s\:R_2}$

➤ Here R is the total reaction force which is given by R₁ + R₂

➤ Hence,

    $\sf{f=\mu_s\:R}$

➤ Here the total frictional force must be less than $\mu_s\:R$

    $\sf{f\leq \mu_s\:R}$

➤ The frictional force here is provided by the centripetal force f

    f = mv²/r

➤ Hence,

    $\sf{\dfrac{mv^{2} }{r } \leq \mu_s\:R----(1)}$

➤ The total normal reaction balances the weight of the car

➤ Hence,

    R = mg

➤ Substitute this in equation 1

    $\sf{\dfrac{mv^{2} }{r } \leq \mu_s\:mg}$

➤ Cancelling m on both sides,

    $\sf{\dfrac{v^{2} }{r } \leq \mu_s\:g}$

   $\sf{v^{2} =\mu_s\:r\:g}$

  $\sf{v=\sqrt{\mu_s\:r\:g}$

➤ Hence the maximum velocity with wich the car can travel is $\sf{\sqrt{\mu_s\:r\:g}$

$\boxed{\bold{V_{max}=\sqrt{\mu_s\:r\:g}} }$