How do you relate the maximum speed of a vehicle along a banked Road with the mass of the vehicle?
Answer
569.4k+ views
Hint: In this question, we need to find the maximum speed of the vehicle at which the vehicle doesn’t skid while moving along a banked road. We basically balance all the factors of the forces exerted on a vehicle.
Complete step by step answer:
A banked turn (or banking turn) is a turn or change of direction in which the vehicle banks or inclines, usually towards the inside of the turn. For a road or railroad this is usually due to the roadbed having a transverse down-slope towards the inside of the curve. The bank angle is the angle at which the vehicle is inclined about its longitudinal axis with respect to the horizontal. Force acting towards the center m vehicle i.e., centripetal force
$F = \dfrac{{m{v^2}}}{R}$
This force is provided by the force of static friction
$F = {f_s}$
$\Rightarrow {f_s} = \dfrac{{m{v^2}}}{r}$
But Static friction is equal or less than the limiting friction
${f_s} \leqslant {f_l}$
$\Rightarrow \dfrac{{m{v^2}}}{r} \leqslant \mu \times N$
Giving the value of N as mg in the above equation we get,
$\dfrac{{m{v^2}}}{r} \leqslant \mu \times mg$
Rearranging and taking the root to get the value of v
$v \leqslant \sqrt {\mu \times r \times g} $
So here we get the value of v as,
$\therefore {v_{\max }} = \sqrt {\mu rg} $
Friction is an opposing force that is set up between the surface of contact, when one body slides or rolls or tends to do so on the surface of another body. Numerical value of static friction is equal to external force which creates the tendency of motion of the body. So, its nature is self-adjusting. Maximum value of static friction is called limiting friction.
${f_s} \to {\text{static friction}}$
${f_s} \to static{\text{ }}friction$
${\text{N : Normal from the ground}}$
Note:Roads are banked to prevent high speed vehicles from skidding. If a vehicle is moving at a speed more than mentioned above, it will skid. If the force of friction is not strong enough, the vehicle will skid. Also, note that friction acts downwards along the road, as the tendency of the vehicle is to skid up.
Complete step by step answer:
A banked turn (or banking turn) is a turn or change of direction in which the vehicle banks or inclines, usually towards the inside of the turn. For a road or railroad this is usually due to the roadbed having a transverse down-slope towards the inside of the curve. The bank angle is the angle at which the vehicle is inclined about its longitudinal axis with respect to the horizontal. Force acting towards the center m vehicle i.e., centripetal force
$F = \dfrac{{m{v^2}}}{R}$
This force is provided by the force of static friction
$F = {f_s}$
$\Rightarrow {f_s} = \dfrac{{m{v^2}}}{r}$
But Static friction is equal or less than the limiting friction
${f_s} \leqslant {f_l}$
$\Rightarrow \dfrac{{m{v^2}}}{r} \leqslant \mu \times N$
Giving the value of N as mg in the above equation we get,
$\dfrac{{m{v^2}}}{r} \leqslant \mu \times mg$
Rearranging and taking the root to get the value of v
$v \leqslant \sqrt {\mu \times r \times g} $
So here we get the value of v as,
$\therefore {v_{\max }} = \sqrt {\mu rg} $
Friction is an opposing force that is set up between the surface of contact, when one body slides or rolls or tends to do so on the surface of another body. Numerical value of static friction is equal to external force which creates the tendency of motion of the body. So, its nature is self-adjusting. Maximum value of static friction is called limiting friction.
${f_s} \to {\text{static friction}}$
${f_s} \to static{\text{ }}friction$
${\text{N : Normal from the ground}}$
Note:Roads are banked to prevent high speed vehicles from skidding. If a vehicle is moving at a speed more than mentioned above, it will skid. If the force of friction is not strong enough, the vehicle will skid. Also, note that friction acts downwards along the road, as the tendency of the vehicle is to skid up.
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