Question

the average speed of vehicle along a circular track is 54 kmph whose radius curvature is 15m. the banking angle required for the curve is
a) tan-¹(1/2)
b) tan-¹(3/2)
c) tan-¹(1/3)
d) tan-¹(2)​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Bankment\:angle=tan^{-1}(\frac{3}{2})}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Avg. \: speed = 54 \: km/h \\ \\ \tt: \implies Radius = 15 \: m \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Inclination \: of \: curve = ?$

• According to given question :

$\tt \circ \: Speed = 54 \times \frac{5}{18} = 15 \: m/s\\ \\ \tt \circ \: Downward \: force = mg \: sin \: \theta \\ \\ \tt \circ \: Upward \: force = \frac{m {v}^{2} }{r}cos \: \theta \\ \\ \bold{For \: not \: skidding} \\ \tt: \implies Downward \: force = Upward \: force \\ \\ \tt: \implies mg \:sin \theta = \frac{m {v}^{2} }{r} cos \: \theta \\ \\ \tt: \implies \frac{sin \: \theta}{cos \: \theta} = \frac{m {v}^{2} }{r \: mg} \\ \\ \tt: \implies tan \: \theta = \frac{ {v}^{2} }{rg} \\ \\ \tt: \implies tan \: \theta = \frac{ {15}^{2} }{15 \times 10} \: \: \: \: \: \: (g = 10 \: {ms}^{2} ) \\ \\ \tt: \implies tan \: \theta = \frac{15 \times 15}{15 \times 10} \\ \\ \tt: \implies tan \: \theta = \frac{3}{2} \\ \\ \green{\tt: \implies \theta = {tan}^{ - 1} ( \frac{3}{2} )} \\ \\ \green{\tt \therefore Bankment \: of \: curve \: is \: {tan}^{ - 1} ( \frac{3}{2} )}$

Answer 2

....

$\tt{\huge{\purple{ ..................... }}}$

QUESTION -

The average speed of vehicle along a circular track is 54 kmph whose radius curvature is 15m.

The banking angle required for the curve is :

A) tan-¹(1/2)

B) tan-¹(3/2)

C) tan-¹(1/3)

D) tan-¹(2)

SOLUTION -

From the above Question, we can gather the following information

.......

The average speed of vehicle along a circular track is 54 kmph whose radius curvature is 15m.

The average Velocity Of the above vehicle = 54 kmph

=> 54 × ( 5 / 18 ) m / s

=> 15 m / s.

The radius of curvature is 15 m / s.

Now,

Let me draw the required figure.

See the above attachment ..

Now, defining the following components of Force Acting On the body when it is bending at the curve..

$F_{Upward } = [ \dfrac{ m { v } ^ 2 }{ r } ] \cos ( \phi ) ............ ( 1 )$

$F _ { Downward } = mg \sin ( \phi ) ............... ( 2 )$

Now we have to calculate the required banking angle for the curve such that the car does not slip.

This means, the system of forces are balanced..

I.e,

The force acting upwards should be equal to the force acting downwards.

$=> [ \dfrac{ m { v } ^ 2 }{ r } ] \cos ( \phi ) = mg \sin ( \phi )$

$=> \dfrac{ m { v } ^ 2 }{ r mg } = \tan ( \phi )$

Now, let us substitute the required values...

$=> \tan ( \phi ) = \dfrac{ { v } ^ 2 }{ r g } => \dfrac{ 225}{15g} => \dfrac{ 15}{g }$

$=> \phi = \tan ^ { -1 } { ( \dfrac{15}{g} ) }$

Thus the required banking angle becomes tan inverse of of ( 15 / g ) .....

ANSWER -

The required banking angle is tan inverse of of ( 15 / g ).

I have kept the answer in terms of g as the value may vary.