Question

A motor cycle racer takes a round with speed 20 m/s in a curvature of radius of R = 40 m, then the leaning angle of motor cycle for safe turn is ( g = 10 m/s^2)​

Answer 1

Explanation:

Given: The speed of motor cycle v= 20 m/s

Radius of curvature R=40m

Let θ be the angle which motorcycle racer makes with the vertical while taking a safe turn then,

tanθ=v^2/Rg= (20)^2/40*10= 1

θ=tan^-1

θ=45

Answer 2

Given:

A motor cycle racer takes a round with speed 20 m/s in a curvature of radius of R = 40 m.

To find:

Leaning angle of cycle ?

Calculation:

• Let the leaning angle be $\theta$.

General expression of leaning angle in banked road is :

$\tan( \theta) = \dfrac{ {v}^{2} }{rg}$

$\implies \: \tan( \theta) = \dfrac{ {20}^{2} }{40 \times 10}$

$\implies \: \tan( \theta) = \dfrac{ 400 }{400}$

$\implies \: \tan( \theta) = 1$

$\implies \: \theta = { \tan}^{ - 1} (1)$

$\implies \: \theta = {45}^{ \circ}$

So, leaning angle with vertical axis is 45°.