A cyclist riding at a speed of 9.8 m/s takes a turn around a circular road of radius 19.6 m. What is his inclination to the vertical?

Question

A cyclist riding at a speed of 9.8 m/s takes a turn around a circular road of radius 19.6 m. What is his inclination to the vertical? (A)  30° (B) 45° (C) 10.5° (D) 26.5°

Tardigrade Admin · Accepted Answer

Angle of incline tan θ=(v2/r g) where v is velocity, r the radius of circular path and g the acceleration due to gravity. Given, v=9.8 m/s, r=19.6 m, g=9.8 m/s 2 ∴ tan θ=(9.8 × 9.8/19.6 × 9.8)=(1/2) ⇒ θ=26.5°

Q. A cyclist riding at a speed of $9.8 m / s$ takes a turn around a circular road of radius $19.6 m$ . What is his inclination to the vertical?

$3 0^{\circ}$

$4 5^{\circ}$

$10. 5^{\circ}$

$26. 5^{\circ}$

Angle of incline $tan θ = \frac{v ^{2}}{r g}$
where $v$ is velocity, $r$ the radius of circular path and $g$ the acceleration due to gravity.
Given, $v = 9.8 m / s, r = 19.6 m, g = 9.8 m / s^{2}$
$∴ tan θ = \frac{9.8 \times 9.8}{19.6 \times 9.8} = \frac{1}{2}$
$\Rightarrow θ = 26. 5^{\circ}$

Q. A cyclist riding at a speed of 9.8m/s takes a turn around a circular road of radius 19.6m. What is his inclination to the vertical?

30∘

45∘

10.5∘

26.5∘