Question

A cyclist is riding a bicycle at a speed of 14√3 m/s takes a turnnaround a circular road of radius 20√3 what is his inclination to the vertical

Answer 1

cyclist rides the bicycle at speed, v = 14√3 m/s .
radius of circular road , r = 20√3 m.

Let inclination angle with vertical by cyclist is A.
a normal reaction acts between floor and bicycle.
then at equilibrium,
NsinA = mv²/r ...........(i)
NcosA = mg.........(ii)

dividing equations (i) by (ii),
tanA = v²/rg
now put v = 14√3 , r = 20√3 and g = 9.8m/s
so , tanA = (14√3)²/(20√3 × 9.8)
= 196 × 3/196√3
= 196√3/196
= √3
hence, tanA = √3 = tan60°
so, A = 60°

Answer 2

Answer:

Step-by-step explanation:

We are given,

v = 14√3 m/sec

And,

r = 20√3 m

Therefore,using the formula

Tan ¤ = v²/rg

We get,

Tan ¤ = (14√3)² / {20√3 x 10}

(i have taken,g = 10 m/s²)

Solving, we get,

Tan ¤ = √3

That is,

¤ = 60•