Question

13.
A cyclist is moving in a circular path of radius 80 m with a velocity of
10 m/s. His angle of bend with the vertical is ( g = 10 m/s²)
(1) tan-(4)
(2) tan (5)
(3) can? (/
(4) tan-(2)​

Answer 1

Answer:

Angle made by the cyclist with the vertical is given as

$\theta = cot^{-1}(8)$

Explanation:

Let the cyclist bend at an angle theta with the vertical

so we vertical component of normal force will balance the weight of the cyclist

$N cos\theta = mg$

Other component of normal force will give centripetal force to the cyclist

So we will have

$N sin\theta = \frac{mv^2}{R}$

now we have

$tan\theta = \frac{v^2}{Rg}$

now we have

$tan\theta = \frac{10^2}{80\times 10}$

$tan\theta = \frac{1}{8}$

$\theta = cot^{-1}(8)$

#Learn

Topic : Circular motion

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Answer 2

The cyclist's angle of bend with the vertical is tan⁻¹ (1/8)

Given:

Radius of the path = r = 80 m

Velocity of the cycle = v = 10 m/s

Explanation:

The bending of the cyclist is given by the formula below:

tan θ = v²/(rg)

Where,

g = Acceleration due to gravity = 10 m/s²

On substituting the values in the formula, we get,

tan θ = (10)²/(80 × 10)

tan θ = 100/800

tan θ = 1/8

∴ θ = tan⁻¹ (1/8)