Question

When a cyclist negotiates a circular path of radius ‘r’ with velocity ‘v’, making an v2
angle  with the horizontal, show that tan theta = v^2/rg

Answer 1

Answer:

for a cyclist to negotiate a circular turn of radius R the condition of angle is given as

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

Explanation:

When cyclist negotiate the circular path by making an angle "theta" with the vertical then component of the normal force will provide the centripetal force for the circular motion

so we have

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

In vertical direction we can use force balance

$N cos\theta = mg$

so from above equations

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

#Learn

Topic : Uniform circular motion

https://brainly.in/question/14394794

Answer 2

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In Bending of A cyclist, it's value is shown as :

$\huge{\boxed{\boxed{\sf{\tan \theta \: = \: \dfrac{v^2}{rg}}}}}$

See attached figure,

When the cyclist bends it forms an angle θ , in the circular motion :

So,

$\large \star {\boxed{\sf{R \sin \theta \: = \: \dfrac{mv^2}{r}}}}$

And where as by the figure,

$\large \star {\boxed{\sf{R \cos \theta \: = \: mg}}}$

Divide both the equations,

$\implies {\sf{\dfrac{R \sin \theta}{R \cos \theta} \: = \: \dfrac{mv^2}{\dfrac{r}{mg}}}} \\ \\ \implies {\sf{\tan \theta \: = \: \dfrac{v^2}{rg}}}$

Where,

• m is mass
• r is radius
• g is gravitational force
• θ is angle of bending of cyclist