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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Circular Motion

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Circular Motion

A scooter and its rider, having a combined mass of 150 kg , are...

A scooter and its rider, having a combined mass of 150 kg150 \text{ kg}, are travelling at 36 km/hr36 \text{ km/hr} and need to negotiate a turn of radius 30 m30 \text{ m}. What should be the proper angle of banking if the required horizontal force for the turn is entirely provided by the normal force from the road? (Take g=10 m/s2g = 10 \text{ m/s}^2)

Options

  1. Atan⁡−1(16)\tan^{-1} \left(\dfrac{1}{6}\right)
  2. Btan⁡−1(13)\tan^{-1} \left(\dfrac{1}{3}\right)
  3. Ctan⁡−1(3)\tan^{-1}(3)
  4. Dtan⁡−1(103)\tan^{-1} \left(\dfrac{10}{3}\right)

Correct answer

B. tan⁡−1(13)\tan^{-1} \left(\dfrac{1}{3}\right)

Step-by-step solution

Given:
Mass m=150 kgm = 150 \text{ kg}
Velocity v=36 km/hr=36×518 m/s=10 m/sv = 36 \text{ km/hr} = 36 \times \dfrac{5}{18} \text{ m/s} = 10 \text{ m/s}
Radius r=30 mr = 30 \text{ m}
Acceleration due to gravity g=10 m/s2g = 10 \text{ m/s}^2

For banking without friction, the necessary centripetal force is provided entirely by the horizontal component of the normal reaction.

Nsin⁡θ=mv2rN \sin \theta = \dfrac{mv^2}{r}

Ncos⁡θ=mgN \cos \theta = mg

Dividing the two equations:

tan⁡θ=v2rg\tan \theta = \dfrac{v^2}{rg}

Substituting the given values:

tan⁡θ=(10)230×10\tan \theta = \dfrac{(10)^2}{30 \times 10}

tan⁡θ=100300=13\tan \theta = \dfrac{100}{300} = \dfrac{1}{3}

$\theta = \tan^{-1} \left(\dfrac{1}{3}\right)
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