Question

A particle moves in a circular path with uniform speed v. When it turns by 90°, find ratio of \(\frac{v}{<\overrightarrow{v}>}\)?

• A. \(\frac{\pi}{\sqrt{2}}\)
• B. \(\frac{2\pi}{\sqrt{2}}\)
• C. \(\frac{\pi}{2}\)
• D. \(2\pi\)

Answer 1

The Correct Option is A

To solve the problem, we need to determine the ratio of the particle's instantaneous speed \( v \) to the average speed \(\langle \overrightarrow{v} \rangle \) for a particle moving in a circular path, after it turns by 90°.

Considering the particle moves with a uniform speed \( v \) in a circular path, it is important to understand the relationship between average speed and instantaneous speed over a particular interval.

When a particle moves in a circular path and turns by 90°, it essentially moves along a quadrant of the circle. The entire circle's circumference is \(2\pi R\), where \(R\) is the radius of the circle.

A 90° turn implies the particle has moved along a quarter of the circle's circumference, i.e., \(\frac{1}{4} \times 2\pi R = \frac{\pi R}{2}\).

The time taken to travel this arc length at uniform speed \( v \) is given by:

\(t = \frac{\text{Arc Length}}{v} = \frac{\pi R}{2v}\)

The resultant displacement of the particle after turning 90° is the length of the straight line (chord) connecting the initial and final points of the arc. This is the hypotenuse of a right-angled triangle with each of the other two sides equal to \( R \), hence given by:

\(D = \sqrt{R^2 + R^2} = \sqrt{2}R\)

Average speed \(\langle \overrightarrow{v} \rangle\) is given by the ratio of the displacement to the time taken:

\(\langle \overrightarrow{v} \rangle = \frac{D}{t} = \frac{\sqrt{2}R}{\frac{\pi R}{2v}} = \frac{2v \sqrt{2}}{\pi}\)

Finally, the ratio of \( v \) to \(\langle \overrightarrow{v} \rangle\) is calculated as follows:

\(\frac{v}{\langle \overrightarrow{v} \rangle} = \frac{v}{\frac{2v \sqrt{2}}{\pi}} = \frac{\pi}{2 \sqrt{2}} \times \frac{1}{v/v} = \frac{\pi}{\sqrt{2}}\)

Therefore, the correct answer is \(\frac{\pi}{\sqrt{2}}\), which corresponds to the first option.