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Chapter 2: Problem 2

A particle is moving along a circular path of radius \(6 \mathrm{~m}\) with a uniform speed of \(8 \mathrm{~ms}^{-1}\). The average acceleration when the particle completes one half of the revolution is (A) \(\frac{16}{3 \pi} \mathrm{ms}^{-2}\) (B) \(\frac{32}{3 \pi} \mathrm{ms}^{-2}\) (C) \(\frac{64}{3 \pi} \mathrm{ms}^{-2}\) (D) None of these

Short Answer

Expert verified
The average acceleration when the particle completes one half of the revolution is \( \frac{64}{3 \pi} \) m/s虏.

Step by step solution

01

Find the time taken to complete half of the revolution

First, let's find the time taken by the particle to complete half of the revolution. To do this, we need to find the length of the path (half of the circumference of the circle) and then divide it by the given speed. Length of half of the circumference = \( \frac{1}{2} * 2 * \pi * r = \pi * r \) Where r = 6 m Now, divide the length by the speed to get the time taken: Time taken = \( \frac{\pi * 6}{8} \) = \( \frac{3 \pi}{4} \) s
02

Calculate the change in velocity

Once the particle completes half of the revolution, its final position will be diametrically opposite to the initial position. Thus, its final velocity vector will be opposite in direction to its initial velocity vector, but with the same magnitude. Since both the initial and final velocities have the same magnitude, the change in velocity will be equal to twice the magnitude of the velocity: Change in velocity = 2 * 8 = 16 m/s
03

Calculate the average acceleration

Now, we will use the formula for average acceleration: Average acceleration = \( \frac{Change \thinspace in \thinspace velocity}{Time \thinspace taken} \) Substitute the values obtained in steps 1 and 2: Average acceleration = \( \frac{16}{\frac{3 \pi}{4}} = \frac{64}{3 \pi} \) m/s虏 Thus, the average acceleration when the particle completes one half of the revolution is: \( \frac{64}{3 \pi} \) m/s虏, which matches option (C).

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Most popular questions from this chapter

A particle can be projected with a given speed in two possible ways so as to make it pass through a point at a distance \(r\) from the point of projection. The product of the times taken to reach this point in the two possible ways is then proportional to (A) \(r\) (B) \(\frac{1}{r}\) (C) \(\frac{1}{r^{2}}\) (D) \(\frac{1}{r^{3}}\)

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