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Chapter 3: Problem 85

A particle is moving along the circular path with a speed \(v\) and tangential acceleration is \(g\) at an instant. If the radius of the circular path be \(r\), then the net acceleration of the particle at that instant is (A) \(\frac{v^{2}}{r}+g\) (B) \(\frac{v^{2}}{r^{2}}+g^{2}\) (C) \(\left[\frac{v^{4}}{r^{2}}+g^{2}\right]^{\frac{1}{2}}\) (D) \(\left[\frac{v^{2}}{r}+g^{2}\right]^{\frac{1}{2}}\)

Short Answer

Expert verified
The short answer to the given problem is: \(a_{net} = \left[\frac{v^4}{r^2}+g^2\right]^{\frac{1}{2}}\) and the correct option is (C).

Step by step solution

01

Calculate the centripetal acceleration

The particle is moving in a circle with radius r and speed v. The centripetal acceleration can be calculated using the formula: \(a_c = \frac{v^2}{r}\)
02

Combining Centripetal and Tangential Acceleration

The net acceleration of the particle can be obtained by considering both centripetal acceleration and tangential acceleration. These two accelerations are orthogonal to each other. Therefore, we can find the net acceleration using Pythagorean theorem: \(a_{net} = \sqrt{a_c^2 + a_t^2}\) Replacing \(a_c\) with the formula from Step 1 and \(a_t\) with g: \(a_{net} = \sqrt{\left(\frac{v^2}{r}\right)^2 + g^2}\)
03

Matching to An Answer

From the provided options, the net acceleration \(a_{net}\) is given exactly by option (C): \(a_{net} = \left[\frac{v^4}{r^2}+g^2\right]^{\frac{1}{2}}\) Hence, the correct answer is (C).

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