Question

A particle of mass $10\,g$ moves along a circle of radius $6.4\,cm$ with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to $8 \times 10^{-4} \, J$ by the end of the second revolution after the beginning of the motion ?

• A. $0.15 \, m / s^2$
• B. $0.18 \, m / s^2$
• C. $0.2 \, m / s^2$
• D. $0.1 \, m / s^2$

Answer 1

The Correct Option is D

Calculation of Tangential Acceleration

The correct option is (D): 0.1 m/s². 

Step-by-Step Calculation:

We are given the following equation for kinetic energy:

\(\frac{1}{2} mv ^{2}=8 \times 10^{-4}\)

Substituting values:

\(\frac{1}{2} \times 10 \times 10^{-3} v ^{2}=8 \times 10^{-4}\)

Simplifying further:

\(v ^{2}=16 \times 10^{-2} \Rightarrow v =0.4 \, \text{m/s}\)

The initial velocity of the particle is:

\(u = 0 \, \text{m/s}\)

Finding Tangential Acceleration

We are asked to find the tangential acceleration at the end of the 2nd revolution. First, we calculate the total distance covered:

\(s = 2(2 \pi r) = 4 \pi r\)

Now, using the equation for motion:

\(v^{2} = 2as\)

Solving for acceleration \(a\):

\(a = \frac{v^{2}}{2s} = \frac{(0.4)^{2}}{2(4 \pi r)}\)

Substituting known values:

\(=\frac{16 \times 10^{-2}}{8 \times 3.14 \times 6.4 \times 10^{-2}}\)

Finally, simplifying:

\(= 0.0995 \, \text{m/s}^{2} \approx 0.1 \, \text{m/s}^{2}\)

Conclusion:

The tangential acceleration at the end of the 2nd revolution is approximately \(0.1 \, \text{m/s}^2\), so the correct option is (D).