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A particle of mass $10 \mathrm{~g}$ moves along a circle of radius $6.4 \mathrm{~cm}$ with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to $8 \times 10^{-4} \mathrm{~J}$ boy the end of the second revolution after the beginning of the motion?
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The correct answer is:
$0.1 \mathrm{~m} / \mathrm{s}^2$
kinetic energy $=8 \times 10^{-4} \mathrm{~J}$
$\begin{aligned}& \text { or, } \frac{1}{2} \mathrm{mv} 2=8 \times 10^{-4} \\& \text { or, } \frac{1}{2} \times 10 \times 10^{-3} \mathrm{v}^2=8 \times 10^{-4} \\& \text { or, } \mathrm{v}^2=16 \times 10^{-2}=\gt\mathrm{v}=0.4 \mathrm{~m} / \mathrm{s}\end{aligned}$
initial velocity of particle, $\mathbf{u}=\mathrm{om} / \mathrm{s}$
we have to find Tangential acceleration at the end of 2 nd revolution.
total distance covered, $s=2(2 \pi r)=4 \pi r$
$\begin{aligned}& \text { so, } v^2=2 \mathrm{as} \\& \mathrm{a}=\frac{\mathrm{v}^2}{2 \mathrm{~s}}=\frac{(0.4)^2}{2(4 \pi r)}
\end{aligned}$
$\begin{aligned} & =\frac{16 \times 10^{-2}}{\left(8 \times 3.14 \times 6.4 \times 10^{-} 2\right)} \\ & =0.0995 \mathrm{~m} / \mathrm{s}^2 \approx 0.1 \mathrm{~m} / \mathrm{s}^2\end{aligned}$
$\begin{aligned}& \text { or, } \frac{1}{2} \mathrm{mv} 2=8 \times 10^{-4} \\& \text { or, } \frac{1}{2} \times 10 \times 10^{-3} \mathrm{v}^2=8 \times 10^{-4} \\& \text { or, } \mathrm{v}^2=16 \times 10^{-2}=\gt\mathrm{v}=0.4 \mathrm{~m} / \mathrm{s}\end{aligned}$
initial velocity of particle, $\mathbf{u}=\mathrm{om} / \mathrm{s}$
we have to find Tangential acceleration at the end of 2 nd revolution.
total distance covered, $s=2(2 \pi r)=4 \pi r$
$\begin{aligned}& \text { so, } v^2=2 \mathrm{as} \\& \mathrm{a}=\frac{\mathrm{v}^2}{2 \mathrm{~s}}=\frac{(0.4)^2}{2(4 \pi r)}
\end{aligned}$
$\begin{aligned} & =\frac{16 \times 10^{-2}}{\left(8 \times 3.14 \times 6.4 \times 10^{-} 2\right)} \\ & =0.0995 \mathrm{~m} / \mathrm{s}^2 \approx 0.1 \mathrm{~m} / \mathrm{s}^2\end{aligned}$
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