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Question

# A 50.0-kg child stands at the rim of a merry-go-round of radius 2.00 m, rotating with an angular speed of 3.00 rad/s. What is the child’s centripetal acceleration?

Solution

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$\textbf{(a)}$ Here given that the mass of the child

\begin{align*} m = 50.0\,\mathrm{kg} \end{align*}

\begin{align*} r = 2.00\,\mathrm{m} \end{align*}

Angular speed of the marry-go-round

\begin{align*} \omega = 3.00\,\mathrm{rad\,s^{-1}} \end{align*}

Now as we know for any object moving in a circular path has an acceleration directed toward the center of the circular path, called a centripetal acceleration and is given by

\begin{align*} a_{\text{c}} = \frac{v^2}{r}\tag{1} \end{align*}

Where

\begin{align*} v = r\,\omega\tag{2} \end{align*}

is the linear speed.

Now from equations (1) and (2), we get

\begin{align*} a_{\text{c}} & = r\,\omega^2\\ & = 2.00\,\mathrm{m} \times \,\left(3.00\,\mathrm{rad\,s^{-1}} \right)^2\\ & = 18.0\,\mathrm{m\,s^{-2}} \end{align*}

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