A child sitting 1.20 m from the center of a merry-go round moves with a speed of$\mathbf{1}\mathbf{.}\mathbf{10}\mathbf{m}\mathbf{/}\mathbf{s}$.Calculate (a) the centripetal acceleration of the child and (b) the net horizontal force exerted on the child$\left(\mathbf{mass}=\mathbf{22}\mathbf{.}\mathbf{5}\mathbf{kg}\right)$.

The child is rotating on the merry-go-round. The centripetal force acts on the child during the circular motion.The centripetal acceleration depends on the speed of the merry-go-round and the distance between the centre of the merry-go-round and the child.

The given data can be listed below as,

• Thedistance between the centre of the merry-go-round and the child is,$r=1.20\mathrm{m}$.
• The speed of the merry-go-round is, $v=1.10\mathrm{m}/\mathrm{s}$.
• The mass of the child is, $m=22.5\mathrm{kg}$.

The centripetal acceleration can be expressed as,

$a_{\mathrm{c}}=\frac{v^2}{r}$

Here, v is the speed of the merry-go-round, r is the distance between the centre of the merry-go-round and the child.

Substitute the values in the above equation.

$\begin{array}{c}{a}_{\mathrm{c}}=\frac{{\left(1.10\mathrm{m}/\mathrm{s}\right)}^2}{1.20\mathrm{m}}\\ \approx 1.01\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, the centripetal acceleration of the child is $1.01\mathrm{m}/{\mathrm{s}}^2$.

The net horizontal force can be expressed as,

$F_{\mathrm{H}}=m{a}_{\mathrm{c}}$

Substitute the values in the above equation.

$\begin{array}{c}{F}_{\mathrm{H}}=22.5\mathrm{kg}\times 1.01\mathrm{m}/{\mathrm{s}}^2\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ =22.725\mathrm{N}\end{array}$

Thus, the net horizontal force acts on the child is $22.725\mathrm{N}$.