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Q1.

Expert-verified
Found in: Page 131

### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

# 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 $\left(\mathbf{mass}=\mathbf{22}\mathbf{.}\mathbf{5}\mathbf{kg}\right)$.

(a) The centripetal acceleration of the child is $1.01\mathrm{m}/{\mathrm{s}}^{2}$.

(b) The net horizontal force exerted on the child is $22.725\mathrm{N}$.

See the step by step solution

## Step 1. Understanding the centripetal acceleration of the child

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.

## Step 2. Identification of given data

The given data can be listed below as,

• The distance 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, $m=22.5\mathrm{kg}$.

## Step 3. (a) Determination of the centripetal acceleration of the child

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 $1.01\mathrm{m}/{\mathrm{s}}^{2}$.

## Step 4. (b) Determination of the net horizontal force exerted on the child

The net horizontal force can be expressed as,

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

Substitute the values in the above equation.

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

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