A child of mass $5 \mathrm{~kg}$ is going round a merry-go-round that makes 1 rotation in $3.14 \mathrm{~s}$. The radius of the merry-go-round is $2 \mathrm{~m}$. The centrifugal force on the child will be

<p>To calculate the centrifugal force acting on the child, we need to find the angular velocity of the merry-go-round and then apply the formula for centrifugal force. </p> <p>The merry-go-round makes 1 rotation in 3.14 seconds, so its angular velocity ($\omega$) can be calculated as:</p> <p>$\omega = \frac{2\pi}{T}$, where $T$ is the time period for one rotation.</p> <p>$\omega = \frac{2\pi}{3.14} = 2 \, \text{radians/s}$</p> <p>Now, the formula for centrifugal force ($F$) is:</p> <p>$F = m \cdot r \cdot \omega^2$, where $m$ is the mass of the child, $r$ is the radius of the merry-go-round, and $\omega$ is the angular velocity.</p> <p>$$F = 5\,\text{kg} \cdot 2\,\text{m} \cdot (2\,\text{radians/s})^2 = 5\,\text{kg} \cdot 2\,\text{m} \cdot 4\,(\text{radians/s})^2$$</p> <p>$F = 40\,\text{N}$</p> <p>Therefore, the centrifugal force on the child is $40\,\text{N}$.</p>