Question 69: Work, Energy and Power & Circular Motion

Question

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

Accepted Answer

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

Answer

50 N

Answer

80 N

Answer

100 N

Answer

40 N

Work, Energy and Power & Circular Motion

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

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