A 15 kg child stands on the edge of a small merry-go-round of radius 2.0 m. If the merry-go- round turns at a steady rate of 11 rpm, calculate the magnitude and direction of the net force on the child.

To calculate the magnitude and direction of the net force on the child, we need to understand the physics principles involved.

First, let's consider the child standing on the merry-go-round. The child can be seen as an object experiencing rotational motion, with the merry-go-round acting as a centripetal force.

The centripetal force required to keep an object moving in a circular path is given by the formula:

Fc = m * ω^2 * r

Where:
- Fc is the centripetal force,
- m is the mass of the object (15 kg in this case),
- ω is the angular velocity (converted from rpm to rad/s),
- and r is the radius of the circular path (2.0 m).

To calculate the angular velocity ω, we need to convert the given rotational speed from rpm to rad/s. We can use the conversion factor:

1 revolution = 2π radians

Given that the merry-go-round turns at 11 rpm, we can calculate ω as follows:

ω = (11 rpm) * (2π rad/1 min) * (1 min/60 sec) = 11 * 2π / 60 rad/s ≈ 1.1547 rad/s

Now substituting the values into the formula:

Fc = (15 kg) * (1.1547 rad/s)^2 * (2.0 m)

Simplifying the equation gives:

Fc ≈ 40.92 N

The magnitude of the net force on the child is approximately 40.92 Newtons.

Since the child is standing on the edge of the merry-go-round, the net force can be considered as the weight of the child (mg) acting downward and the centripetal force (Fc) acting inward towards the center.

Hence, the magnitude of the net force on the child is 40.92 N and its direction is centripetal, directed towards the center of the merry-go-round.