Question

A child is standing at the outer rim of a merry-go-round and being spun by his mom. If the child stands at 2.25 m from the center and is being spun at 12.4 revolutions per minute, find the magnitude of the acceleration of the child (in m/s²).

Answer 1

To find the magnitude of the acceleration of the child on the merry-go-round, we convert the angular velocity to rad/s and apply the centripetal acceleration formula. The result is approximately 3.79 m/s².

Step-by-step explanation:

The student's question involves calculating the magnitude of acceleration of a child standing at the outer rim of a spinning merry-go-round. To solve this problem, we need to first convert the angular velocity from revolutions per minute (rpm) to radians per second (rad/s) then use the formula for centripetal acceleration, which is a = ω^2 * r, where a is the acceleration, ω is the angular velocity in rad/s, and r is the radius of the circular path.

First, we convert 12.4 rpm to rad/s using the conversion factor 1 rev = 2π rad and 1 minute = 60 seconds. So, 12.4 rev/min * (2π rad/rev) * (1 min/60 s) = 1.299 rad/s. Next, we use the formula for centripetal acceleration with r = 2.25 m and ω = 1.299 rad/s to find the acceleration:

a = ω^2 * r = (1.299 rad/s)^2 * 2.25 m = approximately 3.79 m/s².

Therefore, the magnitude of the acceleration of the child is approximately 3.79 m/s².