134k views
4 votes
A merry-go-round makes one complete revolution in 6.6 s. A 35.1 kg child sits on the horizontal floor of the merry-go-round 4.7 m from the center. Find the child's acceleration. The acceleration of gravity is 9.8 m/s². Answer in units of m/s².

User Raphie
by
8.4k points

1 Answer

5 votes

Final answer:

To calculate the child's acceleration on a merry-go-round, you must use the centripetal acceleration formula. Given a radius of 4.7 m and a time of one revolution of 6.6 seconds, the child's centripetal acceleration is approximately 1.353 m/s².

Step-by-step explanation:

To find the child's acceleration on a merry-go-round that makes one complete revolution in 6.6 seconds, we can use the concept of centripetal acceleration, which is given by the formula \( a = \frac{v^2}{r} \), where \( v \) is the linear velocity and \( r \) is the radius of the circular path.

First, we calculate the linear velocity (v) using the circumference of the circular path ( \( C = 2\pi r \) ) and the time for one revolution (T): \( v = \frac{C}{T} = \frac{2\pi r}{T} \).

Then, we plug the linear velocity back into the formula for centripetal acceleration:

\( a = \frac{(2\pi r)^2}{rT^2} = \frac{4\pi^2 r}{T^2} \).

In this case,

\( r = 4.7 \) meters, and

\( T = 6.6 \) seconds, so

\( a = \frac{4\pi^2 \cdot 4.7}{6.6^2} \approx \frac{4\cdot 3.1416^2 \cdot 4.7}{43.56} \approx \frac{58.9046}{43.56} \approx 1.353 \) m/s2.

Thus, the child's centripetal acceleration is approximately 1.353 m/s2.

User Ganzux
by
7.9k points