Cheryl is riding on the edge of a merry-go-round, 3.60 m from the center, which is rotating with an increasing angular speed. Cheryl’s tangential acceleration is 2.30 m/s2. At the instant that Cheryl’s linear speed is 2.80 m/s, what is Cheryl’s total acceleration?
To find Cheryl's total acceleration, we need to consider both tangential acceleration and centripetal acceleration.
The tangential acceleration refers to the rate at which Cheryl's linear speed is changing. In this case, the given tangential acceleration is 2.30 m/s².
The centripetal acceleration refers to the acceleration towards the center of the circular path. It is given by the formula:
a_c = (v²) / r
where a_c is the centripetal acceleration, v is the linear speed, and r is the radius of the circular path.
Now we know that Cheryl's linear speed is 2.80 m/s and the radius of the circular path is 3.60 m.
Let's calculate the centripetal acceleration using the formula:
a_c = (2.80 m/s)² / 3.60 m
a_c = 7.84 m²/s² / 3.60 m
a_c ≈ 2.18 m/s²
So Cheryl's centripetal acceleration is approximately 2.18 m/s².
To find the total acceleration, we add the tangential acceleration and the centripetal acceleration:
a_total = a_t + a_c
a_total = 2.30 m/s² + 2.18 m/s²
a_total ≈ 4.48 m/s²
Therefore, Cheryl's total acceleration is approximately 4.48 m/s².