A young boy swings a yo-yo horizontally above his head so that the yo-yo has a centripetal acceleration of 275 m/s2. If the yo-yo's string is 0.55 m long, what is the yo-yo's tangential speed?

12 ans2784

To find the yo-yo's tangential speed, we can use the relationship between centripetal acceleration and tangential speed.

Centripetal acceleration is given by the formula:

a = v^2 / r

Where:
a = centripetal acceleration
v = tangential speed
r = radius (distance from the center of the circular path to the object)

In this case, we know that the centripetal acceleration is 275 m/s^2 and the radius is 0.55 m. We can rearrange the formula to solve for v:

v = sqrt(a * r)

Substituting the given values, we have:

v = sqrt(275 * 0.55)

Calculating the square root:

v ≈ sqrt(151.25)

v ≈ 12.3 m/s

Therefore, the yo-yo's tangential speed is approximately 12.3 m/s.