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?
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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.