A young boy swings a yo-yo horizontally
above his head so that the yo-yo has a cen-
tripetal acceleration of 240 m/s2.
If the yo-yo’s string is 0.29 m long, what is
the yo-yo’s tangential speed
To find the yo-yo's tangential speed, we need to use the formula for centripetal acceleration:
a = (v^2) / r
Where:
a is the centripetal acceleration
v is the tangential speed
r is the radius or the length of the string
Given:
Centripetal acceleration (a) = 240 m/s^2
Length of the string (r) = 0.29 m
We can rearrange the formula to solve for the tangential speed (v):
v = √(a * r)
Let's substitute the given values into the equation:
v = √(240 m/s^2 * 0.29 m)
v = √(69.6 m^2/s^2)
v ≈ 8.34 m/s
Therefore, the yo-yo's tangential speed is approximately 8.34 m/s.
Use the equation
(acceleration) = V^2/R = 240 m/s^2
and solve for V.