As a young boy swings a yo-yo parallel to the ground and above his head, the yo-yo has a centripetal acceleration of 250m/s^2. If the yo-yo's string is 0.50 m long, what is the yo-yo's tangential speed?
To determine the yo-yo's tangential speed, we can use the relationship between centripetal acceleration and tangential speed.
The formula to calculate centripetal acceleration is:
Centripetal acceleration (a) = (tangential speed)^2 / radius
Given:
Centripetal acceleration (a) = 250 m/s^2
Radius (r) = 0.50 m
We can rearrange the formula to solve for tangential speed:
(tangential speed)^2 = (centripetal acceleration) * (radius)
Substituting the given values:
(tangential speed)^2 = 250 m/s^2 * 0.50 m
(tangential speed)^2 = 125 m^2/s^2
Taking the square root of both sides:
tangential speed = √(125 m^2/s^2)
tangential speed ≈ 11.18 m/s
Therefore, the yo-yo's tangential speed is approximately 11.18 m/s.
To find the yo-yo's tangential speed given its centripetal acceleration and the length of its string, we can apply the concept of circular motion.
First, let's understand the relationship between centripetal acceleration, tangential speed, and the radius of the circular path.
The centripetal acceleration of an object moving in a circle can be calculated using the formula:
a = (v^2) / r,
where:
- a is the centripetal acceleration,
- v is the tangential speed, and
- r is the radius of the circular path.
Now, we are given that the centripetal acceleration is 250 m/s^2 (a = 250 m/s^2) and the length of the string is 0.50 m (r = 0.50 m).
We want to find the tangential speed (v).
Rearranging the formula, we get:
v^2 = a * r.
Plugging in the given values, we have:
v^2 = 250 m/s^2 * 0.50 m.
Multiplying the values, we find:
v^2 = 125 m^2/s^2.
To solve for v, we take the square root of both sides:
v = √(125 m^2/s^2).
Evaluating the square root, we get:
v ≈ 11.18 m/s.
Therefore, the yo-yo's tangential speed is approximately 11.18 m/s.