You plan to throw stones by using a sling of length 1.8 m which you whirl over your head. Suppose you wish to throw a stone a distance of 22 m. What must be the centripetal acceleration of the stone just before its release if it is to reach this distance? Assume that the release height is 2.6 m.
To find the centripetal acceleration of the stone just before its release, we need to use the following equation:
v = √(r * g)
Where v is the release velocity of the stone, r is the radius of circular motion (sling length), and g is the acceleration due to gravity.
First, let's find the release velocity:
v = √(r * g)
Given that the sling length (r) is 1.8 m and the acceleration due to gravity (g) is approximately 9.8 m/s^2, we can substitute these values:
v = √(1.8 * 9.8)
v ≈ √(17.64)
v ≈ 4.2 m/s
Now, let's find the time taken to reach the release height:
h = 2.6 m
Using the equation:
h = (1/2) * g * t^2
Rearranging the equation to solve for t:
t^2 = (2 * h) / g
t^2 = (2 * 2.6) / 9.8
t^2 ≈ 0.533
Taking the square root of both sides:
t ≈ √(0.533)
t ≈ 0.73 s
Next, let's find the horizontal distance covered by the stone in this time:
s = v * t
s ≈ 4.2 * 0.73
s ≈ 3.07 m
Now, to find the centripetal acceleration just before release, we will use the formula:
a = v^2 / r
a = (4.2)^2 / 1.8
a ≈ 9.80 m/s^2
Therefore, the centripetal acceleration of the stone just before its release must be approximately 9.80 m/s^2 to reach a distance of 22 m.