You plan to throw stones by using a sling of length 0.6 m which you whirl over your head. Suppose you wish to throw a stone a distance of 32 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 2.2 m.
time it takes to fall 2.2m
h=1/2 a t^2 solve for t.
distance it goes..
d=vi*t solve for vi.
Now, acceleration= m v^2/r
To determine the centripetal acceleration of the stone just before its release, we can use the formula:
a = (v^2) / r
Where:
a = centripetal acceleration
v = velocity of the stone
r = radius of the circular path
First, we need to find the velocity of the stone just before its release. We know that the stone will travel a distance of 32 m and the sling has a length of 0.6 m. The total length of the path will therefore be 32 m + 0.6 m = 32.6 m.
To find the velocity, we can use the formula for the speed of an object in circular motion:
v = (2πr) / T
Where:
v = velocity
r = radius of the circular path
T = time period of one complete revolution
The radius of our circular path is 0.6 m, and we need to calculate the time period (T). We can use the formula:
T = 2π (sqrt(L / g))
Where:
L = length of the path
g = acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the values, we get:
T = 2π (sqrt(32.6 / 9.8))
Solving this equation, we find the time period (T) of the sling as:
T ≈ 8.486 seconds
Now, using this value of T, we can calculate the velocity (v) as:
v = (2π * 0.6) / 8.486
Solving this equation, we find the velocity (v) just before release to be approximately:
v ≈ 1.414 m/s
Finally, we can calculate the centripetal acceleration (a) using the formula:
a = (v^2) / r
Substituting the values, we get:
a = (1.414^2) / 0.6
Solving this equation, we find the centripetal acceleration (a) to be approximately:
a ≈ 3.323 m/s^2
Therefore, in order for the stone to reach a distance of 32 m, the centripetal acceleration just before its release must be approximately 3.323 m/s^2.