Centripetal Acceleration with distance

[getty102]

Homework Statement

You plan to throw stones by using a sling of length 0.7 m which you whirl over your head. Suppose you wish to throw a stone a distance of 36 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 3.0 m.

Homework Equations

a_c=v²/r

The Attempt at a Solution

I tried to find the velocity in the x-direction but I didn't have an angle to work with. I then solved x=v₀cosθ(t) for t and plugged that into y=y₀+v₀sinθ(t)-g/2(t)². This gave me -y_o=xtanθ-((gx²)/(2v₀²cos²θ). That didn't cancel anything out.
I'm not sure where to go from there.

 

[PeterO]

Homework Statement

You plan to throw stones by using a sling of length 0.7 m which you whirl over your head. Suppose you wish to throw a stone a distance of 36 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 3.0 m.

Homework Equations

a_c=v²/r

The Attempt at a Solution

I tried to find the velocity in the x-direction but I didn't have an angle to work with. I then solved x=v₀cosθ(t) for t and plugged that into y=y₀+v₀sinθ(t)-g/2(t)². This gave me -y_o=xtanθ-((gx²)/(2v₀²cos²θ). That didn't cancel anything out.
I'm not sure where to go from there.

For this sling, it is the speed of the projectile which determines how far it goes. You will either have to be given or assume an angle of release to find that speed.
Once you know the speed you can calculate the centripetal acceleration.

 

[getty102]

I found time t it takes for the stone to fall 3.0m.

t=√(y_o2/g)

then plugged that t into find the velocity it would take the stone to go 36m.

v_0,x=x/t

I plugged that v, and my r, in v²/r to get my centripetal acceleration although it gave me a too large of number. Where am i going wrong?

 

[PeterO]

I found time t it takes for the stone to fall 3.0m.

t=√(y_o2/g)

then plugged that t into find the velocity it would take the stone to go 36m.

v_0,x=x/t

I plugged that v, and my r, in v²/r to get my centripetal acceleration although it gave me a too large of number. Where am i going wrong?

Show me your numbers. I could work it the way you describe.

 

[asteriatic]

To determine the centripetal acceleration of the stone, we must first understand the concept of centripetal acceleration and how it relates to the distance and velocity of the object. Centripetal acceleration is the acceleration towards the center of a circular path, and it is caused by the centripetal force acting on the object.

In this case, the sling is acting as the centripetal force, keeping the stone in a circular path as it is whirled over your head. The distance traveled by the stone is determined by its velocity and the time it takes to travel that distance. We can use the equation ac=v2/r to calculate the centripetal acceleration needed for the stone to reach a distance of 36 m.

To find the velocity of the stone, we can use the equation v=√(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the release height (3.0 m). This gives us a velocity of approximately 7.67 m/s.

Now, we can plug in the values into the equation ac=v2/r to solve for the centripetal acceleration. Rearranging the equation, we get ac=v2/r= (7.67 m/s)^2/0.7 m = 83.6 m/s^2.

Therefore, the centripetal acceleration of the stone just before its release must be 83.6 m/s^2 in order for it to reach a distance of 36 m. It is important to note that this is the minimum required acceleration, and the stone may experience a higher acceleration due to other factors such as air resistance.