Centripetal Acceleration of Stone Problem

In summary, a stone is whirled in a vertical circle with a radius of 1.20 m and a constant speed of 1.50 m/s. The center of the sling is 1.50 m above the ground. The range of the stone is 1.9187 m when released at point A. To find the range at point B, the vector components of the initial position and velocity must be calculated and used with the constant acceleration formula. The time the stone is in the air is found using the initial height and y velocity, and then used to calculate the distance traveled in the x direction. It is important to clearly define the starting points in calculations.
  • #1
The_ArtofScience
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0

Homework Statement



A stone at the end of a sling is whirled in a vertical circle of radius 1.20 m at a constant speed v0 = 1.50 m/s. The center of the sling is 1.50 m above the ground. What is the range of the stone if it is released when the sling is inclined at 30 degrees with the horizontal (a) at A? (b) at B? What is the acceleration of the (c) stone just before it is released at A? (d) just after it is released at A?

http://www.fen.bilkent.edu.tr/~mb/phys101/CH4.pdf
The figure is on page 7. Its problem #57


The Attempt at a Solution



I'm having trouble just thinking of what a set up would be like for this problem. I wish I could give some work here but I'm tottally stumped
 
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  • #2
Hi

probably need to recoginise the answer to c) and d) before attempting a) & b)

for c) and d) think about the type of motion, for c) this is the circular motion, so what is the acceleration. for d) what forces act on the stone after release?

for a) and b) you know initial velocity & position so should be able to calculate range...
 
  • #3
I'm not sure where to begin. First off I don't know what its really asking. I have an idea that the stone at the end of point A will fling at a higher range than point B, but other than that I don't see how to get there

So I think the appropriate eq is xf = x0 + v0t --> 1.20 + 1.50cos(30 deg)t
Solving for t in the y direction 1.50 = 1/2(9.8)t^2 --> t = .55328
Then, xf = 1.20 + 1.50(cos(30 deg))(.55328) = 1.9187 m

I'm not sure what to do with point B. Don't point A and point B both have the same data? But that wouldn't really work out because point B is supposed to have a smaller range than point A
 
  • #4
look like you're on the right track...

this questions consiet of two main parts
- finding vector components
- using constant acceleration formula

first you need to find the vector components of the initial position (on the circle) and velocity

Once you have these you find use the initial hieght & y velocity to findt he time the stone is in the air for, as i think you have done.

Then use the time to find the distance traveled in the x direction

i think in your calulations you need to be a bit more careful defining your starting points... (ie initial position & velcoity) for example the y velocity is not included in your calculation of time
 
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FAQ: Centripetal Acceleration of Stone Problem

What is centripetal acceleration of stone problem?

The centripetal acceleration of stone problem is a physics concept that involves calculating the acceleration of a stone moving in a circular path. It is used to determine the force needed to keep the stone moving in a circular motion, and is an important concept in understanding circular motion and gravity.

How do you calculate centripetal acceleration of a stone?

The formula for calculating centripetal acceleration is a = v²/r, where "a" is the acceleration, "v" is the velocity of the stone, and "r" is the radius of the circular path. This formula can be applied to any object moving in a circular path, including a stone.

What factors affect the centripetal acceleration of a stone?

The centripetal acceleration of a stone is affected by three main factors: the speed of the stone, the radius of the circular path, and the mass of the stone. A higher speed or smaller radius will result in a greater centripetal acceleration, while a larger mass will result in a lower centripetal acceleration.

How does centripetal acceleration of a stone relate to gravity?

The centripetal acceleration of a stone is directly related to the force of gravity acting on the stone. In fact, the force of gravity is what provides the centripetal force needed to keep the stone moving in a circular path. This is because gravity is constantly pulling the stone towards the center of the circular path.

Can the centripetal acceleration of a stone be negative?

No, the centripetal acceleration of a stone cannot be negative. This is because acceleration is a vector quantity, meaning it has both magnitude and direction. In the case of centripetal acceleration, the direction is always towards the center of the circular path, so it cannot be negative. However, the direction can change if the stone's velocity or radius changes, resulting in a change in the direction of the centripetal acceleration.

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