Uniform Circular Motion of a stone

In summary: Can you help?In summary, the stone at the end of the sling is whirled in a vertical circle of radius 1.00 m at a constant speed vi = 2.40 m/s. The center of the string is 1.50 m above the ground.
  • #1
FalconKICK
30
0

Homework Statement


A stone at the end of a sling is whirled in a vertical circle of radius 1.00 m at a constant speed vi = 2.40 m/s as in Figure P4.57. The center of the string is 1.50 m above the ground.

http://xs124.xs.to/xs124/08095/p4-57alt_1_285.gif

I solved these:

(a) What is the range of the stone if it is released when the sling is inclined at 30.0° with the horizontal at A?

1.05577 m

(c) What is the acceleration of the stone just before it is released at A?
Magnitude

5.76 m/s^2

(d) What is the acceleration of the stone just after it is released at A?
Magnitude

9.81 m/s^2

Homework Equations



(b) What is the range of the stone if it is released when the sling is inclined at 30.0° with the horizontal at B?

The Attempt at a Solution



I've tried doing it the same way I did part a. I used

x(t) = x_o+v_o(t)

and

y(t) = y_o+v_o(t)+(.5)(9.8)(t^2)

it didn't work and I've tried everything that I had in my notes.
 
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  • #2
Include the diagram, or at least describe the difference between points A & B.
 
  • #3
Doc Al said:
Include the diagram, or at least describe the difference between points A & B.

Sorry that I didn't include one. I'm trying to find a host. It'll be up shortly.
 
  • #4
FalconKICK said:

The Attempt at a Solution



I've tried doing it the same way I did part a. I used

x(t) = x_o+v_o(t)

and

y(t) = y_o+v_o(t)+(.5)(9.8)(t^2)

it didn't work and I've tried everything that I had in my notes.
What did you use for your initial values of position and speed? The acceleration should be negative.

When they ask for "range", what are they measuring with respect to? Position of the stone when released? Center of the circle?
 
  • #5
Doc Al said:
What did you use for your initial values of position and speed? The acceleration should be negative.

When they ask for "range", what are they measuring with respect to? Position of the stone when released? Center of the circle?

It doesn't specify but I believe it's center of the circle. For X-Component I used 1.2m/s and I used 2.0784m/s for the Y-Component. I'm really confused by this, so I don't really know what to do.
 
  • #6
FalconKICK said:
It doesn't specify but I believe it's center of the circle.
OK.
For X-Component I used 1.2m/s and I used 2.0784m/s for the Y-Component.
Those are the velocity components. What are their signs?

What about initial position?


I'm really confused by this, so I don't really know what to do.
I'd start by figuring out how long the stone is in the air before hitting the ground. Then figure out how far it got horizontally.
 
  • #7
Doc Al said:
OK.

Those are the velocity components. What are their signs?

What about initial position?
I'd start by figuring out how long the stone is in the air before hitting the ground. Then figure out how far it got horizontally.

I solved for time which is .68912 seconds. What do you mean by how far it would go horizontally? I don't know about initial position either.
 
  • #8
FalconKICK said:
I solved for time which is .68912 seconds.
How did you solve for the time?

What do you mean by how far it would go horizontally?
That's what "range" means.

I don't know about initial position either.
What's the position of point B?

How did you solve for the range in part (a)?
 
  • #9
Part a I did it wrong because I subtracted time that I got from 1.5 (height) and for some reason it gave me the correct answer.

Here's how I solved part A.

2.4COS(30) = 2.0784
2.4SIN(30) = 1.2

t =( - 1.2 + sqrt(1.2^(2) + 4(4.9)(1.5) ) / 9.8

t = .444222

1.5 - t = 1.06

I don't really know how to get A either. :( I only have 2 tries left on the site and I feel really stupid right now
 
  • #10
FalconKICK said:
2.4COS(30) = 2.0784
2.4SIN(30) = 1.2
Those are the y and x components of the initial velocity. The sign of the x-component should be negative.

t =( - 1.2 + sqrt(1.2^(2) + 4(4.9)(1.5) ) / 9.8
Where does this come from?

What you are trying to find is the distance the stone lands (from the center of the circle, I presume). You need the initial position and velocity components. Calculate the time it takes to hit the ground (by analyzing vertical motion). Then analyze the horizontal motion.
 
  • #11
Doc Al said:
Those are the y and x components of the initial velocity. The sign of the x-component should be negative.


Where does this come from?

What you are trying to find is the distance the stone lands (from the center of the circle, I presume). You need the initial position and velocity components. Calculate the time it takes to hit the ground (by analyzing vertical motion). Then analyze the horizontal motion.

I know that the height is 1.5 m. I have the velocity components but I do not know how to find the initial position components
 
  • #12
The height of the center of the circle is 1.5 m. Use a little trig to find the initial position components of the stone (which is at the end of the string).
 
  • #13
Doc Al said:
The height of the center of the circle is 1.5 m. Use a little trig to find the initial position components of the stone (which is at the end of the string).

1.5cos(30) = 1.299m
1.5sin(30) = .75m

now what do I do? I don't know how to use the information that I have.
 
  • #14
The length of the string is only 1 m. Once you find the components of the string length, combine that with the location of the center.
 
  • #15
Doc Al said:
The length of the string is only 1 m. Once you find the components of the string length, combine that with the location of the center.

Oh so

1cos(30) + 1.5 = 2.36602 m
1sin(30) + 1.5 = 2 m
 
  • #16
Almost. The location of the center is (0, 1.5m).

Clearly indicate which is the x and which is the y component of the stone's position.
 
  • #17
I did this and still don't get the correct answer. One more try left

the vertical velocity of the stone on release
Vv = 2.4 sin 30 = 1.2m/s

time to reach max height
v = u+at (v=0 at max height, Vv = u)
0 = 1.2 - 9.8t
t = 0.122 sec

max height above release reached
v^2 = u^2 + 2as
0 = 1.2^2 - 2 x 9.8 x s
s = 0.073m

so max height above the ground = 2.08 + 0.073 = 2.153m

time taken to reach ground
s = ut+ 1/2 at^2 (u = 0 as falling from max height)
2.153 = 1/2 x 9.8t^2
t = 0.663 secs

total time in the air = 0.122 + 0.663 secs = 0.785s

horizontal velocity of stone on release = 2.4 cos 30 = 2.08m/s

range = Vh x t = 2.08 x 0.785 = 1.63m
 
  • #18
FalconKICK said:
I did this and still don't get the correct answer. One more try left

the vertical velocity of the stone on release
Vv = 2.4 sin 30 = 1.2m/s

time to reach max height
v = u+at (v=0 at max height, Vv = u)
0 = 1.2 - 9.8t
t = 0.122 sec

max height above release reached
v^2 = u^2 + 2as
0 = 1.2^2 - 2 x 9.8 x s
s = 0.073m

so max height above the ground = 2.08 + 0.073 = 2.153m
Where does 2.08 come from?


time taken to reach ground
s = ut+ 1/2 at^2 (u = 0 as falling from max height)
2.153 = 1/2 x 9.8t^2
t = 0.663 secs

total time in the air = 0.122 + 0.663 secs = 0.785s

horizontal velocity of stone on release = 2.4 cos 30 = 2.08m/s

range = Vh x t = 2.08 x 0.785 = 1.63m
You'll need to correct this part. Also, you found the horizontal distance from the release point, not its distance from the center. (Beats me what they want. Range is usually measured from the starting point, so you may be OK.)

Why not find the time in one step? Using:
Y = Y0 + Vy0*t -.5gt^2

Solve for t where Y = 0.
 
  • #19
Doc Al said:
Where does 2.08 come from?
2.4COS(30)

You'll need to correct this part. Also, you found the horizontal distance from the release point, not its distance from the center. (Beats me what they want. Range is usually measured from the starting point, so you may be OK.)

Why not find the time in one step? Using:
Y = Y0 + Vy0*t -.5gt^2

Solve for t where Y = 0.

I'm not sure either I think I'm supposed to add 1.5 to whatever answer I get not really sure.
 
  • #20
FalconKICK said:
2.4COS(30)
That's speed, not height.
 
  • #21
Doc Al said:
That's speed, not height.

Hmm...i have no idea what I'm doing then.
 
  • #22
Doc Al said:
That's speed, not height.

Should I replace it with 1Cos(30) + 1.5m?
 
  • #23
FalconKICK said:
Should I replace it with 1Cos(30) + 1.5m?
The height of point A is 1.5m + 1Sin(30).
 
  • #24
I solved it yesterday on my last try. It's actually pretty easy, but I need some help with Circular Motion so I'll need to talk to my teacher about it.
 

Related to Uniform Circular Motion of a stone

1. What is uniform circular motion?

Uniform circular motion is the motion of an object in a circular path at a constant speed. This means that the object is moving at the same speed and in the same direction throughout the entire circular path.

2. Why does a stone in uniform circular motion not have a constant velocity?

Although the speed of the stone is constant, its velocity is constantly changing because velocity is a vector quantity that takes into account both speed and direction. In circular motion, the direction of the velocity is constantly changing as the object moves along the circular path.

3. How is centripetal force related to uniform circular motion?

Centripetal force is the force that acts towards the center of a circular path, keeping an object in uniform circular motion. Without this force, the object would continue moving in a straight line tangent to the circle.

4. Can an object in uniform circular motion ever have a centripetal acceleration of zero?

No, an object in uniform circular motion will always have a non-zero centripetal acceleration. This is because the direction of the velocity is constantly changing, which means there is always a change in the object's acceleration.

5. What is the relationship between the radius of the circular path and the speed of the object in uniform circular motion?

The speed of an object in uniform circular motion is directly proportional to the radius of the circular path. This means that as the radius increases, the speed of the object also increases, and vice versa.

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