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Radial Acceleration of ball

  • Thread starter twoods
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  • #1
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Hi!
I am reviewing for the Physics GRE and am perplexed by this problem:

Homework Statement


A ball swings in a vertical circle at the end of a rope 1.5m long. When the ball is 36.9 degrees past the lowest point on its way up, its total acceleration is (-22.5, 20.1)m/s^2. At this instant (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball.


Homework Equations


a_c = v^2/R
sin(angle) = opposite/hyp.
cos(angle) = adjacent/hyp.

The Attempt at a Solution


See attached images.

I agree with the solutions I have attached for part (a), but not for (b) or (c). Especially how the solution manual determines the components of the radial acceleration in part (b) and direction of velocity in part (c).
 

Attachments

Answers and Replies

  • #2
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Reading this has made me question my own knowledge. For part b) I've always been told to work out the square root of the sum of each component squared. In this case, that will be [itex]\sqrt{22.5^{2}+20.2^{2}}[/itex] which is about 30.2 m s^-2. I've never heard of the method stated in the solution.

For part c) the direction at that given point is tangent to the circle like you said and the tangent makes the angle 36.9 with the horizontal, and you've have used the same method to work out the velocity as stated in the solution and that's correct.
 
  • #3
1,065
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Any vector can be replaced(resolved) to minimum of 2 vectors.
The book resolved the horizontal to tangential and radial
The vertical acceleration too is resolved to radial and tangential.
You can also resolve the radial to horizontal and vertical componets.

In your solution which is wrong,
1. You resolve ar1 to vertical only. 3=2+1. Never 3=2
2. Only one radial acceleration. You can resolve this acceleration to vertical AND horizontal. Remember the word AND when resolving vectors.
 
  • #4
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Thank y'all for the replies!

Azizlwl, I solved for the hypotenuse of the "acceleration triangles" - so the radial parts - not the vertical parts - and added them together.

Does this not work?
 
  • #5
Is the solution wrong? How is the angle of the rope related to the tangential acceleration anyway?


Edit: Whoops I didn't read the question.
 
  • #6
Reading this has made me question my own knowledge. For part b) I've always been told to work out the square root of the sum of each component squared. In this case, that will be [itex]\sqrt{22.5^{2}+20.2^{2}}[/itex] which is about 30.2 m s^-2. I've never heard of the method stated in the solution.

For part c) the direction at that given point is tangent to the circle like you said and the tangent makes the angle 36.9 with the horizontal, and you've have used the same method to work out the velocity as stated in the solution and that's correct.
The radial acceleration has to point towards the center of the circle. If you use Pythagoras Theorem the resulting acceleration will probably point somewhere else.

If you have to use Pythagoras Theorem then you'll need to find the values of tangential acceleration and the net acceleration before you can find the radial acceleration. And obviously you don't have tangential acceleration.

See the right diagram:
http://buphy.bu.edu/~duffy/PY105/15g.GIF
 

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