Magitude of accelaration

  • Thread starter rasikan
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  • #1
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Homework Statement


A particle moves along a circular path having a radius of 2.0 m. At an instant when the speed of the particle is equal to 3.0 m/s and changing at the rate of 5.0 m/s2, what is the magnitude of the total acceleration of the particle?



Homework Equations


magitude= V^2/R




The Attempt at a Solution



But i dont know how to slove with this equation
 

Answers and Replies

  • #2
Dick
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Acceleration is a vector. There will be a component tangent to the circle and a normal component. Which one is given to you in the problem statement and which do you have to find using your equation?
 
  • #3
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Dick,
v=3m/s
a=5 m/s^2

I guess I can find the magnitude when v=3 m/s . is that
right?


Acceleration is a vector. There will be a component tangent to the circle and a normal component. Which one is given to you in the problem statement and which do you have to find using your equation?
 
  • #4
Dick
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Ok, so what are the two components of the acceleration.
 
  • #5
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the two components are ax and ay
 
  • #6
Dick
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The components are normal to the circle and tangent to the circle, so x,y coordinates are probably not appropriate. I would say the tangent component is 5 m/s^2. Why would I say that?
 
  • #7
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coz that is a constant acceleration?
 
  • #8
Dick
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No, because they told me. If an object is moving on a circle and it's speed is changing at 5 m/s^2, in what direction is that acceleration pointed?
 
  • #9
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this is my calculation

a=Sq root of (at)^2+(ar)^2

at= 5m/s^2
ar=v^2/r= 9/2=4.5
a=sq root(5^2+4.5^2)
=6.7 m/s^2

is this is right??
 
  • #10
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towards its center
No, because they told me. If an object is moving on a circle and it's speed is changing at 5 m/s^2, in what direction is that acceleration pointed?
 
  • #11
Dick
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towards its center
Your calculation is fine. What does "towards its center" mean? You labelled it as at in your calculation?
 
  • #12
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u mean my answer is right??
 
  • #13
Dick
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Yeah. Does that surprise you? If you're clear on the two components of the acceleration then I think you can go on to the next problem...
 
  • #14
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yaa i got it now dick thanks for you help
 

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