Solve Centripetal Acceleration at 0.050m from CD-ROM Center

  • Thread starter scw287
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In summary, the computer is reading data from a rotating CD-ROM. At a point that is 0.030 m from the center of the disc, the centripetal acceleration is 118 m/s2. At a point that is 0.050 m from the center of the disc, the centripetal acceleration is .720 m/s2.
  • #1
scw287
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Homework Statement



A computer is reading data from a rotating CD-ROM. At a point that is 0.030 m from the center of the disc, the centripetal acceleration is 118 m/s2. What is the centripetal acceleration at a point that is 0.050 m from the center of the disc?

Homework Equations



Ac=v^2/r-centripual acceleration
possibly v=2pier/T

The Attempt at a Solution


Aa=Va^2/Ra
Ab=Vb^2/Rb
substituted Va=1.667*r*a, 1.667*ra<<<1.667 I got by dividing .050m/.030m to get the ratio of the radius'

I set up the equations like this:

118^2/.5^2=(1.667*118 m/s^2*V)^2/(1.667*.30)
>>crossed multiplied .25(196.706V)^2=13924(.5001)
solved for V and got V=.720<<<this is wrong

I think I'm on the right track I think there's an error on the way I set up my equations before substituting in the numbers.
 
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  • #2
What is Va?

How does V ratio with radius?
 
  • #3
Not quite sure what you are doing, but here's a hint that might help: Rewrite the formula for centripetal acceleration in terms of angular velocity (omega) instead of tangential velocity (v). (How are those two quantities related?)
 
  • #4
Va means velocity times the acceleration
 
  • #5
Doc Al said:
Not quite sure what you are doing, but here's a hint that might help: Rewrite the formula for centripetal acceleration in terms of angular velocity (omega) instead of tangential velocity (v). (How are those two quantities related?)

I haven't learned angular velocity yet so I'm not sure how to relate the two. I was trying to use the formula for centrepitual acceleration A=v^2/r
I then set up the equation for the first distance from the center of the CD and then set up a second equation for the second distance from the center. I used the ratio .05/.03=1.667 to tie the two distances together. I then subtitued all the known and solved for the velocity.
 
  • #6
scw287 said:
Va means velocity times the acceleration

No it doesn't, not as you've used it:

Aa=Va^2/Ra
Ab=Vb^2/Rb


You've used a and b as the first and second points. Call them V1 and V2 if you like to be clearer.

Whatever notation you use, my hint was to calculate the linear velocity (Va or V1 or whatever), and then use what you know about how the velocity varies with radius to solve the problem.
 
  • #7
use the expression for centripetal acceleration and see how it is related to the radius. thry to write it in terms of something that these two points have in common (like period of the motion)

then solve for the period. since the period is the same for both, equate the two get the answer.
 
  • #8
So I'm trying to learn how to do it using the angular velocity way. I'm not familiar with this method and I was hoping if you could check my work.a = omega^2 r

So a is proportional to r.

a(r) = a(@ r') (r / r')

plug the numbers in and i get

a(.050)=118(.030)(.050/.030)

=118<<i don't think this answer is right, I have a feeling there is something wrong with the left side maybe I'm not supposed to multiply a*the radius?
 
  • #9
a is proportional to r but not as "constant = a*r"

look at the equation you wrote first a=omega^2 r
then you get "omega^2(which is a constant) = ...)
 
  • #10
esalihm said:
a is proportional to r but not as "constant = a*r"

look at the equation you wrote first a=omega^2 r
then you get "omega^2(which is a constant) = ...)

Its equal to the angular velocity so would it be 118*(.050/.030)
 
  • #11
scw287 said:
So I'm trying to learn how to do it using the angular velocity way. I'm not familiar with this method and I was hoping if you could check my work.


a = omega^2 r

So a is proportional to r.

a(r) = a(@ r') (r / r')
Exactly right.
plug the numbers in and i get

a(.050)=118(.030)(.050/.030)
Careful with your notation. Express it like this:
a_2 = a_1 (r_2/r_1)

=118<<i don't think this answer is right, I have a feeling there is something wrong with the left side maybe I'm not supposed to multiply a*the radius?
See my note above.
 
  • #12
scw287 said:
Its equal to the angular velocity so would it be 118*(.050/.030)
Exactly...
 
  • #13
Doc Al said:
Exactly right.

Careful with your notation. Express it like this:
a_2 = a_1 (r_2/r_1)


See my note above.
that is much easier to understand!
 

1. What is the formula for calculating centripetal acceleration?

The formula for calculating centripetal acceleration is a = v2/r, where a is the centripetal acceleration, v is the velocity, and r is the radius of the circular motion.

2. Why is centripetal acceleration important?

Centripetal acceleration is important because it is responsible for keeping an object moving in a circular path. Without it, the object would continue in a straight line tangent to the circle.

3. How does centripetal acceleration change with distance from the center?

According to the formula, centripetal acceleration is inversely proportional to the radius. This means that as the distance from the center increases, the centripetal acceleration decreases.

4. How does the speed of an object affect centripetal acceleration?

The speed of an object has a direct relationship with centripetal acceleration. This means that as the speed increases, so does the centripetal acceleration.

5. How do you solve for centripetal acceleration at a specific distance from the center?

To solve for centripetal acceleration at a specific distance from the center, you will need to know the velocity and radius of the circular motion. Then, you can use the formula a = v2/r to calculate the centripetal acceleration at that distance.

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