Newton's 2nd Law for a Particle in Uniform Circular motion

  • Thread starter clintau
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  • #1
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1. A 4.0kg object is attached to a vertical rod by 2 strings. Object roatates in a horizontal circle at a constand speed 6.00 m/s. Find the tention in the upper and lower string. The strings are 2 meters long. the distance between the attachment points for each string is 3 meters.



2. F=ma=m(v^2/r)



3. I know that you must sum the forces in the x and y direction but cant seem to get rolling on this
 

Answers and Replies

  • #2
G01
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Yes, you must sum the forces from both strings. The sum of the forces is a net centripetal force. What does this mean it has to be equal to?

HINT: What are all centripetal forces equal to?
 
  • #3
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i don't know.
 
  • #4
rock.freak667
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Draw a vector diagram...then split the tension into it's components.
 
  • #5
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Y direction
Fupper= T*cos(theta)+mg=0
Flower= T*cos(theta)-mg=0

yes no?
 
  • #6
G01
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i don't know.

Yes, you do. You said it in your first post.

F_c=mv^2/r

So, if:

[tex]\Sigma F_{radial direction} = F_c[/tex] and [tex]F_c=\frac{mv^2}{r}[/tex] then...

Also,

You need to sum the radial components (x components) of the forces, since only those will contribute to the net centripetal force. To find these it would help to draw a free body diagram as clintau said. How far can you get now?
 
  • #7
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so Fc=4*6^2/r=109.09

That gives the force in the x direction....What about the Y?
 
  • #8
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how would you find the tensions in the strings?
 
  • #9
rock.freak667
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[tex]\Sigma F_{vertical}=0[/tex]
 
  • #10
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Fcx is the T in the upper string. How do you get the T in the lower?

Tupper-mg? gives app 69N. Answer should be 56.2

what is wrong?
 
  • #11
rl.bhat
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Y direction
Fupper= T*cos(theta)+mg=0
Flower= T*cos(theta)-mg=0

yes no?

Instead of taking componets of tention on Y direction, take the component of mg along the tensions.
So T(upper) = T1 + mgcos(theta)
T(lower) = T2 - mgsin(theta)
Now T(upper), T(lower) and Fc are in equilibrium. Apply the Lami's theorem.
[T1 + mgcos(theta)]/sin(pi/2 + theta) = [T2 - mgsin(theta)]/sin(pi/2 + theta) = [mV^2/r]/sin(pi -2*theta).
Now solve for T1 and T2.
 

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