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Uniform Circular motion

  • Thread starter unknown_2
  • Start date
  • #1
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This problem has been discussed before, but one of the parts were not discussed.

Homework Statement


Your calculation is actually a derivation of the centripetal acceleration. To see this, express the acceleration of the particle in terms of its position r(t).

r(t) = [tex]Rcos(\omega t)\textbf{i} + Rsin(\omega t)\textbf{j}[/tex]

Homework Equations



[tex]v = \omega R [/tex]
[tex]a = \frac{v^2}{R} [/tex]
[tex]a = \omega v [/tex]

The Attempt at a Solution



[tex]v = \frac{d}{dt}r(t) \times \omega[/tex]

btw, this is a Mastering engineering question, so i submitted and it says that:
The correct answer does not depend on the variables and functions: d, d, r.

is there another way to display this or am i missing something?

thx,
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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Acceleration = V^2/R = w^2*R. In a uniform circular motion w and R are constant.
 
  • #3
29
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Acceleration = V^2/R = w^2*R. In a uniform circular motion w and R are constant.
omg....too stressed out about finals...i think i got it =_=

please correct me if i'm wrong (again)...

a(t) = [tex] -\omega^{2}r(t)[/tex]
 
  • #4
rl.bhat
Homework Helper
4,433
5
omg....too stressed out about finals...i think i got it =_=

please correct me if i'm wrong (again)...

a(t) = [tex] -\omega^{2}r(t)[/tex]
If you express in the vector form, it is correct.
 

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