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Rotational Motion

  • Thread starter Jason03
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  • #1
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Heres the problem Im workin on

http://img294.imageshack.us/img294/9166/rotatesm3.jpg [Broken]

Now in order to find the magnitude of accleration Im assuming I would need to find the two components of acceleration first, Centripetal and Tangential......

I found tangetial by the formula

[tex] a_{t} = r\alpha [/tex]


I converted the 2 revolutions to

[tex] 4\pi radians [/tex]

but im not exactly sure how to get tangential...i need angualr velocity......

how to I get angular velocity from revolutions?
 
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Answers and Replies

  • #2
Doc Al
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You'll need to do a little kinematics. You have the angle in radians,the angular acceleration, and the initial angular speed--use kinematic formulas to find the final angular speed. Hint: Angle is the analog to distance, just like angular speed and acceleration are the analogs to linear speed and acceleration.
 
  • #3
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ok how does this look

[tex] \omega^2 = \omega_{o} + 2\alpha(\theta-\theta_{o})[/tex]

[tex]= 0 + 2(.8)(4\pi radians)[/tex]

[tex]\omega = 4.48 rad/s[/tex]



[tex]a_{n} = r\omega^2 = (.6)(20.16) = 27.18 [/tex]

[tex] a_{t} = r\theta = (.6)(.8) = .480 [/tex]

[tex]A = \sqrt{(9.480)^2 + (27.18)^2} = 27.18 [/tex]
 
Last edited:
  • #4
Doc Al
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Your method looks good, but check your arithmetic here:
[tex]a_{n} = r\omega^2 = (.6)(20.16) = 27.18 [/tex]
 
  • #5
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ohhh thanks....im having trouble reading my calculators display!

that changes the answer to 12.09 rad/s^2
 
  • #6
Doc Al
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that changes the answer to 12.09 rad/s^2
That looks better, but be careful with units. The acceleration is in m/s^2.
 

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