Solving Rotational Motion Problem: Find Acceleration

AI Thread Summary
To find the acceleration in a rotational motion problem, both centripetal and tangential components need to be calculated. The tangential acceleration is derived using the formula a_t = rα, where α is angular acceleration. Angular velocity can be determined from revolutions by converting them to radians and applying kinematic equations. The calculations for centripetal acceleration involve using a_n = rω², while ensuring unit consistency is crucial. The discussion emphasizes careful arithmetic and unit conversion to arrive at accurate results.
Jason03
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Heres the problem I am workin on

http://img294.imageshack.us/img294/9166/rotatesm3.jpg

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

I found tangetial by the formula

a_{t} = r\alpha


I converted the 2 revolutions to

4\pi radians

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

how to I get angular velocity from revolutions?
 
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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.
 
ok how does this look

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

= 0 + 2(.8)(4\pi radians)

\omega = 4.48 rad/s



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

a_{t} = r\theta = (.6)(.8) = .480

A = \sqrt{(9.480)^2 + (27.18)^2} = 27.18
 
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Your method looks good, but check your arithmetic here:
Jason03 said:
a_{n} = r\omega^2 = (.6)(20.16) = 27.18
 
ohhh thanks...im having trouble reading my calculators display!

that changes the answer to 12.09 rad/s^2
 
Jason03 said:
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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