Angular Acceleration

Posted on 2006-04-16
Medium Priority
Last Modified: 2006-11-18
A) Prove that when an object starts from rest and rotates about a fixed axis wit a constant angular acceleration, the radial acceleration of a point in the object is directly proportional to its angular displacment.

    This is how I did this part:

      a_rad = w^2*r, where w = omega.
      w^2 = w_0^2 + 2a(theta - theta0)

      a_rad = 2ar(theta - theta0)

Where a = alpha, r = radius... This part seems correct

B) Through what angle has the object turned at the instant when the resultant accleration of a point makes an angle of 36.9 degrees with the radial acceleration.

I'm not sure how to do this part...
Question by:BrianGEFF719
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Expert Comment

ID: 16466329
Resultant acceleration = tangential accelleration + radial accelleration
a_tan = r*a
atan(a_tan/a_rad)=36.9 degrees
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Author Comment

ID: 16466353
How did you determine that the tangential accleration was the y component and the radial was the x component? I thoguht the same thing arctan(y/x), but I didnt know which was which

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Accepted Solution

ozo earned 2000 total points
ID: 16466382
if tangential accelleration = 0,  the resultant accleration of a point makes an angle of 0 degrees with the radial acceleration.
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Author Comment

ID: 16466604

Theta = 1/(2 * tan(36.9))

is this correct?
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Author Comment

ID: 16466607
or .375 radians?


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