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Another angular acceleration problem. Converting radians to meters

  • #1
[SOLVED] Another angular acceleration problem. Converting radians to meters...

Homework Statement


A gyroscope flywheel of radius 3.13 cm is accelerated from rest at 15.0 rad/s2 until its angular speed is 2760 rev/min.

(a) What is the tangential acceleration of a point on the rim of the flywheel during this spin-up process?
m/s2

(b) What is the radial acceleration of this point when the flywheel is spinning at full speed?
m/s2

(c) Through what distance does a point on the rim move during the spin-up?
m


Homework Equations


X=Xo + Vot +1/2(a)t(squared)
V= Vo +at
V(squared) + Vo (squared)= 2a(x)
Replace x with radians, V with angular velocity (W), and a with angular accelation



The Attempt at a Solution


Working out a and b wasn't difficult, but finding c proved difficult. I first converted the radius to meters. I calculated both parts a and b. In part a I converted 2760rev/min to 289.03rad/s and converted the radius given in cm to m. Then I multiplied both and got a small tangential acceleration of .4695. Next for part b I calculated the radial acceleration by using the 289.03rad/s(squared) and multiplying it by the radius. For part c I found the hypotenuse acceleration of both of these by pythagorean's theorem and then I tried to use the basic kinematics equations converted to angular values for my answer and ended up getting 2784.54 radians. So after that I tried to divide it by 2times pie to convert it over to meters getting 443.17meters, but for some reason I still got it wrong. Did I convert it wrong?
 

Answers and Replies

  • #2
In this problem Part C is what I am having difficultly getting. I need to get the answer in meters, but the angular kinematics equations only give me radians what should I do?
 
  • #3
Nevermind this one I found out how to do. .5((2760 times 2pi/60) times 2614.69 Part B answer)squared. Divided by tangential acceleration of .4695
 

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