Toy Gyroscope Problem: Finding Angular Speed and Upward Force | Homework Help

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The discussion revolves around solving a physics problem involving a toy gyroscope, specifically finding the upward force exerted by the pivot and the angular speed of the rotor. Key equations include angular momentum and the relationship between torque and angular velocity. The upward force is determined to be equal to the total weight of the gyroscope, while the angular speed was initially miscalculated due to a conversion error from radians per second to revolutions per minute. Participants shared their struggles with the problem and emphasized the importance of correctly applying formulas and unit conversions. The conversation highlights common challenges in understanding rotational dynamics in physics.
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Homework Statement



The rotor (flywheel) of a toy gyroscope has mass 0.140 kilograms. Its moment of inertia about its axis is 1.20x10^{-4} kilogram meters squared. The mass of the frame is 0.0250 kilograms. The gyroscope is supported on a single pivot with its center of mass a horizontal distance 4.00 centimeters from the pivot. The gyroscope is precessing in a horizontal plane at the rate of one revolution in 2.20 seconds.

yf_Figure_10_49.jpg


Find the upward force exerted by the pivot.

Find the angular speed omega at which the rotor is spinning about its axis, expressed in revolutions per minute.

Homework Equations



angular momentum and acceleration equations eg:
w = O / t
w= w0 + at
O = w0t + ½ at^2
w^2 = w0^2 + 2aO


The Attempt at a Solution



having a lot of trouble determining what information is relevant in this question...very little idea how to go about it sorry.

i think it has to do with L = Iw, with I being 1.2x10^4, but I'm unsure of how to get the angular velocity, and after that the force they are asking for.
 
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PCB250??
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now guess why I am here...
 
You will have to change (Big Omega) O from rev/2.2s into rad/s : O = 2pi/2.2s = 2.856 rad/s.
The upward force is equal to the weight because the gyroscope is not moving sideways, up or down so Fx = 0, and Fy = 0. Hence n = w = (m_rotor + m_frame)*9.8.

We know that O = T/L (where Torque: T = radius * w and L = I * w (small omega))

I is given with 1.2E^-4 kg*m^2.

There's an equation that tells you that L = I*w here so you just have to find the L from the equations above and then isolate w(small omega) from L = I * w(small omega).

then you'll get some number but you will have to answer in rev/min and remember that O is given in rev/2.2s but you changed it to 2.856rad/s

I have to admit that I'm in trouble finding the answer in rev/min if somebody could point out the right way to do so.
 
well evanQ, guess who had two attempts left when doing that. then i got the right answer! but it wasnt it revolutions per minute. so i had to convert it...and i MULTIPLIED BY 60 INSTEAD OF DIVIDED! ahhh! soooooo annoyed.

anywho, that's all in the past...now there another one to worry about. where we get marks taken off for getting multichoice wrong, nooooooo
 
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