Calculating Angular Momentum and Precession of a Toy Gyroscope

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The discussion focuses on calculating the angular momentum and precession of a toy gyroscope with a mass of 2.5 kg and a radius of 2 cm, mounted on a 7 cm axle. The angular momentum of the disk about its axle is determined to be 0.148 kg·m²/s, while the torque due to the weight of the gyroscope is calculated as 1.72 N·m. The participants discuss the relationship between torque and angular momentum to find the precession period, emphasizing the need to use the equation torque/(Iω) to derive the angular velocity of precession. The final calculations for part d, involving doubled dimensions, hinge on the results from part c.

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cuz937100
A toy gyroscope consists of a M = 2.5 kg disk with a radius of R = 2 cm mounted at the end of a light axle r = 7 cm long. The gyroscope spins at f = 47 rev/s. It is held horizontal at one end of its axle and precesses about its stand as shown.

a) What is the angular momentum of the disk about its axle?
b) What is the torque about the point of support due to the weight of the gyro?
c) How long does it take the gyroscope to precess once around?
d) If the radius of the disk and the length of the axle were both doubled (R = 4 cm, r = 14 cm), and the rate at which the disk spins were kept the same, how long would it take the disk to precess once?

I have figured out parts a and b (a.0.148,b.1.72), and I thought that part c would just be the inverse of (part b/part a)(Torque/Angular Momentum) due to the fact that angular precesion is just a frequency meaning the inverse would be the period. This did not work and I can not figure out part d without part c. Can someone please help me with this problem? There is a pic of it attached...
 

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cuz937100 said:
A toy gyroscope consists of a M = 2.5 kg disk with a radius of R = 2 cm ...
HOLY CRAP!




cuz937100 said:
... I thought that part c would just be the inverse of ... (Torque/Angular Momentum) due to the fact that angular precesion is just a frequency meaning the inverse would be the period.
It seems to be just an issue of a factor of 2π. τ/L = ANGULAR frequency of precession. Recall:

f = ω/2π

=>

T = 2π/ω

That's my suggestion.
 
Ok, I understand what you're saying, but how do I apply it to the problem. The help for the problem says

C.HELP: Use the answers from parts (a) and (b).
HELP: The precession rate is the torque divided by the component of the angular momentum that is precessing.
 
Anymore suggestions? I really need help with this one, I have an exam over this section this week...
 
Use the equation torque/(Iw). This will give you your angular velocity of precession. Now you need to take 2pi divided by your result to get it in terms of angular frequency. Glad I could help cuz
 
lol, thanks a lot masamune, i already got it though...And I got it before you did ANDREW...
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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