Find Magnitude & Direction of Force on Gyroscope Shaft w/ 5kg Mass

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Homework Help Overview

The problem involves a gyroscope system with a mass of 5kg, where participants are tasked with finding the magnitude and direction of the forces exerted on the shaft under various conditions of rotation. The context includes considerations of rotational motion and forces acting on a spinning object.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations related to torque and forces, with one attempting to derive expressions for different rotational scenarios. Questions arise regarding the sufficiency of information provided in the problem and interpretations of the rotation described.

Discussion Status

The discussion is ongoing, with some participants expressing confusion about the problem's wording and suggesting that it may be misinterpreted. There is an exploration of different interpretations of the shaft's rotation and its implications for the forces involved.

Contextual Notes

Some participants note potential missing information that could affect the analysis, and there are discussions about the assumptions made regarding the direction of rotation and the setup of the gyroscope system.

John O' Meara
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A demonstration gyro wheel is constructed by removing the tire from a bicycle wheel 1.0m in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects .2m at each side of the wheel and a man holds the ends of the shaft in his hands. The mass of the system is 5kg and its entire mass may be assumed to be located at its rim. The shaft is horizontal and the wheel is spinning about the shaft at 5 rev/s. Find the magnitude and direction of the force each hand exerts on the shaft under the following conditions:
(a) The shaft is at rest
(b) The shaft is rotating in a horizontal plane about its center at .04 rev/s
(c) The shaft is rotating in a horizontal plane about its center at .20 rev/s
(d) At what rate must the shaft rotate in order that it may be supported at one end only?
My attempt (a) w = 5 rev/s = 10*PI rad/s. moment of inertia I=m*r^2 = 5*(.5)^2. Let the force in each hand =p; torque = r*p therefore the torque (T) = 4.9N.m
(b) If the shaft rotates at .04 rev/s, then the wheel rotates at 5.04 rev/s
(d)the requirement is that the resultant vertical force = zero i.e., p - m*g = 0 therefore p (the normal reaction force) =5kg*9.8m/s/s. T=I*(alpha) = (alpha) = T/I=9.8/(5*.25)=7.84; also t=dL/dt. Any help in doing parts a,b,c and d would be welcome, many thanks.
 
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also t=dL/dt, should read T=dL/dt. I don't think that enough information has been given in the question! what do you think?
 
Does anyone know anything about this question?
 
John O' Meara said:
Does anyone know anything about this question?
I think the question is easily misinterpreted, and it appers to me you have misinterpreted at least part of it. The rotations of the shaft in a horizontal plane are intended to mean rotations about a vertical axis through the center of the horizontal shaft, not rotations about the long axis of the shaft. In other words, the shaft remains horizontal, but you are changing the direction of the long axis of the shaft in that plane.
 

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