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A gyroscope question

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.
 
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?
 

OlderDan

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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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