A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel 0.650 m in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects 0.200 m at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is 8.00 kg; 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 5.00 rev/s. Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at 0.050 rev/s; (c) when the shaft is rotating in a horizontal plane about its center at 0.300 rev/s. (d) At what rate must the shaft rotate in order that it may be supported at one end only?

It is given that the diameter of wheel as D = 0.650m , shaft length as d = 0.200m , mass of the system as m = 8.00 kg and the angular speed of the wheel as $\mathrm{\omega }=5.00\mathrm{rev}/\mathrm{s}$or $\mathrm{\omega }=31.4rad/\mathrm{s}$.

The weight of the wheel is depends on the force exerted by the hands. So the weight of the wheel can be written as $W=F_L+F_R$ and the net torque is given by $\tau =d\left(F_R-F_L\right)$.

The torque is related to the angular speed by $\mathrm{\tau }=\mathrm{\Omega }/\mathrm{\omega }$then,

$\begin{array}{r}d\left(F_R-F_L\right)=\mathrm{\Omega }/\omega \\ F_R-F_L=\frac{\mathrm{\Omega }/\omega }{d}\end{array}$

Substitute $F_L=mg-F_R$in the above equation

$\begin{array}{r}{F}_R-\left(mg-F_R\right)=\frac{\mathrm{\Omega }/\omega }{d}\\ 2F_R-mg=\frac{\mathrm{\Omega }/\omega }{d}\\ 2F_R=\frac{\mathrm{\Omega }/\omega }{d}+mg\\ F_R=\frac{1}{2}\left(\frac{\mathrm{\Omega }/\omega }{d}+mg\right)\end{array}$

Similarly, find $F_L$ as follows:

$\begin{array}{r}{F}_L=mg-\frac{1}{2}\left(\frac{\mathrm{\Omega }/\omega }{d}+mg\right)\\ =\frac{1}{2}\left(-\frac{\mathrm{\Omega }/\omega }{d}+mg\right)\\ =\frac{1}{2}\left(mg-\frac{\mathrm{\Omega }/\omega }{d}\right)\end{array}$

Since, the moment of inertia about the axis through the center is$I=m{R}^2$ then,
$F_R=\frac{1}{2}\left(\frac{\mathrm{\Omega }m{R}^2\omega }{d}+mg\right)\text{and}{F}_L=\frac{1}{2}\left(mg-\frac{\mathrm{\Omega }m{R}^2\omega }{d}\right)$

Here, the shaft does not rotate implies $\mathrm{\Omega }=0$ then,

$\begin{array}{r}{F}_R=\frac{1}{2}\left(\frac{\mathrm{\Omega }m{R}^2\omega }{d}+mg\right)\\ =\frac{1}{2}\left(0+\left(8.00\mathrm{kg}\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right)\\ =39.2\mathrm{N}\end{array}$

Similarly find $F_L$as follows:

$\begin{array}{r}{F}_L=\frac{1}{2}\left(mg-\frac{\mathrm{\Omega }m{R}^2\omega }{d}\right)\\ =\frac{1}{2}\left(\left(8.00\mathrm{kg}\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)-0\right)\\ =39.2\mathrm{N}\end{array}$

Therefore, the shaft is rest at ${\mathrm{F}}_{\mathrm{R}}=39.2\mathrm{N}$and $F_L=39.2N$.

It is given that $\mathrm{\Omega }=0.050\mathrm{rev}/\mathrm{s}$that implies $\mathrm{\Omega }=0.314\mathrm{rad}/\mathrm{s}$.

Substitute the known values in $F_R$and simplify.

$\begin{array}{r}{F}_R=\frac{1}{2}\left(\frac{\mathrm{\Omega }m{R}^2\omega }{d}+mg\right)\\ =\frac{1}{2}\left(\frac{\left(8.00\mathrm{kg}\right)(0.314\mathrm{rad}/\mathrm{s})\left(0.325\mathrm{m}{)}^2\right(31.4\mathrm{rad}/\mathrm{s})}{0.200\mathrm{m}}+\left(8.00\mathrm{kg}\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right)\\ =\frac{\left(8.00\mathrm{kg}\right)}{2}\left(\frac{(0.314\mathrm{rad}/\mathrm{s})\left(0.325\mathrm{m}{)}^2\right(31.4\mathrm{rad}/\mathrm{s})}{0.200\mathrm{m}}+\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right)\\ =60.0\mathrm{N}\end{array}$

Similarly, Substitute the known values in $F_L$and simplify.

$\begin{array}{r}{F}_L=\frac{1}{2}\left(\frac{\mathrm{\Omega }m{R}^2\omega }{d}+mg\right)\\ =\frac{1}{2}\left(-\frac{\left(8.00\mathrm{kg}\right)(0.314\mathrm{rad}/\mathrm{s})\left(0.325\mathrm{m}{)}^2\right(31.4\mathrm{rad}/\mathrm{s})}{0.200\mathrm{m}}+\left(8.00\mathrm{kg}\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right)\\ =\frac{\left(8.00\mathrm{kg}\right)}{2}\left(-\frac{(0.314\mathrm{rad}/\mathrm{s})\left(0.325\mathrm{m}{)}^2\right(31.4\mathrm{rad}/\mathrm{s})}{0.200\mathrm{m}}+\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right)\\ =18.4\mathrm{N}\end{array}$

Therefore, the required values are ${\mathrm{F}}_{\mathrm{R}}=60.0\mathrm{N}\mathrm{and}{\mathrm{F}}_{\mathrm{L}}=18.4\mathrm{N}$.

It is given that $\mathrm{\Omega }=0.300\mathrm{rev}/\mathrm{s}$that implies$\mathrm{\Omega }=1.89\mathrm{rad}/\mathrm{s}$.

Substitute the known values in $F_R$and simplify.

$\begin{array}{r}{F}_R=\frac{1}{2}\left(\frac{\mathrm{\Omega }m{R}^2\omega }{d}+mg\right)\\ =\frac{1}{2}\left(\frac{\left(8.00\mathrm{kg}\right)(1.89\mathrm{rad}/\mathrm{s})\left(0.325\mathrm{m}{)}^2\right(31.4\mathrm{rad}/\mathrm{s})}{0.200\mathrm{m}}+\left(8.00\mathrm{kg}\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right)\\ =\frac{\left(8.00\mathrm{kg}\right)}{2}\left(\frac{(1.89\mathrm{rad}/\mathrm{s})\left(0.325\mathrm{m}{)}^2\right(31.4\mathrm{rad}/\mathrm{s})}{0.200\mathrm{m}}+\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right)\\ =165\mathrm{N}\end{array}$

Similarly, Substitute the known values in $F_L$and simplify.

$\begin{array}{r}{F}_L=\frac{1}{2}\left(\frac{\mathrm{\Omega }m{R}^2\omega }{d}+mg\right)\\ =\frac{1}{2}\left(-\frac{\left(8.00\mathrm{kg}\right)(1.89\mathrm{rad}/\mathrm{s})\left(0.325\mathrm{m}{)}^2\right(31.4\mathrm{rad}/\mathrm{s})}{0.200\mathrm{m}}+\left(8.00\mathrm{kg}\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right)\\ =\frac{\left(8.00\mathrm{kg}\right)}{2}\left(-\frac{(1.89\mathrm{rad}/\mathrm{s})\left(0.325\mathrm{m}{)}^2\right(31.4\mathrm{rad}/\mathrm{s})}{0.200\mathrm{m}}+\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\right)\\ =-86.2\mathrm{N}\end{array}$

Therefore, the required values are ${\mathrm{F}}_{\mathrm{R}}=165\mathrm{N}\mathrm{and}{\mathrm{F}}_{\mathrm{L}}=-86.2\mathrm{N}$and .

Let one force $F_R$be zero then $F_L=mg$.

Consider the equation $\Omega =\frac{wd}{I\omega }$which can be written as $\Omega =\frac{mgd}{m{R}^2\omega }$.

Substitute all the known values and simplify.

$\begin{array}{r}\mathrm{\Omega }=\frac{mgd}{m{R}^2\omega }\\ =\frac{gd}{R^2\omega }\\ =\frac{\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(0.200\mathrm{m}\right)}{\left(0.325\mathrm{m}{)}^2\right(31.4\mathrm{rad}/\mathrm{s})}\\ =0.591\mathrm{rad}/\mathrm{s}\end{array}$

Therefore, the shaft rotate in a rate at $\mathrm{\Omega }=0.591\mathrm{rad}/\mathrm{s}$.