91影视

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 $\mathrm{蝇}=5.00\mathrm{rev}/\mathrm{s}$or $\mathrm{蝇}=31.4rad/\mathrm{s}$.

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 $蟿=d\left(F_R-F_L\right)$.

The torque is related to the angular speed by $\mathrm{蟿}=\mathrm{惟}/\mathrm{蝇}$then,

$\begin{array}{r}d\left(F_R鈭�F_L\right)=\mathrm{惟}/蝇\\ F_R鈭�F_L=\frac{\mathrm{惟}/蝇}{d}\end{array}$

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

$\begin{array}{r}{F}_R鈭�\left(mg鈭�F_R\right)=\frac{\mathrm{惟}/蝇}{d}\\ 2F_R鈭�mg=\frac{\mathrm{惟}/蝇}{d}\\ 2F_R=\frac{\mathrm{惟}/蝇}{d}+mg\\ F_R=\frac{1}{2}\left(\frac{\mathrm{惟}/蝇}{d}+mg\right)\end{array}$

Similarly, find $F_L$ as follows:

$\begin{array}{r}{F}_L=mg鈭�\frac{1}{2}\left(\frac{\mathrm{惟}/蝇}{d}+mg\right)\\ =\frac{1}{2}\left(鈭�\frac{\mathrm{惟}/蝇}{d}+mg\right)\\ =\frac{1}{2}\left(mg鈭�\frac{\mathrm{惟}/蝇}{d}\right)\end{array}$

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

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

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

Similarly find $F_L$as follows:

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

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

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

Substitute the known values in $F_R$and simplify.

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

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

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

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 $\mathrm{惟}=0.300\mathrm{rev}/\mathrm{s}$that implies$\mathrm{惟}=1.89\mathrm{rad}/\mathrm{s}$.

Substitute the known values in $F_R$and simplify.

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

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

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

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 $惟=\frac{wd}{I蝇}$which can be written as $惟=\frac{mgd}{m{R}^2蝇}$.

Substitute all the known values and simplify.

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

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