91影视

We have the given data:

The radius of the wheel (R) =0.250 m.

Mass of the wheel (m) =9.20 Kg.

Horizontal Force exerted on the cord (F) =40.0 N.

The angular acceleration is given by,

$伪=\frac{蟿}{l}......\left(1\right)$.

Where the torque exerted by the force is,

$蟿=RF鈰�鈰�\left(2\right)$,

and the moment of inertia of a solid wheel about an axis through its center is,

$I=\frac{1}{2}m{R}^2鈰�鈰�\left(3\right)$

Using and in the equation and Substituting values, we get,

$$\begin{gathered} 伪=\frac{RF}{{\displaystyle \frac{1}{2}}m{R}^2} \\ =\frac{2F}{mR} \\ =\frac{2\left(40.0\right)}{\left(9.20\right)\left(0.250\right)} \\ =34.8 \\ 鈭�伪=34.8rad/s \end{gathered}$$

Now, consider the linear acceleration of the part of the cord pulled from the wheel.

The relation between angular acceleration and linear acceleration is given by,

$a=伪R$.

Therefore,

$$\begin{gathered} a=(34.8)鈰�(0.250) \\ a=8.70m/s^2 \end{gathered}$$

Hence, the angular acceleration of the wheel is, $a=8.70m/s^2$.

The force exerted by the axle on the wheel keeps the wheel in place.

The horizontal force on the wheel is$F_x$ and vertical force is its weight $F_y$.

Then the total force F is given by,

$F=\sqrt{F_x^2+F_y^2}$.

The force exerted by the axle balances the force F and it has the same magnitude as force F, but points in opposite direction.

$F_{axle}=F=\sqrt{F_x^2+F_y^2}$.

Substituting the values and $F_y=mg$, we get,

$$\begin{gathered} F_{axle}=\sqrt{F_x^2+{\left(mg\right)}^2} \\ =\sqrt{{\left(40.0\right)}^2+\left(\right(9.20).{(9.80))}^2} \\ =98.6N \end{gathered}$$

$鈭�F_{axle}=98.6N$.

Now, taking the direction of the pull to be up in the positive x- direction and positive y-direction, we get the axle force as:

$F_{axle}=-F=-F_x{e}_x+F_y{e}_y$.

The angle that the force makes with the negative x- axis is,

$\tan 胃=\frac{F_y}{F_x}=\frac{mg}{F_x}$.

Substituting values, we get,

$$\begin{gathered} \tan 胃=\frac{\left(9.20\right)\left(9.80\right)}{40.0} \\ \tan 胃=2.254 \\ 胃={\tan }^{-1}\left(2.254\right) \\ 胃=66.1掳 \end{gathered}$$

If $F_x$were acting upward, part (a) of the calculation would not change since this would not change the torque.

Part (b) would change since $F_x$and$F_y$ would add differently.