A cord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. A steady horizontal pull of 40.0 N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axle exerts on the wheel. (c) Which of the answer in part (a) and (b) would change if the pull were upward instead of horizontal?

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,

$\alpha =\frac{\tau }{l}......\left(1\right)$.

Where the torque exerted by the force is,

$\tau =RF\cdots \cdots \left(2\right)$,

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

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

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

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

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,

$a=\alpha R$.

Therefore,

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

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,

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

$\therefore F_{axle}=98.6N$.

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,

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

Substituting values, we get,

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

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.