The radius of the spool is $r = 0.066 mathrm{~m}$. The mass of the spool is $m = 1.8 mathrm{~kg}$. The angular acceleration is $alpha = 120 mathrm{~rad/s^{2}}$.

$textbf{Required: }$

Finding the required tension to give the spool the angular acceleration.

Since the only force that acts on the meterstick is the weight. Then the direction of the net force is downward. As the textbook mentions that the torque is given by the product of force and distance

$$
begin{align*}
tau &= r ~ T ~ sin left( theta right) \
&= r ~ T ~ sin left( 90^{circ} right) \
&= r ~ T
end{align*}
$$

In order to evaluate the angular acceleration of the spool, we use the following relation:

$$
begin{align*}
alpha &= dfrac{ tau }{ I } \
tau &= alpha ~ I \
end{align*}
$$

Rearrange and solve for the acting force:

$$
begin{align*}
T &= dfrac{ tau }{ r ~ sin left( theta right) } \
&= dfrac{ alpha ~ I }{r ~ ~ sin left( 90^{circ} right) } \
&= dfrac{ 15 mathrm{~N cdot m} }{ 0.25 mathrm{~m} } \
&= 60 mathrm{~N}
end{align*}
$$

In order to evaluate the moment of inertia of the spool, we use the following relation:

$$
begin{align*}
I &= dfrac{1}{2} ~ m ~ r^{2} \
end{align*}
$$

Substituting from the previous calculation, then we get

$$
begin{align*}
T &= dfrac{ alpha ~ I }{ r } \
&= dfrac{ alpha ~ dfrac{1}{2} ~ m ~ r^{2} }{ r } \
&= dfrac{ alpha ~ m ~ r }{ 2} \
&= dfrac{ 120 mathrm{~rad/s^{2}} times 1.8 mathrm{~kg} times 0.066 mathrm{~m} }{ 2 } \
&= 7.128 mathrm{~N}
end{align*}
$$

So, the required tension to give the spool the angular acceleration is $7.128 mathrm{~N}$.

The radius of the wheel is $r = 0.35 mathrm{~m}$. The mass of the wheel is $m = 0.75 mathrm{~kg}$. The torque that applied to the wheel is $tau = 0.97 mathrm{~N cdot m}$.

$textbf{Required: }$

Finding the angular acceleration of the wheel.

In order to evaluate the angular acceleration of the spool, we use the following relation:

$$
begin{align*}
alpha &= dfrac{ tau }{ I } \
end{align*}
$$

In order to evaluate the moment of inertia of the wheel as hoop, we use the following relation:

$$
begin{align*}
I &= m ~ r^{2} \
end{align*}
$$

Substituting from the previous calculation, then we get

$$
begin{align*}
alpha &= dfrac{ tau }{ I } \
&= dfrac{ tau }{ m ~ r^{2} } \
&= dfrac{ 0.97 mathrm{~N cdot m} }{ 0.75 mathrm{~kg} times left( 0.35 mathrm{~m} right)^{2} } \
&= 10.558 mathrm{~rad/s^{2}}
end{align*}
$$

So, the angular acceleration of the wheel is $10.558 mathrm{~rad/s^{2}}$.