The angular speed of the CD is $omega = 44 mathrm{~rad/s}$. The tangential speed of the CD is $v = 1.25 mathrm{~m/s}$.

$textbf{Required: }$

Finding the distance from the axis of rotation (radius).

As the textbook mentions, the tangential speed of a rotating object is given by

$$
begin{align*}
v &= r ~ omega \
end{align*}
$$

Rearrange and solve for the distance from the axis of rotation:

$$
begin{align*}
r &= dfrac{ v }{ omega } \
&= dfrac{ 1.25 mathrm{~m/s} }{ 44 mathrm{~rad/s} } \
&= 0.028 mathrm{~m} \
&= 2.841 mathrm{~cm}
end{align*}
$$

So, the distance from the axis of rotation (radius) is $2.841 mathrm{~cm}$.

The length of the hour’s hand is $r = 8.2 mathrm{~cm}$. The time taken to cover one revolution by the hour hand is $Delta t = 24 mathrm{~hr}$.

$textbf{Required: }$

Finding the tangential speed of the tip of the hour’s hand.

As the textbook mentions, the tangential speed of a rotating object is given by

$$
begin{align*}
v &= r ~ omega \
end{align*}
$$

In order to evaluate the angular speed, we use the following relation:

$$
begin{align*}
omega &= dfrac{ 2 pi }{ T } \
end{align*}
$$

Substituting from the previous calculations, then we get

$$
begin{align*}
v &= r ~ omega \
&= r ~ dfrac{ 2 pi }{ T } \
&= 8.2 times 10^{-2} mathrm{~m} times dfrac{ 2 pi }{ 24 times 60 times 60 mathrm{~s} } \
&= 5.963 times 10^{-6} mathrm{~m/s}
end{align*}
$$

So, the tangential speed of the tip of the hour’s hand is $5.963 times 10^{-6} mathrm{~m/s}$.