To find the angular speed ($\omega$) of the revolving door, we can use the relationship between tangential speed ($v$) and angular speed:

$v = r \cdot \omega$

where:

• $v = 1.9 \, \text{m/s}$ (tangential speed),
• $r = 0.76 \, \text{m}$ (distance from the center of the door to the woman).

Rearranging for $\omega$:

$\omega = \frac{v}{r}$

Substitute the values:

$\omega = \frac{1.9 \, \text{m/s}}{0.76 \, \text{m}} \approx 2.5 \, \text{rad/s}$

Answer:

The angular speed of the door is approximately $2.5 \, \text{rad/s}$.

Would you like further details or have any questions?

Here are five related questions for further exploration:

1. What would be the angular speed if the tangential speed doubled?
2. How would the angular speed change if the woman was closer to the center?
3. What is the tangential acceleration if she increases her speed uniformly?
4. How much time would it take for one full revolution?
5. How would the angular velocity be affected if the door had resistance?

Tip: Remember that angular speed can be derived by dividing tangential speed by the radius when objects move in a circular path.