Question

a 51.0 kg stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. if the plane’s speed at the lowest point of the circle is 94.0 m/s, what is the minimum radius of the circle so that the acceleration at the point will not exceed 4.00g? express your answer with the appropriate units.

Explanation:

Step1: Recall centripetal acceleration formula

The centripetal acceleration \( a_c \) is given by \( a_c = \frac{v^2}{R} \), where \( v \) is the speed and \( R \) is the radius of the circular path. We want \( a_c \leq 4.00g \), so we'll use \( a_c = 4.00g \) to find the minimum \( R \) (since a larger \( R \) would give a smaller \( a_c \), so the minimum \( R \) corresponds to the maximum allowed \( a_c \)).

Step2: Substitute \( a_c = 4.00g \) into the formula

We know \( g = 9.8 \, \text{m/s}^2 \), \( v = 94.0 \, \text{m/s} \), and \( a_c = 4.00g \). Substituting into \( a_c = \frac{v^2}{R} \), we get \( 4.00g = \frac{v^2}{R} \).

Step3: Solve for \( R \)

Rearranging the formula for \( R \), we have \( R = \frac{v^2}{4.00g} \).

Step4: Plug in the values

Substitute \( v = 94.0 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \) into the formula:
\( R = \frac{(94.0)^2}{4.00 \times 9.8} \)
First, calculate \( (94.0)^2 = 8836 \). Then, calculate \( 4.00 \times 9.8 = 39.2 \). Now, divide: \( R = \frac{8836}{39.2} \approx 225.41 \, \text{m} \).

Answer:

The minimum radius \( R \) is approximately \( \boldsymbol{225} \, \text{m} \) (or more precisely \( 225.41 \, \text{m} \)) with units of meters.