Q56P - University Physics with Modern Physics

Question

A 50.0-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. (a) If the plane’s speed at the lowest point of the circle is 95.0m/s , what is the minimum radius of the circle so that the acceleration at this point will not exceed  4.00 g? (b) What is the apparent weight of the pilot at the lowest point of the pullout?

Step-by-Step Solution

Answer

(a) The minimum radius of the circle is, 230.23 m.

(b) The apparent weight of the pilot is, 2450 N.

Step 1: Identification of given data

The given data can be listed below as, 

• The mass of the pilot is,  $\mathrm{m}=50.0\hspace{0.33em}\mathrm{kg}$.
• Velocity of the plain is,  $\mathrm{v}=95.0\hspace{0.33em}\mathrm{m}/\mathrm{s}$.

Step 2: Significance of Centripetal force

Force acting on the body due to which it moves on a circular path is known as centripetal force. It is always directed to the center of the circular path.

Step 3: Determination of minimum radius of the circle.

The path is circular, the expression for centripetal force can be expressed as,

$F=\frac{m{v}^2}{r}$ 

 

 

Here m is the mass of the pilot, v  is the velocity of plane, and r is the radius of the circular path.

 

Substitute 50 kg for m , 95.0 m/s  for v  in the above equation.

 

$F=\frac{\left(50 kg\right){\left(95.0 m/s\right)}^2}{r}$ 

 

The expression for the force using Newton’s second law can be expressed as,

 

$F=m a$ 

 

Here m is the mass, and a is the acceleration.

 

Substitute $\frac{\left(50 kg\right){\left(95.0 m.s\right)}^2}{r}$ for F , $4.00\hspace{0.33em}(9.8\hspace{0.33em}m/s^2)$ for a , and 50 kg for m in the above equation.

$$\begin{gathered} \frac{\left(50 \mathrm{kg}\right){\left(95.0 \mathrm{m}.\mathrm{s}\right)}^2}{\mathrm{r}}=\left(50 \mathrm{kg}\right)\times 4.00\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{r}=\frac{{\left(95.0 \mathrm{m}/\mathrm{s}\right)}^2}{4.00\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)} \\ =230.23 \mathrm{m} \end{gathered}$$ 

Hence, required radius is, 230.23 m.

Step 4: Determination of apparent weight of the pilot at the lowest point of pullout.

The expression for the weight at lowest point can be expressed as,

$N_{bottom}=m g+\frac{m v^2}{r}$ 

 

 

Substitute 50 kg for m , $9.8m/s^2$ for g ,and 95.0 m/s  for v , 230.23 m for r in the above equation.

$$\begin{gathered} {\mathrm{N}}_{\mathrm{bottom}}=\left(50 \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)+\frac{\left(50 \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)}{230.23 \mathrm{m}} \\ =2450 \mathrm{N} \end{gathered}$$ 

 

 Hence, the required weight is, 2450 N.