Let's break down the problem step-by-step: --- ## **Given Data** - **Radius of turn, \( R \) = 15 m** - **Mass of car, \( M \) = 798 kg** - **Coefficient of static friction, \( \mu = .85 \)** - **Gravitational acceleration, \( g = 9.81 \, \text{m/s}^2 \)** - **Tail fin lift coefficient, \( C_L = 2. \)** - **Fin width = 2 m, Length = 30 cm = .3 m** - **\( C_L/C_D = 35 \) (lift-to-drag ratio)** --- ## **(a) Maximum Speed Without Slipping** #### **Step 1: Forces Involved** - The car will slip if the frictional force is less than the required centripetal force. - **Maximum static friction force:** \[ f_{max} = \mu N \] where \( N \) is the normal force. #### **Step 2: Normal Force Calculation** - **Without tail fin:** \( N = Mg \) - **With tail fin:** The tail fin provides additional downward force (lift), \( F_{\text{lift}} \), so \[ N = Mg + F_{\text{lift}} \] #### **Step 3: Centripetal Force Requirement** \[ \frac{Mv^2}{R} \leq f_{max} \] #### **Step 4: With Tail Fin (Lift Downward)** \[ f_{max} = \mu (Mg + F_{\text{lift}}) \] \[ \frac{Mv^2}{R} = \mu (Mg + F_{\text{lift}}) \] \[ v_{max} = \sqrt{ \mu R \left( g + \frac{F_{\text{lift}}}{M} \right) } \] #### **Step 5: Calculate \( F_{\text{lift}} \)** \[ F_{\text{lift}} = \frac{1}{2} \rho v^2 C_L A \] - Air density (\( \rho \)), assume \( = 1.225 \, \text{kg/m}^3 \) (standard). - Area \( A = 2\,\text{m} \times .3\,\text{m} = .6\,\text{m}^2 \). \[ F_{\text{lift}} = \frac{1}{2} \times 1.225 \times v^2 \times 2. \times .6 = .735 v^2 \] #### **Step 6: Substitute \( F_{\text{lift}} \) into \( v_{max} \)** \[ v_{max} = \sqrt{ \mu R \left( g + \frac{.735 v_{max}^2}{M} \right) } \] Let \( x = v_{max}^2 \): \[ x = \mu R \left( g + \frac{.735 x}{M} \right ) \] \[ x = \mu R g + \mu R \frac{.735 x}{M} \] \[ x - \mu R \frac{.735 x}{M} = \mu R g \] \[ x \left( 1 - \frac{ \mu R \times .735 }{ M } \right ) = \mu R g \] \[ x = \frac{ \mu R g }{ 1 - \frac{ \mu R \times .735 }{ M } } \] **Plug in values:** - \( \mu = .85 \) - \( R = 15 \) - \( g = 9.81 \) - \( M = 798 \) Calculate denominator: \[ \frac{\mu R \times .735}{M} = \frac{.85 \times 15 \times .735}{798} = \frac{9.3675}{798} \approx .01174 \] So, \[ x = \frac{.85 \times 15 \times 9.81}{1 - .01174} \] \[ .85 \times 15 = 12.75 \] \[ 12.75 \times 9.81 = 125.0075 \] \[ x = \frac{125.0075}{.98826} \approx 126.56 \] \[ v_{max} = \sqrt{126.56} \approx 11.26 \, \text{m/s} \] --- ### **If the Tail Fin Were Not Present** Then \( F_{\text{lift}} = \): \[ v_{max} = \sqrt{ \mu g R } = \sqrt{ .85 \times 9.81 \times 15 } \] \[ .85 \times 9.81 = 8.3385 \] \[ 8.3385 \times 15 = 125.0775 \] \[ v_{max} = \sqrt{125.08} \approx 11.19\, \text{m/s} \] --- #### **Summary (Part a):** - **With tail fin:** \( v_{max} = 11.26\, \text{m/s} \) - **Without tail fin:** \( v_{max} = 11.19\, \text{m/s} \) --- ## **(b) Power to Overcome Drag Due to the Fin** - Lift-to-drag ratio: \( \frac{C_L}{C_D} = 35 \implies C_D = \frac{C_L}{35} = \frac{2}{35} \approx .057 \) - Drag force: \[ F_D = \frac{1}{2} \rho v^2 C_D A \] \[ = .5 \times 1.225 \times (11.26)^2 \times .057 \times .6 \] First, \( (11.26)^2 = 126.77 \) \[ F_D = .5 \times 1.225 \times 126.77 \times .057 \times .6 \] \[ = .6125 \times 126.77 = 77.65 \] \[ 77.65 \times .057 = 4.426 \] \[ 4.426 \times .6 = 2.656 \, \text{N} \] - **Power required:** \[ P = F_D \times v \] \[ = 2.656 \times 11.26 \approx 29.92\, \text{W} \] --- ## **Final Answers** ### **(a) Maximum speed:** - **With tail fin:** \( \boxed{11.26 \, \text{m/s}} \) - **Without tail fin:** \( \boxed{11.19 \, \text{m/s}} \) ### **(b) Power to overcome drag due to the fin:** - \( \boxed{29.9 \, \text{W}} \) (rounded to 1 decimal place)