Centripetal Force Problem – Lightning McQueen

Problem Statement

• Calculate the centripetal force required to keep Lightning McQueen (massive car character) moving in a circle on a flat, horizontal racetrack.

• Scenario specifics:

  • Car mass: 800 kg

  • Speed: 50 m s⁻¹

  • Track radius: 200 m

Key Physics Concept: Centripetal Force

• Centripetal force (symbol $F_c$) is the net force that keeps an object moving in uniform circular motion, always directed toward the center of the circle.

• It prevents the object from flying off tangentially due to inertia.

• Magnitude is determined by $F_c = \frac{m v^2}{r}$ where

  • $m$ = mass of object (kg)

  • $v$ = linear velocity (m s⁻¹)

  • $r$ = radius of circular path (m)

Given Data (Organized Table-Style)

• $m = 800\,\text{kg}$

• $v = 50\,\text{m}\,\text{s}^{-1}$

• $r = 200\,\text{m}$

Step-by-Step Calculation

1. Write the formula:
   $F_c = \frac{m v^2}{r}$

2. Substitute numerical values:
   $F_c = \frac{800\,(50)^2}{200}$

3. Simplify efficiently (cancel zeros):

   • Cancel one factor of 100 by dividing numerator and denominator: $(800 \rightarrow 8; 200 \rightarrow 2)$

   • Expression becomes $F_c = \frac{8\,(50)^2}{2}$

4. Compute squared velocity:
   $(50)^2 = 2{,}500$

5. Divide then multiply (or multiply then divide):

   • $\frac{8}{2} = 4$

   • $4 \times 2{,}500 = 10{,}000$

6. Attach proper SI unit (newton, N):
   $F_c = 10{,}000\,\text{N}$

Final Result

• Lightning McQueen experiences a centripetal force of $10{,}000\,\text{N}$ directed toward the center of the racetrack to maintain his circular path at 50 m s⁻¹.

Additional Insights & Real-World Relevance

• Racing Safety: Engineers design tires, suspension, and track banking to supply at least this much lateral force; otherwise, the car would skid outward.

• Proportionalities:

  • $F_c \propto v^2$ (doubling speed quadruples required force)

  • $F_c \propto \frac{1}{r}$ (tighter turns demand more force)

• Ethical/Practical Implication: Understanding centripetal requirements helps prevent accidents and ensures spectator and driver safety.

• Connection to Earlier Coursework: Builds on Newton’s 2nd law (sum of forces equals mass times acceleration), where centripetal acceleration is $a_c = \frac{v^2}{r}$.

• Hypothetical Extension: If McQueen sped up to 100 m s⁻¹ on the same track, required force would become $\frac{800\,(100)^2}{200} = 40{,}000\,\text{N}$ (quadruple), illustrating the quadratic speed dependence.

Problem Statement

• Calculate the centripetal force required to keep a rollercoaster car moving through a vertical loop.

• Scenario specifics:

  • Car mass: 500 kg

  • Speed at the bottom of the loop: 20 m s⁻¹

  • Loop radius: 10 m

Key Physics Concept: Centripetal Force

• Centripetal force (symbol $F_c$) is the net force that keeps an object moving in uniform circular motion, always directed toward the center of the circle.

• It prevents the object from flying off tangentially due to inertia.

• Magnitude is determined by $F_c = \frac{m v^2}{r}$ where

  • $m$ = mass of object (kg)

  • $v$ = linear velocity (m s⁻¹)

  • $r$ = radius of circular path (m)

Given Data (Organized Table-Style)

• $m = 500\,\text{kg}$

• $v = 20\,\text{m}\,\text{s}^{-1}$

• $r = 10\,\text{m}$

Step-by-Step Calculation

1. Write the formula:

   $F_c = \frac{m v^2}{r}$

2. Substitute numerical values:

   $F_c = \frac{500\,(20)^2}{10}$

3. Compute squared velocity:

   $(20)^2 = 400$

4. Substitute squared velocity into the formula:

   $F_c = \frac{500 \times 400}{10}$

5. Perform multiplication then division:

   $500 \times 400 = 200{,}000$
   $\frac{200{,}000}{10} = 20{,}000$

6. Attach proper SI unit (newton, N):

   $F_c = 20{,}000\,\text{N}$

Final Result

• The rollercoaster car experiences a centripetal force of $20{,}000\,\text{N}$ directed toward the center of the loop to maintain its circular path at 20 m s⁻¹.

Additional Insights & Real-World Relevance

• Rollercoaster Design: Understanding the maximum centripetal force (often at the bottom of the loop where speed is highest) is critical for structural integrity and passenger safety.

• Proportionalities:

  • $F_c \propto v^2$ (doubling speed quadruples required force)

  • $F_c \propto \frac{1}{r}$ (tighter turns demand more force)

• Ethical/Practical Implication: Proper design based on these calculations prevents ride malfunctions and ensures thrill without undue risk.

• Connection to Earlier Coursework: Builds on Newton’s 2nd law (sum of forces equals mass times acceleration), where centripetal acceleration is $a_c = \frac{v^2}{r}$.

• Hypothetical Extension: If the rollercoaster car doubled its speed to 40 m s⁻¹ at the bottom of the same 10m loop, the required force would become $\frac{500\,(40)^2}{10} = 80{,}000\,\text{N}$ (quadruple), illustrating the quadratic speed dependence.