Sample Problems - Vertical Circles

Page 1: Vertical Circles Sample Problem

Problem 1: Ball on a String

  • Given Data:

    • Mass of the ball, m = 0.025 kg

    • Length of the string, L = 0.6 m

    • Number of revolutions = 10

    • Time taken, t = 6 s

  • a. Tension at the Bottom of the Circle

    • Free Body Diagram (FBD):

      • Forces acting on the ball:

        • Tension (T) acting upwards

        • Gravitational force (mg) acting downwards

    • Net Force Equation:

      • F_net = T - mg = m * a_c

      • Where a_c = centripetal acceleration = v^2/r

    • Calculating Velocity:

      • v = distance/time = (2π * radius * revolutions)/time

      • v = (2π * 0.6 m * 10)/6 s

      • v = 6.2832 m/s

    • Centripetal Acceleration:

      • r = length of string = 0.6 m

      • a_c = v^2/r = (6.2832 m/s)^2 / 0.6 m = 65.38 m/s²

    • Tension Calculation:

      • T = mg + ma_c

      • T = (0.025 kg)(9.81 m/s²) + (0.025 kg)(65.38 m/s²)

      • T = 0.24525 N + 1.6345 N

      • T = 1.87975 N

  • b. Tension at the Top of the Circle

    • Free Body Diagram (FBD):

      • Forces acting on the ball:

        • Tension (T) acting downwards

        • Gravitational force (mg) also acting downwards

    • Net Force Equation:

      • F_net = mg + T = m * a_c

    • Solve for Tension:

      • T = m * a_c - mg

      • T = (0.025 kg)(65.38 m/s²) - (0.025 kg)(9.81 m/s²)

      • T = 1.6345 N - 0.24525 N

      • T = 1.38925 N

Page 2: Additional Problems

Problem 2: Roller Coaster Passenger

  • Given Data:

    • Mass of passenger, m = 60 kg

    • Radius of loop, r = 20 m

    • Speed, v = 21 m/s

  • a. Normal Force at Bottom and Top of Loop

    • Bottom of the Loop:

      • FBD:

        • Forces acting: Normal force (F_N) acting upwards and weight acting downwards.

      • Net Force Equation:

        • F_N - mg = m * a_c

        • a_c = v²/r = (21 m/s)² / 20 m

        • a_c = 22.05 m/s²

        • F_N = m * g + m * a_c

        • F_N = 60 kg * 9.81 m/s² + 60 kg * 22.05 m/s²

        • F_N = 588.6 N + 1323 N = 1911.6 N

    • Top of the Loop:

      • FBD:

        • Forces acting: Normal force (F_N) acting downwards and weight acting downwards.

      • Net Force Equation:

        • mg + F_N = m * a_c

        • F_N = m * a_c - mg

        • F_N = (60 kg * 22.05 m/s²) - (60 kg * 9.81 m/s²)

        • F_N = 1323 N - 588.6 N = 734.4 N

  • b. G's Experienced at Bottom and Top of Loop

    • Bottom of Loop:

      • G's = F_N / (mg)

      • G's = 1911.6 N / 588.6 N = 3.25 g's

    • Top of Loop:

      • G's = 734.4 N / 588.6 N = 1.25 g's

  • c. Minimum Safe Velocity

    • At the top of the loop, to ensure at least some tension:

      • Need to ensure that gravitational force can provide centripetal force.

      • mg = m * v²/r

      • v² = g * r

      • v = sqrt(g * r) = sqrt(9.81 m/s² * 20 m) = 14.14 m/s

Problem 3: Driver in a Car on a Hill

  • Given Data:

    • Mass of driver, m = 60 kg

    • Radius of hill, r = 25 m

  • Feeling Weightless:

    • You feel weightless when the normal force is 0.

    • Conditions at the Top of the Hill:

      • FBD:

        • Weight acting downwards and normal force acting downwards.

      • Net Force Equation:

        • mg = m * a_c

        • a_c = v²/r

        • Setting normal force to zero:

        • mg = m * v²/r

        • g = v²/r

        • v² = g * r

        • v = sqrt(g * r) = sqrt(9.81 m/s² * 25 m) = 15.65 m/s

    • Therefore, to feel weightless, the driver must be traveling at 15.65 m/s.

80/20 Rule in Physics

The 80/20 rule, or Pareto Principle, can be applied to understanding key concepts in physics such as uniform circular motion, work, power, energy, momentum, and impulse. This principle highlights that a minority of causes (20%) often lead to a majority of effects (80%).

Key Concepts

  1. Uniform Circular Motion

    • Involves an object moving in a circle at a constant speed.

    • Key Formula: Centripetal acceleration ( a_c = \frac{v^2}{r} )

    • Most problems can be simplified by focusing on the tension and acceleration rather than deriving complex paths.

  2. Work and Energy

    • Work is the energy transferred by forces acting over a distance.

    • Key Formula: Work ( W = F \cdot d \cdot cos(\theta) )

    • A few specific forces (like gravitational or frictional forces) often account for most energy changes in a system.

  3. Power

    • Power is the rate at which work is done.

    • Key Formula: Power ( P = \frac{W}{t} )

    • Understanding maximum power output or peak scenarios can lead to insights for efficiency in systems.

  4. Momentum and Impulse

    • Momentum is the product of mass and velocity.

    • Key Formula: ( p = m \cdot v )

    • Impulse is the change in momentum.

    • Key Formula: Impulse ( J = F \cdot t = \Delta p )

    • Focus on key impacts (like collision situations) to understand most momentum scenarios effectively.

Conclusion

By identifying and concentrating on the significant elements that lead to major outcomes in uniform circular motion, work, energy, momentum, and impulse, students can simplify complex concepts and focus on the most impactful principles.