Problem Set - Circular Motion Lesson 4

The world around us is filled with motion, and understanding its various forms is fundamental to grasping the laws of physics. Circular motion, a specific type of movement where an object traces a circular path, plays a crucial role in many real-world phenomena, from the orbit of planets to the spinning of a washing machine. Let's dive into the intricacies of circular motion, dissecting the concepts through a comprehensive problem set tailored for Lesson 4, designed to enhance your understanding and problem-solving skills.

Understanding the Fundamentals of Circular Motion

Before tackling the problems, it’s vital to solidify our understanding of the underlying principles. Circular motion occurs when an object moves along a circular path. Key concepts include:

• Angular Displacement (θ): Measured in radians, it represents the angle through which an object has moved on its circular path.
• Angular Velocity (ω): The rate of change of angular displacement with respect to time, indicating how fast an object is rotating. Its unit is radians per second (rad/s).
• Angular Acceleration (α): The rate of change of angular velocity, signifying how quickly the rotational speed is changing. Measured in radians per second squared (rad/s²).
• Centripetal Acceleration (ac): Acceleration directed towards the center of the circular path, essential for maintaining circular motion. It's calculated as ac = v²/r or ac = ω²r, where v is the linear velocity and r is the radius of the circle.
• Centripetal Force (Fc): The force responsible for causing centripetal acceleration. It’s always directed towards the center of the circle and calculated as Fc = mv²/r or Fc = mω²r, where m is the mass of the object.
• Period (T): The time taken for one complete revolution around the circular path.
• Frequency (f): The number of revolutions per unit time, which is the inverse of the period (f = 1/T).

Understanding the relationship between these concepts is crucial for solving circular motion problems effectively.

Problem Set: Lesson 4 - Circular Motion

Now, let's apply these concepts through a series of carefully designed problems. Each problem is followed by a detailed solution, providing step-by-step guidance.

Problem 1: A car is traveling around a circular track with a radius of 200 meters at a constant speed of 25 m/s. Calculate the car's centripetal acceleration and the centripetal force acting on the car if its mass is 1500 kg.

Solution:

1. Identify the knowns:

   • Radius (r) = 200 m
   • Velocity (v) = 25 m/s
   • Mass (m) = 1500 kg

2. Calculate centripetal acceleration (ac):

   • ac = v²/r
   • ac = (25 m/s)² / 200 m
   • ac = 625 m²/s² / 200 m
   • ac = 3.125 m/s²

3. Calculate centripetal force (Fc):

   • Fc = mv²/r or Fc = mac
   • Fc = 1500 kg * 3.125 m/s²
   • Fc = 4687.5 N

Answer: The centripetal acceleration is 3.125 m/s², and the centripetal force acting on the car is 4687.5 N.

Problem 2: A Ferris wheel with a radius of 15 meters completes one revolution in 20 seconds. Determine the angular velocity of the Ferris wheel and the linear speed of a passenger sitting on the rim.

Solution:

1. Identify the knowns:

   • Radius (r) = 15 m
   • Period (T) = 20 s

2. Calculate angular velocity (ω):

   • Angular velocity ω = 2π / T (since one revolution is 2π radians)
   • ω = 2π / 20 s
   • ω = π / 10 rad/s
   • ω ≈ 0.314 rad/s

3. Calculate linear speed (v):

   • Linear speed v = rω
   • v = 15 m * 0.314 rad/s
   • v ≈ 4.71 m/s

Answer: The angular velocity of the Ferris wheel is approximately 0.314 rad/s, and the linear speed of a passenger is approximately 4.71 m/s.

Problem 3: A small object with a mass of 0.5 kg is attached to a string and whirled in a horizontal circle with a radius of 0.8 meters. If the object completes 2 revolutions per second, calculate the tension in the string.

Solution:

1. Identify the knowns:

   • Mass (m) = 0.5 kg
   • Radius (r) = 0.8 m
   • Frequency (f) = 2 Hz (revolutions per second)

2. Calculate angular velocity (ω):

   • ω = 2πf
   • ω = 2π * 2 Hz
   • ω = 4π rad/s
   • ω ≈ 12.57 rad/s

3. Calculate centripetal force (Fc), which is equal to the tension in the string:

   • Fc = mω²r
   • Fc = 0.5 kg * (12.57 rad/s)² * 0.8 m
   • Fc = 0.5 kg * 158.0049 rad²/s² * 0.8 m
   • Fc ≈ 63.2 N

Answer: The tension in the string is approximately 63.2 N.

Problem 4: A satellite orbits the Earth at a distance of 20,000 km from the Earth's center. If the satellite's orbital period is 12 hours, determine its orbital speed and centripetal acceleration. (Assume Earth's radius is negligible compared to the orbital radius.)

Solution:

1. Identify the knowns:

   • Radius (r) = 20,000 km = 20,000,000 m
   • Period (T) = 12 hours = 12 * 3600 s = 43,200 s

2. Calculate orbital speed (v):

   • v = 2πr / T
   • v = (2 * π * 20,000,000 m) / 43,200 s
   • v ≈ 2908.9 m/s

3. Calculate centripetal acceleration (ac):

   • ac = v²/r
   • ac = (2908.9 m/s)² / 20,000,000 m
   • ac ≈ 0.423 m/s²

Answer: The satellite's orbital speed is approximately 2908.9 m/s, and its centripetal acceleration is approximately 0.423 m/s².

Problem 5: A roller coaster car with a mass of 800 kg goes over a circular hill with a radius of 30 meters. What is the maximum speed the car can have at the top of the hill without losing contact with the track?

Solution:

1. Understanding the concept: At the point where the car is about to lose contact, the normal force (the force exerted by the track on the car) is zero. Therefore, the centripetal force is provided solely by the gravitational force.

2. Identify the knowns:

   • Mass (m) = 800 kg
   • Radius (r) = 30 m
   • Acceleration due to gravity (g) = 9.8 m/s²

3. Set the centripetal force equal to the gravitational force:

   • Fc = Fg
   • mv²/r = mg

4. Solve for velocity (v):

   • v² = gr
   • v = √(gr)
   • v = √(9.8 m/s² * 30 m)
   • v = √294 m²/s²
   • v ≈ 17.15 m/s

Answer: The maximum speed the car can have at the top of the hill without losing contact is approximately 17.15 m/s.

Problem 6: A bicycle wheel with a diameter of 70 cm rotates at a constant angular velocity of 5 rad/s. Determine the linear speed of a point on the tire and the number of revolutions the wheel makes in 1 minute.

Solution:

1. Identify the knowns:

   • Diameter = 70 cm, so Radius (r) = 35 cm = 0.35 m
   • Angular velocity (ω) = 5 rad/s

2. Calculate linear speed (v):

   • v = rω
   • v = 0.35 m * 5 rad/s
   • v = 1.75 m/s

3. Calculate the number of revolutions per second (frequency, f):

   • ω = 2πf
   • f = ω / 2π
   • f = 5 rad/s / (2π rad/revolution)
   • f ≈ 0.796 revolutions/s

4. Calculate the number of revolutions per minute:

   • Revolutions per minute = f * 60 s/minute
   • Revolutions per minute = 0.796 revolutions/s * 60 s/minute
   • Revolutions per minute ≈ 47.76 revolutions/minute

Answer: The linear speed of a point on the tire is 1.75 m/s, and the wheel makes approximately 47.76 revolutions in 1 minute.

Problem 7: A centrifuge with a radius of 20 cm spins at a rate of 10,000 rpm (revolutions per minute). Calculate the centripetal acceleration experienced by a particle at the edge of the centrifuge.

Solution:

1. Identify the knowns:

   • Radius (r) = 20 cm = 0.2 m
   • Rotation rate = 10,000 rpm

2. Convert rpm to rad/s:

   • 1 revolution = 2π radians
   • 1 minute = 60 seconds
   • ω = (10,000 revolutions/minute) * (2π radians/revolution) / (60 seconds/minute)
   • ω ≈ 1047.2 rad/s

3. Calculate centripetal acceleration (ac):

   • ac = rω²
   • ac = 0.2 m * (1047.2 rad/s)²
   • ac ≈ 219,325 m/s²

Answer: The centripetal acceleration experienced by a particle at the edge of the centrifuge is approximately 219,325 m/s². This is a very high acceleration, which is why centrifuges are used to separate substances based on density.

Problem 8: A car is traveling at a constant speed of 30 m/s around a banked curve with a radius of 150 meters. What is the optimal banking angle for this curve, assuming no friction?

Solution:

1. Understanding the concept: The optimal banking angle is the angle at which the horizontal component of the normal force provides the necessary centripetal force, eliminating the need for friction to prevent the car from sliding.

2. Relevant equation: tan(θ) = v²/ (gr), where θ is the banking angle, v is the speed, g is the acceleration due to gravity, and r is the radius of the curve.

3. Identify the knowns:

   • Velocity (v) = 30 m/s
   • Radius (r) = 150 m
   • Acceleration due to gravity (g) = 9.8 m/s²

4. Calculate the banking angle (θ):

   • tan(θ) = v² / (gr)
   • tan(θ) = (30 m/s)² / (9.8 m/s² * 150 m)
   • tan(θ) = 900 m²/s² / 1470 m²/s²
   • tan(θ) ≈ 0.6122
   • θ = arctan(0.6122)
   • θ ≈ 31.54 degrees

Answer: The optimal banking angle for the curve is approximately 31.54 degrees.

Problem 9: A point on the edge of a rotating disk with a radius of 0.4 meters has a tangential speed of 2 m/s. If the disk starts from rest and accelerates uniformly to this speed in 5 seconds, find the angular acceleration of the disk.

Solution:

1. Identify the knowns:

   • Radius (r) = 0.4 m
   • Final tangential speed (vf) = 2 m/s
   • Initial tangential speed (vi) = 0 m/s (starts from rest)
   • Time (t) = 5 s

2. Calculate the final angular velocity (ωf):

   • vf = rωf
   • ωf = vf / r
   • ωf = 2 m/s / 0.4 m
   • ωf = 5 rad/s

3. Calculate the initial angular velocity (ωi):

   • Since the disk starts from rest, ωi = 0 rad/s

4. Calculate the angular acceleration (α):

   • α = (ωf - ωi) / t
   • α = (5 rad/s - 0 rad/s) / 5 s
   • α = 1 rad/s²

Answer: The angular acceleration of the disk is 1 rad/s².

Problem 10: A playground merry-go-round has a radius of 2.5 meters and a moment of inertia of 300 kg·m². A child applies a tangential force of 50 N to the edge of the merry-go-round. Assuming no friction, what is the angular acceleration of the merry-go-round?

Solution:

1. Understanding the concept: The torque applied to the merry-go-round causes angular acceleration. Torque (τ) is related to the force (F) and radius (r) by τ = rF. Angular acceleration (α) is related to torque (τ) and moment of inertia (I) by τ = Iα.

2. Identify the knowns:

   • Radius (r) = 2.5 m
   • Moment of inertia (I) = 300 kg·m²
   • Tangential force (F) = 50 N

3. Calculate the torque (τ):

   • τ = rF
   • τ = 2.5 m * 50 N
   • τ = 125 N·m

4. Calculate the angular acceleration (α):

   • τ = Iα
   • α = τ / I
   • α = 125 N·m / 300 kg·m²
   • α ≈ 0.417 rad/s²

Answer: The angular acceleration of the merry-go-round is approximately 0.417 rad/s².

Practical Applications and Real-World Examples

Circular motion isn't just a theoretical concept; it's a fundamental aspect of our world:

• Planetary Orbits: The Earth's orbit around the Sun, as well as the orbits of other planets and satellites, are prime examples of circular (or elliptical, which can be approximated as circular) motion.
• Rotating Machinery: Many machines, such as motors, turbines, and gears, rely on circular motion to function.
• Vehicular Motion: Cars turning corners, airplanes banking, and even the wheels of a bicycle involve circular motion principles.
• Amusement Park Rides: Roller coasters, Ferris wheels, and carousels all utilize circular motion to create thrilling experiences.
• Centrifuges: Used in laboratories and industries to separate substances based on density, centrifuges rely on centripetal force generated by high-speed circular motion.

Common Mistakes to Avoid

When solving circular motion problems, be aware of these common pitfalls:

• Confusing Angular and Linear Quantities: Remember to use the appropriate formulas that relate angular and linear velocity, acceleration, and displacement.
• Incorrectly Identifying the Source of Centripetal Force: Always analyze the situation to determine which force (or combination of forces) is providing the necessary centripetal force.
• Forgetting to Convert Units: Ensure all quantities are in consistent units (e.g., meters, seconds, radians) before performing calculations.
• Assuming Constant Speed in Non-Uniform Circular Motion: Be mindful that circular motion can be non-uniform, meaning the speed is not constant, and therefore, angular acceleration is present.

Conclusion

Mastering circular motion requires a solid understanding of the fundamental concepts and plenty of practice. This problem set, coupled with a careful review of the underlying principles, should provide a strong foundation for tackling a wide range of circular motion problems. By understanding the applications and avoiding common mistakes, you can confidently navigate the world of rotational dynamics. Remember to always visualize the problem, identify the knowns and unknowns, and apply the appropriate formulas to arrive at the correct solution. Happy problem-solving!