Problem Set - Circular Motion Lesson 4

Centripetal force and centripetal acceleration are concepts that often cause confusion, but mastering them is crucial to understanding circular motion. This problem set focuses on applying these principles, alongside related concepts like tangential velocity and period, to a variety of scenarios.

Understanding Circular Motion

Circular motion, at its core, describes the movement of an object along a circular path. This seemingly simple motion becomes complex when we delve into the forces and accelerations acting upon the object. Unlike linear motion where velocity and acceleration are often in the same direction, circular motion introduces the concept of centripetal acceleration, which is always directed towards the center of the circle, constantly changing the object's direction and thus maintaining its circular path.

The key to solving circular motion problems lies in understanding the relationship between these quantities:

• Tangential Velocity (v): The instantaneous linear speed of the object along the circular path. It's a scalar quantity, simply representing the speed at a given point.
• Angular Velocity (ω): The rate at which the object's angular position changes, measured in radians per second (rad/s). It describes how quickly the object is rotating.
• Radius (r): The distance from the center of the circular path to the object.
• Centripetal Acceleration (ac): The acceleration directed towards the center of the circle, responsible for changing the direction of the object's velocity.
• Centripetal Force (Fc): The net force directed towards the center of the circle, causing the centripetal acceleration.

Let's explore a variety of problems that will help solidify your understanding of these concepts.

Problem Set: Circular Motion - Lesson 4

Here are a set of problems designed to test your knowledge of circular motion principles. Each problem builds upon the previous one, gradually increasing in complexity. Remember to apply the concepts of centripetal force, centripetal acceleration, tangential velocity, and period when solving these problems.

Problem 1: The Rotating Toy Plane

A toy plane is attached to a string and whirled in a horizontal circle with a radius of 1.5 meters. If the plane completes 20 revolutions in 1 minute, determine:

a) The period of the motion.

b) The tangential velocity of the plane.

c) The centripetal acceleration of the plane.

d) If the mass of the plane is 0.2 kg, what is the centripetal force acting on it?

Problem 2: Car on a Curved Road

A car with a mass of 1200 kg is traveling on a curved road with a radius of curvature of 50 meters. The coefficient of static friction between the tires and the road is 0.8.

a) What is the maximum speed the car can travel without skidding?

b) What force provides the centripetal force in this scenario?

Problem 3: The Gravitron Ride

A Gravitron ride at an amusement park has a radius of 4 meters and spins at a rate of 24 revolutions per minute.

a) What is the tangential speed of a rider?

b) What is the centripetal acceleration experienced by a rider?

c) What is the minimum coefficient of static friction required between the rider and the wall to prevent them from slipping down?

Problem 4: The Conical Pendulum

A conical pendulum consists of a mass of 0.5 kg attached to a string of length 1 meter. The mass rotates in a horizontal circle with a constant speed, and the string makes an angle of 30 degrees with the vertical.

a) Draw a free body diagram showing all the forces acting on the mass.

b) Determine the tension in the string.

c) Determine the speed of the mass.

d) Determine the period of the motion.

Problem 5: Satellite in Orbit

A satellite orbits the Earth at a height of 400 km above the Earth's surface. The radius of the Earth is 6371 km, and the mass of the Earth is 5.972 × 10^24 kg.

a) What is the orbital radius of the satellite?

b) What is the orbital speed of the satellite?

c) What is the period of the satellite's orbit?

d) What provides the centripetal force that keeps the satellite in orbit?

Problem 6: Ball on a String (Vertical Circle)

A ball with a mass of 0.3 kg is attached to a string of length 0.8 meters and swung in a vertical circle.

a) What is the minimum speed the ball must have at the top of the circle to maintain circular motion?

b) If the speed of the ball at the bottom of the circle is 6 m/s, what is the tension in the string at the bottom of the circle?

c) What is the tension in the string at the top of the circle if the speed at the top is twice the minimum speed calculated in part (a)?

Problem 7: Banked Curve

A curved road with a radius of 80 meters is banked at an angle of 15 degrees.

a) For what speed is the curve ideally banked (i.e., no friction required)?

b) If a car travels at a speed faster than the ideal speed, what force prevents it from sliding up the banked curve?

c) Explain the role of the banking angle in reducing the reliance on friction for safe turning.

Problem 8: Rotating Space Station

A cylindrical space station with a radius of 50 meters is designed to simulate Earth's gravity by rotating about its axis.

a) What angular velocity is required to simulate Earth's gravity (9.8 m/s^2) at the outer edge of the space station?

b) What is the tangential velocity at the outer edge of the space station?

c) Explain the sensation of "artificial gravity" experienced by someone inside the rotating space station.

Problem 9: Designing a Loop-the-Loop

You are designing a loop-the-loop for a roller coaster. The radius of the loop is 12 meters.

a) What is the minimum height from which the roller coaster must start to successfully complete the loop, assuming negligible friction?

b) Explain the energy transformations that occur as the roller coaster travels from the starting point to the top of the loop.

c) How would friction affect the minimum starting height required?

Problem 10: The Centrifuge

A centrifuge is used to separate blood samples. The sample tubes are placed in holders that are located 15 cm from the axis of rotation. The centrifuge spins at a rate of 3000 revolutions per minute.

a) What is the centripetal acceleration experienced by the blood cells?

b) What is the purpose of using a centrifuge to separate blood components?

c) How does the density of the blood components affect their separation in the centrifuge?

Solutions and Explanations

Now, let's walk through the solutions to each of the problems, with detailed explanations to enhance your understanding.

Solution 1: The Rotating Toy Plane

a) Period (T): The period is the time it takes for one complete revolution. Since the plane completes 20 revolutions in 60 seconds, the period is:

T = Time / Number of Revolutions = 60 s / 20 = 3 s

b) Tangential Velocity (v): The tangential velocity is the distance traveled in one revolution (circumference) divided by the period.

v = (2πr) / T = (2π * 1.5 m) / 3 s ≈ 3.14 m/s

c) Centripetal Acceleration (ac): The centripetal acceleration can be calculated using the formula:

ac = v^2 / r = (3.14 m/s)^2 / 1.5 m ≈ 6.58 m/s^2

d) Centripetal Force (Fc): The centripetal force is calculated using Newton's second law:

Fc = m * ac = 0.2 kg * 6.58 m/s^2 ≈ 1.32 N

Solution 2: Car on a Curved Road

a) Maximum Speed (v): The maximum speed is determined by the condition where the centripetal force required for the car to turn is equal to the maximum static friction force. The static friction force provides the centripetal force.

Fc = Ff (friction) mv^2 / r = μmg v^2 = μgr v = √(μgr) = √(0.8 * 9.8 m/s^2 * 50 m) ≈ 19.8 m/s

b) Centripetal Force Provider: The static friction between the tires and the road provides the necessary centripetal force. Crucially, it's static friction because the tires are not slipping relative to the road surface.

Solution 3: The Gravitron Ride

a) Tangential Speed (v): First, convert the revolutions per minute to revolutions per second: 24 rev/min = 24/60 rev/s = 0.4 rev/s. Then, calculate the tangential speed:

v = 2πr * frequency = 2π * 4 m * 0.4 rev/s ≈ 10.05 m/s

b) Centripetal Acceleration (ac):

ac = v^2 / r = (10.05 m/s)^2 / 4 m ≈ 25.25 m/s^2

c) Minimum Coefficient of Static Friction (μ): The centripetal force is provided by the normal force from the wall (N). To prevent slipping, the force of gravity (mg) must be less than or equal to the force of static friction (μN). Since N = mac:

mg ≤ μN = μmac g ≤ μac μ ≥ g / ac = 9.8 m/s^2 / 25.25 m/s^2 ≈ 0.39

Solution 4: The Conical Pendulum

a) Free Body Diagram: The forces acting on the mass are: * Tension (T) in the string, acting along the string. * Weight (mg), acting vertically downwards.

b) Tension in the String (T): The tension has vertical (Ty) and horizontal (Tx) components.

*   Ty = Tcos(30°) = mg
*   T = mg / cos(30°) = (0.5 kg * 9.8 m/s^2) / cos(30°) ≈ 5.66 N

c) Speed of the Mass (v): The horizontal component of tension provides the centripetal force.

*   Tx = Tsin(30°) = mv^2 / r
*   First, find the radius: r = Lsin(30°) = 1 m * sin(30°) = 0.5 m
*   v^2 = (Tsin(30°) * r) / m = (5.66 N * sin(30°) * 0.5 m) / 0.5 kg
*   v = √(2.83 m^2/s^2) ≈ 1.68 m/s

d) Period of the Motion (T):

*   T = 2πr / v = 2π * 0.5 m / 1.68 m/s ≈ 1.87 s

Solution 5: Satellite in Orbit

a) Orbital Radius (r):

r = Radius of Earth + Altitude = 6371 km + 400 km = 6771 km = 6.771 x 10^6 m

b) Orbital Speed (v): The gravitational force provides the centripetal force.

• GMm/r^2 = mv^2/r (where G is the gravitational constant, 6.674 x 10^-11 Nm^2/kg^2, and M is the mass of the Earth)
• v = √(GM/r) = √((6.674 x 10^-11 Nm^2/kg^2 * 5.972 x 10^24 kg) / 6.771 x 10^6 m) ≈ 7672 m/s

c) Period of the Orbit (T):

• T = 2πr / v = 2π * 6.771 x 10^6 m / 7672 m/s ≈ 5540 s (approximately 92 minutes)

d) Centripetal Force Provider: The gravitational force between the Earth and the satellite provides the centripetal force.

Solution 6: Ball on a String (Vertical Circle)

a) Minimum Speed at the Top (v_top): At the top of the circle, the tension and weight both point downwards, contributing to the centripetal force. The minimum speed occurs when the tension is zero.

*  mg = mv_top^2 / r
*  v_top = √(gr) = √(9.8 m/s^2 * 0.8 m) ≈ 2.8 m/s

b) Tension at the Bottom (T_bottom): At the bottom of the circle, the tension points upwards, and the weight points downwards. The net force (T_bottom - mg) provides the centripetal force.

*  T_bottom - mg = mv_bottom^2 / r
*  T_bottom = mg + mv_bottom^2 / r = (0.3 kg * 9.8 m/s^2) + (0.3 kg * (6 m/s)^2 / 0.8 m) ≈ 9.99 N

c) Tension at the Top (T_top): Given v_top = 2 * 2.8 m/s = 5.6 m/s

*   T_top + mg = mv_top^2 / r
*   T_top = mv_top^2 / r - mg = (0.3 kg * (5.6 m/s)^2 / 0.8 m) - (0.3 kg * 9.8 m/s^2) = 8.13 N

Solution 7: Banked Curve

a) Ideal Speed (v): For an ideally banked curve, the horizontal component of the normal force provides the centripetal force, and the vertical component balances the weight.

• tan(θ) = v^2 / (gr)
• v = √(gr * tan(θ)) = √(9.8 m/s^2 * 80 m * tan(15°)) ≈ 14.4 m/s

b) Force Preventing Sliding Up: If the car travels faster than the ideal speed, static friction prevents it from sliding up the banked curve. The friction force acts down the slope.

c) Role of Banking Angle: The banking angle allows a component of the normal force to contribute to the centripetal force required for turning. This reduces the reliance on friction, making the turn safer, especially in wet or icy conditions where friction is reduced.

Solution 8: Rotating Space Station

a) Angular Velocity (ω): Artificial gravity (ag) is equal to the centripetal acceleration.

*   ag = rω^2
*   ω = √(ag / r) = √(9.8 m/s^2 / 50 m) ≈ 0.44 rad/s

b) Tangential Velocity (v):

*   v = rω = 50 m * 0.44 rad/s ≈ 22 m/s

c) Sensation of Artificial Gravity: The sensation of "artificial gravity" arises from the normal force exerted by the floor of the rotating space station on the person's feet. This normal force provides the centripetal force necessary to keep the person moving in a circle. The person feels this normal force as their weight. It's important to note this artificial gravity simulates weight, but not the other effects of gravity like atmospheric pressure.

Solution 9: Designing a Loop-the-Loop

a) Minimum Starting Height (h): At the top of the loop, the minimum speed is v = √(gr). We can use conservation of energy:

*   Potential Energy (PE) at the start = Kinetic Energy (KE) at the top + PE at the top
*   mgh = (1/2)mv^2 + mg(2r) = (1/2)m(gr) + 2mgr
*   h = (1/2)r + 2r = (5/2)r = (5/2) * 12 m = 30 m

b) Energy Transformations: * At the starting point: All energy is potential energy (PE). * As the roller coaster descends: PE is converted into kinetic energy (KE). * At the bottom of the loop: Most of the energy is KE. * As the roller coaster ascends the loop: KE is converted back into PE. * At the top of the loop: A portion of the initial PE has transformed into KE, and the remaining energy is in the form of PE.

c) Effect of Friction: Friction would dissipate some of the mechanical energy (PE and KE) as heat. Therefore, the minimum starting height required would be higher than 30 meters to compensate for the energy lost due to friction.

Solution 10: The Centrifuge

a) Centripetal Acceleration (ac): First, convert revolutions per minute to radians per second. 3000 rev/min = 3000/60 rev/s = 50 rev/s. 1 revolution = 2π radians. Therefore, ω = 50 * 2π = 100π rad/s. The radius is 15 cm = 0.15 m.

• ac = rω^2 = 0.15 m * (100π rad/s)^2 ≈ 14800 m/s^2

b) Purpose of Centrifuge: A centrifuge separates components of a mixture based on their density. The high centripetal acceleration causes denser components to sediment (settle) at the bottom of the tube, while less dense components remain closer to the top.

c) Effect of Density: Denser components experience a greater force due to the centripetal acceleration (F = ma = ρVa, where ρ is density and V is volume), causing them to sediment faster and further than less dense components. This difference in sedimentation rate allows for separation.

Key Takeaways

This problem set illustrates the fundamental principles of circular motion. By working through these problems, you should have gained a deeper understanding of:

• The relationship between tangential velocity, angular velocity, radius, centripetal acceleration, and centripetal force.
• The importance of identifying the force that provides the centripetal force in different scenarios.
• How to apply Newton's second law to solve circular motion problems.
• The concept of artificial gravity and how it is created.
• Energy transformations in circular motion scenarios.

Remember to practice applying these concepts to various problems to solidify your understanding. Circular motion is a fundamental concept in physics with numerous real-world applications, so mastering it is essential for further studies in physics and engineering.