Textbook Question
The normal force equals the magnitude of the gravitational force as a roller-coaster car crosses the top of a 40-m-diameter loop-the-loop. What is the car's speed at the top?
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Ch 08: Dynamics II: Motion in a Plane
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 8, Problem 24a
A 500 g ball moves in a vertical circle on a 102-cm-long string. If the speed at the top is 4.0 m/s, then the speed at the bottom will be 7.5 m/s. (You'll learn how to show this in Chapter 10.) What is the gravitational force acting on the ball?
Verified step by step guidance1
Convert the mass of the ball from grams to kilograms, as the SI unit for mass is kilograms. Use the conversion: \( 1 \text{ g} = 0.001 \text{ kg} \). Thus, \( m = 500 \text{ g} = 0.5 \text{ kg} \).
Recall the formula for gravitational force: \( F_g = m \cdot g \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \) on Earth's surface).
Substitute the values into the formula: \( F_g = 0.5 \cdot 9.8 \).
Simplify the expression to calculate the gravitational force acting on the ball. This will give the force in newtons (N).
The gravitational force calculated represents the constant downward force acting on the ball due to Earth's gravity, regardless of its position in the vertical circle.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Gravitational Force
Gravitational force is the attractive force between two masses, described by Newton's law of universal gravitation. For an object near the Earth's surface, this force can be calculated using the formula F = mg, where m is the mass of the object and g is the acceleration due to gravity (approximately 9.81 m/s²). In this scenario, the gravitational force acting on the ball can be determined by multiplying its mass (0.5 kg) by g.
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Gravitational Forces in 2D
Centripetal Force
Centripetal force is the net force required to keep an object moving in a circular path, directed towards the center of the circle. It is essential for maintaining circular motion and can be calculated using the formula F_c = mv²/r, where m is the mass, v is the speed, and r is the radius of the circular path. At different points in the vertical circle, the gravitational force and tension in the string contribute to the centripetal force.
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Energy Conservation in Circular Motion
In circular motion, mechanical energy is conserved if only conservative forces (like gravity) are acting. The total mechanical energy at the top and bottom of the circle can be analyzed using the principles of kinetic and potential energy. At the top, the ball has both kinetic energy (due to its speed) and potential energy (due to its height), while at the bottom, the potential energy is lower, and the kinetic energy is higher, allowing for the calculation of forces and speeds at different points.
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Related Practice
Textbook Question
A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have at the top without flying off the road?
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Textbook Question
A 500 g ball moves in a vertical circle on a 102-cm-long string. If the speed at the top is 4.0 m/s, then the speed at the bottom will be 7.5 m/s. (You'll learn how to show this in Chapter 10.) What is the tension in the string when the ball is at the top?
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Textbook Question
A new car is tested on a 200-m-diameter track. If the car speeds up at a steady 1.5 m/s2, how long after starting is the magnitude of its centripetal acceleration equal to the tangential acceleration?
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Textbook Question
A heavy ball with a weight of 100 N (m = 10.2 kg) is hung from the ceiling of a lecture hall on a 4.5-m-long rope. The ball is pulled to one side and released to swing as a pendulum, reaching a speed of 5.5 m/s as it passes through the lowest point. What is the tension in the rope at that point?
Textbook Question
The weight of passengers on a roller coaster increases by 50% as the car goes through a dip with a 30 m radius of curvature. What is the car's speed at the bottom of the dip?
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