Ch 05: Applying Newton's Laws

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 05: Applying Newton's Laws

Problem 45b

Chapter 5, Problem 45b

A small remote-controlled car with mass $1.60$ kg moves at a constant speed of $v = 12.0$ m/s in a track formed by a vertical circle inside a hollow metal cylinder that has a radius of $5.00$ m (Fig. E $5.45$ ). What is the magnitude of the normal force exerted on the car by the walls of the cylinder at point $B$ (top of the track)?

Verified step by step guidance

Step 1: Identify the forces acting on the car at point B (top of the track). At this point, the forces include the gravitational force (mg, acting downward) and the normal force exerted by the walls of the cylinder (N, also acting downward). Both forces contribute to the centripetal force required to keep the car moving in a circular path.

Step 2: Write the equation for the net centripetal force at point B. The centripetal force is provided by the sum of the gravitational force and the normal force. Using Newton's second law for circular motion: F_c = m * υ² / r, where m is the mass of the car, υ is its speed, and r is the radius of the circular path.

Step 3: Substitute the expressions for the forces into the centripetal force equation. The net centripetal force is the sum of the gravitational force and the normal force: m * υ² / r = N + mg.

Step 4: Rearrange the equation to solve for the normal force (N). Isolate N on one side of the equation: N = m * υ² / r - mg.

Step 5: Substitute the given values into the equation to prepare for calculation. Use m = 1.60 kg, υ = 12.0 m/s, r = 5.00 m, and g = 9.80 m/s². Plug these values into the formula N = m * υ² / r - mg to compute the magnitude of the normal force.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Centripetal Force

Centripetal force is the net force acting on an object moving in a circular path, directed towards the center of the circle. For an object like the remote-controlled car, this force is necessary to keep it moving in a circular trajectory. It can be calculated using the formula F_c = m*v^2/r, where m is the mass, v is the velocity, and r is the radius of the circular path.

Normal Force

The normal force is the force exerted by a surface to support the weight of an object resting on it, acting perpendicular to the surface. In the context of the car at the top of the vertical circle, the normal force counteracts both the gravitational force acting on the car and provides the necessary centripetal force to maintain circular motion.

Gravitational Force

Gravitational force is the attractive force between two masses, calculated using Newton's law of universal gravitation. For the car, this force acts downward and is given by F_g = m*g, where m is the mass of the car and g is the acceleration due to gravity (approximately 9.81 m/s²). At the top of the circular path, this force influences the net force acting on the car.

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Textbook Question

A small remote-controlled car with mass $1.60$ kg moves at a constant speed of $v equals 12.0$ m/s in a track formed by a vertical circle inside a hollow metal cylinder that has a radius of $5.00$ m (Fig. E $5.45$ ). What is the magnitude of the normal force exerted on the car by the walls of the cylinder at point $A$ (bottom of the track)?

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