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U.S. nickels have a mass of \(5.00 \mathrm{g}\) and are \(1.95 \mathrm{mm}\) thick. If you stack 3 nickels on a table, how far above the table is their center of gravity?

Short Answer

Expert verified
The center of gravity for the stack of nickels is 2.925 mm above the table.

Step by step solution

01

Calculate the Total Height

The thickness of each nickel is presented as \(1.95 \, \mathrm{mm}\). Since there are 3 nickels, the total height would be the thickness of one nickel times 3. So total height becomes \(1.95 \, \mathrm{mm} \times 3 = 5.85 \, \mathrm{mm}\)
02

Find the Centre of Gravity

The centre of gravity is mid-way up the stack. Hence it is half of the total height of the stack of nickels from the base. Therefore the centre of gravity is \( \frac{5.85 \, \mathrm{mm}}{2} = 2.925 \, \mathrm{mm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Physics Problem Solving
Understanding physics problems often starts with identifying the key concepts and relationships involved in the question. In our example, we focus on the center of gravity calculation for stacked objects. Problem-solving in physics typically involves a step-by-step approach that breaks down complex problems into manageable parts.

Solving this problem first requires calculating the total height of the nickel stack. Since the stack is made up of identical items, we can simply multiply the thickness of a single object by the number of objects. This leads to our next step: identifying the center of gravity. With the understanding that the center of gravity for a uniform stack is in the middle, we simply divide the total height by two.

For more complex problems, drawing a diagram or visualizing the situation can be immensely helpful. Clear notation and systematic calculations can also reduce mistakes. Remember that understanding why each step is taken is just as important as performing the step correctly - it ensures that the problem-solving skills are transferable to different types of physics problems.
Mass and Gravity Relationship
The relationship between mass and gravity is fundamental in physics, particularly when discussing the center of gravity. The mass of an object affects how gravity acts upon it, and conversely, gravity influences how we calculate the mass's impact on different physical systems. For instance, in our scenario, the mass of each nickel is constant, but it's the gravitational pull that helps us understand how the stack behaves as a single entity.

Each nickel in the stack can be seen as a point mass, individually subject to Earth's gravity. However, when stacked, the gravitational force acts on the center of gravity of the stack as if all its mass were concentrated at that point. It’s crucial to comprehend that regardless of how the mass is distributed, gravity will act as if the stack's total mass is located at the center of gravity.

Hence, calculations that involve centers of gravity always have to account for how mass is distributed, as this determines the precise location of the center of gravity. Understanding this concept ensures students can approach similar problems involving irregular shapes or different mass distributions with confidence.
Centre of Gravity Concept
The centre of gravity is a pivotal concept in mechanics, denoting the point in a body or system where the weight is evenly distributed in all directions. In other words, it's the average location of the weight of an object. For symmetrical objects of uniform density, like the stack of nickels in our example, finding the center of gravity is straightforward—it's the geometrical center.

For irregular objects or systems with uneven mass distribution, however, the centre of gravity may not align with the geometrical center. Determining the center of gravity in such cases requires a more detailed analysis of the object's shape and mass distribution. This could involve calculus or experimental methods for complex shapes.

Understanding where the center of gravity lies is essential in various fields, from engineering to sports science, as it affects stability and balance. For students, grasping this concept could mean the difference between merely solving a particular problem and gaining a deeper comprehension of physical systems and their behaviors.

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Most popular questions from this chapter

The grand jeté is a classic ballet maneuver in which a dancer executes a horizontal leap while moving her arms and legs up and then down. At the center of the leap, the arms and legs are gracefully extended, as we see in Figure \(\mathrm{P} 7.81 \mathrm{a}\). The goal of the leap is to create the illusion of flight. As the dancer moves through the air, he or she is in free fall. In Chapter \(3,\) we saw that this leads to projectile motion. But what part of the dancer follows the usual parabolic path? It won't come as a surprise to learn that it's the center of gravity. But when you watch a dancer leap through the air, you don't watch her center of gravity, you watch her head. If the translational motion of her head is horizontal-not parabolicthis creates the illusion that she is flying through the air, held up by unseen forces. Figure \(\mathrm{P} 7.81 \mathrm{b}\) illustrates how the dancer creates this illusion. While in the air, she changes the position of her center of gravity relative to her body by moving her arms and legs up, then down. Her center of gravity moves in a parabolic path, but her head moves in a straight line. It's not flight, but it will appear that way, at least for a moment. In addition to changing her center of gravity, a dancer may change her moment of inertia. Consider her moment of inertia about a vertical axis through the center of her body. When she raises her arms and legs, this A. Increases her moment of inertia. B. Decreases her moment of inertia. C. Does not change her moment of inertia.

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