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A uniform rod is 2.00 \(\mathrm{m}\) long and has 1.80 \(\mathrm{kg} . \mathrm{A}\) \(2.40-\) -kg clamp is attached to the rod. How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1.20 \(\mathrm{m}\) from the left-hand end of the rod?

Short Answer

Expert verified
The center of gravity of the clamp should be 1.35 m from the left end.

Step by step solution

01

Identify Given Values

We know the length of the rod is 2.00 m, the mass of the rod is 1.80 kg, and the mass of the clamp is 2.40 kg. The combined center of gravity should be 1.20 m from the left end of the rod.
02

Determine Center of Gravity for the Rod

For a uniform rod, the center of gravity is at its midpoint. Thus the center of gravity for the rod is \[ \frac{2.00}{2} = 1.00 \text{ m from the left end.} \]
03

Set Up the Equation for Center of Gravity

The equation for the center of gravity \( x_{cg} \) of a system is given by:\[ x_{cg} = \frac{m_{rod} \times x_{rod} + m_{clamp} \times x_{clamp}}{m_{rod} + m_{clamp}} \]where \( x_{rod} = 1.00 \text{ m} \) and \( x_{clamp} \) is the unknown distance from the left end of the rod where the clamp's center of gravity is.
04

Substitute Known Values into the Equation

Substitute the known quantities into the equation:\[ 1.20 = \frac{1.80 \times 1.00 + 2.40 \times x_{clamp}}{1.80 + 2.40} \] This simplifies to: \[ 1.20 = \frac{1.80 + 2.40 \times x_{clamp}}{4.20} \]
05

Solve for x_clamp

Clear the fraction by multiplying both sides by 4.20:\[ 1.20 \times 4.20 = 1.80 + 2.40 \times x_{clamp} \]\[ 5.04 = 1.80 + 2.40 \times x_{clamp} \]Subtract 1.80 from both sides:\[ 3.24 = 2.40 \times x_{clamp} \]Now, divide by 2.40:\[ x_{clamp} = \frac{3.24}{2.40} = 1.35 \text{ m} \]
06

Conclusion

The distance from the left end of the rod to the center of gravity of the clamp should be 1.35 m.

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Key Concepts

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

Uniform Rod
A uniform rod is a simple and important concept in physics. It is a straight, elongated object where mass is distributed evenly along its entire length. In this case, the uniformity ensures that the center of gravity is located precisely at the midpoint of the rod. This means that if the rod is 2.00 meters long, its center of gravity is exactly at 1.00 meter from either end.
Understanding the center of gravity is crucial because it determines how the rod will balance and behave under various conditions. In calculations, assuming uniformity simplifies the process since the center of gravity is always the midpoint, allowing us to quickly proceed with further problem-solving steps.
For students, visualizing the rod balancing on a point below its midpoint can help solidify the understanding of the concept of uniformity and center of gravity.
Composite Object
When dealing with a composite object, we're handling a scenario where multiple objects are combined. In this exercise, a uniform rod and a clamp represent such a system. The mass and dimensions of each part impact the overall center of gravity of the combined system.
The center of gravity is calculated using all the individual masses and their respective distances from a reference point. The goal is to determine a single point where the whole system can perfectly balance.
For example, here we combine the mass of the rod and clamp. They are treated as point masses at their respective centers of gravity, and the entire system acts as a single mass located at the calculated center of gravity. This highlights how each piece contributes to the balance of the whole system.
Distance Calculation
Distance calculation is an important aspect when determining the center of gravity in a composite system. In this exercise, it involves finding the precise location for the clamp's center of gravity to achieve the desired balance: to have the center of gravity of the entire system at 1.20 meters from the left end of the rod.
The fundamental equation used to compute the center of gravity is \[ x_{cg} = \frac{m_{rod} \times x_{rod} + m_{clamp} \times x_{clamp}}{m_{rod} + m_{clamp}} \]
By inserting the known values, students can solve for the unknown distance, in this case, the distance of the clamp's center of gravity. This process teaches the importance of balancing forces and distances in physical systems, making it a core learning objective in physics courses.
Physics Problem Solving
Physics problem solving often involves applying theoretical principles to practical scenarios. Here, students use information about mass, distance, and balance to determine the center of gravity within a composite object.
The solution requires a clear and logical series of steps: identifying known values, using appropriate equations, and applying algebraic manipulation to solve for unknowns.
Problem-solving in physics not only helps in understanding specific topics like mechanics but also hones analytical and critical thinking skills. It encourages students to break down complex scenarios into simpler, manageable parts and find solutions using established scientific methods. This exercise, therefore, is an excellent opportunity to develop these vital skills through practical application.

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Most popular questions from this chapter

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