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Chapter 8: Problem 9

Three point masses are placed on the \(x\) -axis: \(200 \mathrm{~g}\) at \(x=0,500 \mathrm{~g}\) at \(x=30 \mathrm{~cm}\), and \(400 \mathrm{~g}\) at \(x=70 \mathrm{~cm} .\) Find their center of mass. We can make the calculation with respect to any point, but since all the data is measured from the \(x=0\) origin, that point will do nicely. $$ x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(0.20 \mathrm{~kg})+(0.30 \mathrm{~m})(0.50 \mathrm{~kg})+(0.70 \mathrm{~m})(0.40 \mathrm{~kg})}{(0.20+0.50+0.40) \mathrm{kg}}=0.39 \mathrm{~m} $$ The center of mass is located at a distance of \(0.39 \mathrm{~m}\), in the positive \(x\) -direction, from the origin. The \(y\) - and \(z\) -coordinates of the center of mass are zero.

Short Answer

Expert verified
The center of mass is located at \(0.39\) m on the x-axis from the origin.

Step by step solution

01

Identify Masses and Positions

The given point masses are 200 g at \(x = 0\) cm, 500 g at \(x = 30\) cm, and 400 g at \(x = 70\) cm. Convert these to kg and meters: \(m_1 = 0.20\) kg at \(x_1 = 0\) m, \(m_2 = 0.50\) kg at \(x_2 = 0.30\) m, and \(m_3 = 0.40\) kg at \(x_3 = 0.70\) m.
02

Calculate Weighted Sum of Positions

Compute the sum \(\sum x_i m_i = (0)\times(0.20) + (0.30)\times(0.50) + (0.70)\times(0.40)\). This calculation results in \(0 + 0.15 + 0.28 = 0.43\) kg路m.
03

Calculate Total Mass

The total mass is the sum \(\sum m_i = 0.20 + 0.50 + 0.40 = 1.10\) kg.
04

Compute Center of Mass on x-axis

Divide the weighted sum of positions by the total mass for the center of mass position on the x-axis: \(x_{\text{cm}} = \frac{0.43}{1.10} \approx 0.39\) m.

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Key Concepts

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

Point Masses
Point masses are a simplified concept used in physics to make calculations manageable and straightforward. Imagine each object as a point, with all its mass concentrated in a single place. This approach helps us ignore the complexities of objects' shapes and sizes. In the problem you're tackling, we have three point masses:
  • A 200 g mass located at position 0 on the x-axis.
  • A 500 g mass located at position 30 cm on the x-axis.
  • A 400 g mass located at position 70 cm on the x-axis.
All these masses are treated as if all their weight is focused entirely at one point along the x-axis. This is why such calculations are often easier and handier in theoretical and practical physics. The challenge arises when trying to find a balance point for these masses, which leads us to the concept of the center of mass.
X-Axis Calculation
The x-axis calculation is essential to finding where the center of mass resides. Let's break it down:First, any mass and position must be consistent in units鈥攗sually, kilograms and meters are ideal. Converting:
  • 200 g becomes 0.20 kg at 0 m.
  • 500 g becomes 0.50 kg at 0.30 m.
  • 400 g becomes 0.40 kg at 0.70 m.
To find the x-coordinate of the center of mass, we calculate a weighted average of all positions along the x-axis. This means multiplying each point's mass by its position and then dividing the total by the sum of all masses. Formulaically, this is expressed as:\[ x_{\text{cm}} = \frac{\sum x_i m_i}{\sum m_i} \]Here, \(\sum x_i m_i\) computes to 0.43 kg鈰卪, and \(\sum m_i\) is 1.10 kg, resulting eventually in \(x_{\text{cm}} \approx 0.39\) m. This means all point masses balance at 0.39 meters along the axis.
Physics Problem Solving
Physics problem solving involves step-by-step analysis and calculation, and calculating the center of mass is a perfect example. We've simplified complex bodily interactions into basic, solvable components: 1. **Identify and convert measurements**: Start by recognizing what quantities and units you're working with.
  • Grams to kilograms.
  • Centimeters to meters for position.
2. **Weighted sum of positions**:
  • Multiply each mass by its position.
  • Add up these products to get a cumulative value.
3. **Calculate the total mass**: Sum of all the masses in the system gives a total amount. 4. **Compute the coordinate for center of mass**: Divide the weighted sum by total mass to find where the center of mass lies on the x-axis. This method hinges on understanding basic algebra and physics principles. By systematically following these steps, even complex arrangements of masses can be deciphered and handled in a concise manner. Mathematics serves as the core tool in translating a story of mass distribution into the precise language of numbers.

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Most popular questions from this chapter

An \(8.0\) -g bullet is fired horizontally into a \(9.00\) -kg cube of wood, which is at rest on a frictionless air table. The bullet lodges in the wood. The cube is free to move and has a speed of \(40 \mathrm{~cm} / \mathrm{s}\) after impact. Find the initial velocity of the bullet. This is an example of a completely inelastic collision for which momentum is conserved, although \(\mathrm{KE}\) is not. Consider the system (cube \(+\) bullet). The velocity, and hence the momentum, of the cube before impact is zero. Take the bullet's initial motion to be positive in the positive \(x\) -direction. The momentum conservation law tells us that Momentum of system before impact = Momentum of system after impact $$ \begin{aligned} \text { (Momentum of bullet) }+\text { (Momentum of cube) } &=(\text { Momentum of bullet }+\text { Cube }) \\ m_{B} v_{B x}+m_{C} v_{C x} &=\left(m_{B}+m_{C}\right) v_{x} \\ (0.0080 \mathrm{~kg}) v_{B x}+0 &=(9.008 \mathrm{~kg})(0.40 \mathrm{~m} / \mathrm{s}) \end{aligned} $$ Solving yields \(v_{B x}=0.45 \mathrm{~km} / \mathrm{s}\) and so \(\overrightarrow{\mathbf{v}}_{B}=0.45 \mathrm{~km} / \mathrm{s}\) - POSITIVE \(X\) -DIRECTION.

Two identical balls traveling parallel to the \(x\) -axis have speeds of \(30 \mathrm{~cm} / \mathrm{s}\) and are oppositely directed. They collide off center perfectly elastically. After the collision, one ball is moving at an angle of \(30^{\circ}\) above the \(+x\) -axis. Find its speed and the velocity of the other ball.

A system consists of the following masses in the \(x y\) -plane: \(4.0 \mathrm{~kg}\) at coordinates \((x=0, y=5.0 \mathrm{~m})\), \(7.0 \mathrm{~kg}\) at \((3.0 \mathrm{~m}, 8.0 \mathrm{~m})\), and \(5.0 \mathrm{~kg}\) at \((-3.0 \mathrm{~m},-6.0 \mathrm{~m})\). Find the position of its center of mass. $$ \begin{array}{l} x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(4.0 \mathrm{~kg})+(3.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-3.0 \mathrm{~m})(5.0 \mathrm{~kg})}{(4.0+7.0+5.0) \mathrm{kg}}=0.38 \mathrm{~m} \\ y_{\mathrm{cm}}=\frac{\sum y_{i} m_{i}}{\sum m_{i}}=\frac{(5.0 \mathrm{~m})(4.0 \mathrm{~kg})+(8.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-6.0 \mathrm{~m})(5.0 \mathrm{~kg})}{16 \mathrm{~kg}}=2.9 \mathrm{~m} \end{array} $$ and \(z_{\mathrm{cm}}=0\). These distances are, of course, measured from the origin \((0,0,0,\), .

Two identical railroad cars sit on a horizontal track, with a distance \(D\) between their two centers of mass. By means of a cable between them, a winch on one is used to pull the two together. (a) Describe their relative motion. ( \(b\) ) Repeat the analysis if the mass of one car is now three times that of the other. Keep in mind that the velocity of the center of mass of a system can only be changed by an external force. Here the forces due to the cable acting on the two cars are internal to the two-car system. The net external force on the system is zero, and so its center of mass does not move, even though each car travels toward the other. Taking the origin of coordinates at the center of mass, $$ x_{\mathrm{cm}}=0=\frac{\sum m_{i} x_{i}}{\sum m_{i}}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}} $$ where \(x_{1}\) and \(x_{2}\) are the positions of the centers of mass of the two cars. (a) If \(m_{1}=m_{2}\), this equation becomes $$ 0=\frac{x_{1}+x_{2}}{2} \quad \text { or } \quad x_{1}=-x_{2} $$ The two cars approach the center of mass, which is originally midway between the two cars (that is, \(D / 2\) from each), in such a way that their centers of mass are always equidistant from it. (b) If \(m_{1}=3 m_{2}\), then we have $$ 0=\frac{3 m_{2} x_{1}+m_{2} x_{2}}{3 m_{2}+m_{2}}=\frac{3 x_{1}+x_{2}}{4} $$ from which \(x_{1}=-x_{2} / 3\). Since \(m_{1}>m_{2}\), it must be that \(x_{1}

The nucleus of an atom has a mass of \(3.80 \times 10^{-25} \mathrm{~kg}\) and is at rest. The nucleus is radioactive and suddenly ejects a particle of mass \(6.6 \times 10^{-27} \mathrm{~kg}\) and speed \(1.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\). Find the recoil speed of the nucleus that is left behind. The particle flies off in one direction, the nucleus recoils away in the opposite direction, and momentum is conserved. Take the direction of the ejected particle as positive. We are given, \(m_{n i}=3.80 \times 10^{-25} \mathrm{~kg}\), \(m_{\mathrm{p}}=6.6 \times 10^{-27} \mathrm{~kg}, m_{n f}=m_{n i}-m_{p}=3.73 \times 10^{-25} \mathrm{~kg}\), and \(v_{p f}=1.5 \times 10^{7} \mathrm{~m} / \mathrm{s} ;\) find the final speed of the nucleus, \(\boldsymbol{v}_{n f}\) Solving leads to $$ -v_{n f}=\frac{\left(6.6 \times 10^{-27} \mathrm{~kg}\right)\left(1.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\right)}{3.73 \times 10^{-25} \mathrm{~kg}}=\frac{10.0 \times 10^{-20}}{3.73 \times 10^{-25}}=2.7 \times 10^{5} \mathrm{~m} / \mathrm{s} $$ The fact that this is negative tells us that the velocity vector of the nucleus points in the negative direction, opposite to the velocity of the particle, which we took to be positive.
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