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Chapter 8: Problem 29

Four masses are positioned in the \(x y\) -plane as follows: \(300 \mathrm{~g}\) at \((x=0, y=2.0 \mathrm{~m}), 500 \mathrm{~g}\) at \((-20 \mathrm{~m},-3.0 \mathrm{~m})\), \(700 \mathrm{~g}\) at \((50 \mathrm{~cm}, 30 \mathrm{~cm})\), and \(900 \mathrm{~g}\) at \((-80 \mathrm{~cm}, 150 \mathrm{~cm})\). Find their center of mass.

Short Answer

Expert verified
The center of mass is at (-4.54 m, 0.72 m).

Step by step solution

01

Convert Units to Meters

Convert the masses and positions to the same units to ensure consistency. Since the coordinates are given in meters and centimeters, convert all coordinates to meters. - Mass 1 (300 g) at coordinates (0, 2.0 m) - Mass 2 (500 g) at coordinates (-20 m, -3.0 m) - Mass 3 (700 g) at coordinates (50 cm, 30 cm) which converts to (0.5 m, 0.3 m) - Mass 4 (900 g) at coordinates (-80 cm, 150 cm) which converts to (-0.8 m, 1.5 m)
02

Calculate Weighted Position Sums

To find the center of mass, calculate the weighted sum of the positions for both the x and y coordinates. The formula for the x-coordinate of the center of mass is:\[ x_{cm} = \frac{\sum m_i x_i}{\sum m_i} \]Calculate the sum of products of mass and x-coordinate, and similarly for the y-coordinate:- \( \sum m_i x_i = (300 \times 0) + (500 \times (-20)) + (700 \times 0.5) + (900 \times (-0.8)) = -10,900 \) g*m- \( \sum m_i y_i = (300 \times 2.0) + (500 \times (-3.0)) + (700 \times 0.3) + (900 \times 1.5) = 1720 \) g*m
03

Calculate Total Mass

Sum the masses of all objects to use in the center of mass formula:\[ \sum m_i = 300 + 500 + 700 + 900 = 2400 \] g
04

Calculate Center of Mass Coordinates

Use the formula for the center of mass to find the x and y coordinates:- \( x_{cm} = \frac{-10,900 \, \text{g*m}}{2400 \, \text{g}} = -4.54 \, \text{m} \)- \( y_{cm} = \frac{1720 \, \text{g*m}}{2400 \, \text{g}} = 0.72 \, \text{m} \)
05

Conclusion

The center of mass of the system is located at coordinates:- \( x = -4.54 \, \text{m} \)- \( y = 0.72 \, \text{m} \)

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Key Concepts

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

Weighted Sum
When trying to find the center of mass, the idea of a weighted sum is essential. A weighted sum takes into account not just the values being added, but also assigns them weights to indicate their significance. In the exercise, the position of each mass is multiplied by the mass itself. This gives a product, which is then added up for each coordinate to find a total sum for x and y coordinates.
This method ensures that heavier masses have more impact on the position of the center of mass.
  • Use the formula: \( x_{cm} = \frac{\sum m_i x_i}{\sum m_i} \)
  • Ensure all positions are multiplied correctly with corresponding masses.
This approach is critical because it enables us to account for the distribution of matter, reflecting the true balance point.
Coordinate Conversion
In order to correctly apply the math for the center of mass, all coordinates must be in the same unit. In the given problem, positions are given in both meters and centimeters, requiring a conversion for consistency.
Usually, converting all measurements to meters can simplify calculations, since it's a standard unit of length in science and engineering.To convert centimeters to meters, simply divide by 100. For instance:
  • Convert 50 cm to meters by calculating \( \frac{50}{100} = 0.5 \) m.
  • Convert -80 cm by calculating \( \frac{-80}{100} = -0.8 \) m.
This conversion ensures all distances are comparable, making the calculations straightforward and error-free. Remember, whenever there are mixed units, always take time to convert.
Mass Distribution
Knowing how mass is spread out in relation to chosen coordinates is crucial for determining the center of mass. In our exercise scenario, the masses are distributed unevenly across the plane. Understanding this helps us figure out how each mass affects the system's balance. Some key points to remember about mass distribution include:
  • Heavier masses have a greater influence on the center of mass due to their larger contributions in the weighted sum.
  • The position of a mass in a coordinate system impacts how it shifts the center of mass. A far-away heavy mass greatly influences the center compared with a light mass nearby.
Always analyze the distribution to predict how it might influence the end results during or before performing calculations.
Position Calculation
Once you have the total weighted sums and the total mass, you can proceed to calculate the center of mass position for the system. This step consolidates all data into a single point location in the space.The formulae used are:
  • \( x_{cm} = \frac{-10,900 \, \text{g*m}}{2400 \, \text{g}} = -4.54 \, \text{m} \)
  • \( y_{cm} = \frac{1720 \, \text{g*m}}{2400 \, \text{g}} = 0.72 \, \text{m} \)
This calculation involves dividing the sum of products by the total mass for each coordinate direction. The results are the coordinates of the center of mass. Remember, the calculated point, \((-4.54, 0.72)\), tells us where the equilibrium of this mass system lies, should it stand free in space.

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Most popular questions from this chapter

Two balls of equal mass approach the coordinate origin, one moving downward along the \(y\) -axis at \(2.00 \mathrm{~m} / \mathrm{s}\) and the other moving to the right along the \(-x\) -axis at \(3.00 \mathrm{~m} / \mathrm{s}\). After they collide, one ball moves out to the right along the \(+x\) -axis at \(1.20 \mathrm{~m} / \mathrm{s}\). Find the scalar \(x\) and \(y\) velocity components of the other ball. This is a two-dimensional collision and momentum must be conserved independently in each perpendicular direction, \(x\) and \(y\). Take \(u p\) and to the right as positive. Accordingly, keeping in mind that before impact only one ball had an \(x\) -component of velocity, Or $$ \begin{aligned} \text { (Momentum before) }_{x} &=\text { (Momentum after) }_{x} \\ m(3.00 \mathrm{~m} / \mathrm{s})+0 &=m(1.20 \mathrm{~m} / \mathrm{s})+m v_{x} \end{aligned} $$ Here \(v_{\mathrm{x}}\) is the unknown \(x\) -component of velocity of the second ball acquired on impact. Since we know that the first ball lost some of its \(x\) -momentum, the second ball must have gained it. Moreover, (Momentum before) \(_{y}=(\text { Momentum after })_{y}\) or $$ 0+m(-2.00 \mathrm{~m} / \mathrm{s})=0+m v_{y} $$ Here \(v_{\mathrm{y}}\) is the \(y\) -component of velocity of the second ball. (Why the minus sign?) Solving each equation, after cancelling the mass we find that \(v_{x}=1.80 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=-2.00 \mathrm{~m} / \mathrm{s}\)

Two girls (masses \(m_{1}\) and \(m_{2}\) ) are on roller skates and stand at rest, close to each other and face to face. Girl- 1 pushes squarely against girl- 2 and sends her moving backward. Assuming the girls move freely on their skates, write an expression for the speed with which girl- 1 moves. We take the two girls to comprise the system under consideration. The problem states that girl- 2 moves "backward," so let that be the negative direction; therefore, the "forward" direction is positive. There is no resultant external force on the system (the push of one girl on the other is an internal force), and so momentum is conserved: $$ \begin{aligned} \text { Momentum before } &=\text { Momentum after } \\ 0 &=m_{1} v_{1}+m_{2} v_{2} \end{aligned} $$ from which \(\quad v_{1}=-\frac{m_{2}}{m_{1}} v_{2}\) Girl- 1 recoils with this speed. Notice that if \(m_{2} / m_{1}\) is very large, \(v_{1}\) is much larger than \(v_{2}\). The velocity of girl- \(1, \overrightarrow{\mathbf{v}}_{1}\), points in the positive forward direction. The velocity of girl- \(2, \overrightarrow{\mathbf{v}}_{2}\), points in the negative backward direction. If we put numbers into the equation, \(v_{2}\) would have to be negative and \(v_{1}\) would come out positive.

Two identical balls undergo a collision at the origin of coordinates. Before collision their scalar velocity components are \(\left(u_{x}=40 \mathrm{~cm} / \mathrm{s}, u_{y}=0\right)\) and \(\left(u_{x}=-30 \mathrm{~cm} / \mathrm{s}, u_{y}=20 \mathrm{~cm} / \mathrm{s}\right) .\) After collision, the first ball (the one moving along the \(x\) -axis) is standing still. Find the scalar velocity components of the second ball. [Hint: After the collision, the moving ball must have all of the momentum of the system.]

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.

A 90 -g ball moving at \(100 \mathrm{~cm} / \mathrm{s}\) collides head-on with a stationary 10 -g ball. Determine the speed of each after impact if \((a)\) they stick together, \((b)\) the collision is perfectly elastic, (c) the coefficient of restitution is \(0.90\).
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