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Classical Mechanics

Moment of Inertia

Concept Quizzes

Rotational Form of Newton's Second Law
Moment of Inertia of Point Masses
Moment of Inertia of Mass Distributions
Center of mass of a collection of points
Center of a mass distribution

Challenge Quizzes

Moment of Inertia: Level 3-5 Challenges

Center of mass of a collection of points

Three particles of respective masses $m_1=12.0\text{ kg},$ $m_2=25.0\text{ kg}$ and $m_3=38.0\text{ kg}$ form an equilateral triangle of side length $a=140\text{ cm}.$ If we locate $m_1$ at the origin on the $xy$ -plane, and put $m_2$ to the right of $m_1$ on the $x$ -axis, as shown in the above figure, what are the approximate coordinates of the center of mass of this system?

The value of $\sqrt{3}$ is $1.732.$

(97.3\text{ cm},37.0\text{ cm})

(82.1\text{ cm},37.0\text{ cm})

(82.1\text{ cm},61.3\text{ cm})

(97.3\text{ cm},61.3\text{ cm})

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by Brilliant Staff

Three particles are on the $xy$ -plane, as shown in the above figure. The masses of the three particles are $m_1=2.0\text{ kg},$ $m_2=5.0\text{ kg}$ and $m_3=9.0\text{ kg}.$ If the scales on the axes are set by $x_s=4.0\text{ m}$ and $y_s=6.0\text{ m},$ what are the $xy$ -coordinates of the system's center of mass?

\displaystyle{ \left(\frac{19}{8}\text{ m},\frac{16}{69}\text{ m}\right)}

\displaystyle{ \left(\frac{8}{19}\text{ m},\frac{69}{16}\text{ m}\right)}

\displaystyle{ \left(\frac{8}{19}\text{ m},\frac{16}{69}\text{ m}\right)}

\displaystyle{ \left(\frac{19}{8}\text{ m},\frac{69}{16}\text{ m}\right)}

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by Brilliant Staff

Consider a uniform bar of length $3L$ and mass $m_3=5m.$ Two balls are hanging on strings with negligible mass from the two ends of the bar, and their masses are $m_1=m$ and $m_2=3m.$ The lengths of the string on which the balls are hanging are $L$ and $2L,$ respectively, as shown in the above figure. What is the center of mass of this system relative to the midpoint of the bar?

\displaystyle{ \left(\frac{1}{3}L,-\frac{7}{9}L\right)}

\displaystyle{ \left(\frac{7}{10}L,-\frac{3}{8}L\right)}

\displaystyle{ \left(\frac{3}{8}L,-\frac{7}{10}L\right)}

\displaystyle{ \left(\frac{7}{10}L,-\frac{7}{8}L\right)}

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by Brilliant Staff

Three particles $A,$ $B$ and $C$ are on the $xy$ -plane. Their masses are $m_A=2.00\text{ kg},$ $m_B=4.00\text{ kg}$ and $m_C=3.00\text{ kg},$ and the coordinates of $A$ and $B$ are $(-1.40\text{ m},0.48\text{ m})$ and $(0.70\text{ m},-0.72\text{ m}),$ respectively. If the coordinates of the center of mass of the three-particle system is $(-0.50\text{ m},-0.70\text{ m}),$ what is the coordinates of particle $C ?$

(-1.25\text{ m},-1.46\text{ m})

(-1.25\text{ m},-1.90\text{ m})

(-1.50\text{ m},-1.46\text{ m})

(-1.50\text{ m},-1.90\text{ m})

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by Brilliant Staff

Three particles of respective masses $m_1=13.0\text{ kg},$ $m_2=29.0\text{ kg}$ and $m_3=37.0\text{ kg}$ form an equilateral triangle of side length $a=140\text{ cm}.$ If we locate $m_1$ at the origin on the $xy$ -plane, and put $m_2$ to the right of $m_1$ on the $x$ -axis, as shown in the above figure, what are the approximate coordinates of the center of mass of this system?

The value of $\sqrt{3}$ is $1.732.$

(98.2\text{ cm},56.7\text{ cm})

(84.2\text{ cm},56.7\text{ cm})

(84.2\text{ cm},34.2\text{ cm})

(98.2\text{ cm},34.2\text{ cm})

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by Brilliant Staff