Classical Mechanics
# Moment of Inertia

$m_1=12.0\text{ kg},$ $m_2=25.0\text{ kg}$ 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?

Three particles of respective massesThe value of $\sqrt{3}$ is $1.732.$

$xy$-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?

Three particles are on the$3L$ 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?

Consider a uniform bar of length$m_1=13.0\text{ kg},$ $m_2=29.0\text{ kg}$ 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?

Three particles of respective massesThe value of $\sqrt{3}$ is $1.732.$

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