AP Physics C: Mechanics 2.1 Systems and Center of Mass Questions

Analyse a system's mass distribution by locating its center of mass from discrete components, composite bodies, continuous density, coordinates, and symmetry.

Exam points

calculate center-of-mass position or coordinates for a system of discrete masses
replace uniform component bodies by masses at their own centers to analyse a composite system
integrate a continuous mass distribution and use symmetry to locate its center of mass

Covered syllabus points

2.1.A—Describe the properties and interactions of a system 2.1.B—Describe the location of a system’s center of mass with respect to the system’s constituent parts

[Maximum number: 2]

Object A is a long, thin, uniform rod of mass M and length 2 L that is free to rotate about a pivot of negligible friction at its left end, as shown above.

Determine the following for the given coordinate system shown in the figure.

The x-coordinate of the center of mass of object B

The y-coordinate of the center of mass of object B

Object B has a rotational inertia of $I_{\mathrm{B}}$ about its pivot.

i. For correctly substituting into an equation for the center of mass of an object in the 1 point horizontal direction

X_{C M}=\frac{\sum m_{i} x_{i}}{\sum m_{i}}=\frac{\left[\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)+\left(\frac{M}{2}\right)(L)\right]}{\left(\frac{M}{2}+\frac{M}{2}\right)}=\frac{\left(\frac{M L}{4}+\frac{M L}{2}\right)}{M}=\frac{3}{4} L

For correctly substituting into an equation for the center of mass of an object in the vertical 1 point direction

Y_{C M}=\frac{\sum m_{i} y_{i}}{\sum m_{i}}=\frac{\left[\left(\frac{M}{2}\right)(L)+\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)\right]}{\left(\frac{M}{2}+\frac{M}{2}\right)}=\frac{\left(\frac{M L}{2}+\frac{M L}{4}\right)}{M}=\frac{3}{4} L

Total for part (b) 2 points

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Covered syllabus points

Question 1