Systems and Center of Mass — Practice Problem Solutions

1. At the 50 cm mark — the exact midpoint. A uniform stick has its mass spread evenly, so by symmetry, the center of mass sits right at the geometric center with no calculation needed.

2. At 4 m — the midpoint. Because the two masses are equal, the weighted average happens to reduce to the plain average here; that's a special case, not a general rule.

3. $x_{cm}=\dfrac{(1)(0)+(3)(20)}{1+3}=\dfrac{60}{4}=15\text{ cm}$.

4. The center of mass lies somewhere along the pin's central vertical axis of symmetry — but not necessarily at the geometric midpoint of its height, since a bowling pin isn't uniformly shaped along its length. Because more of its mass is concentrated in the wider base, the center of mass sits below the halfway point of the pin's height, shifted toward that heavier bottom section.

5. Correct: $x_{cm}=\dfrac{(4)(0)+(2)(12)}{4+2}=\dfrac{24}{6}=4\text{ m}$. The student's mistake was taking the plain average of the two positions ($\frac{0+12}{2}=6\text{ m}$) instead of weighting by mass — since the 4 kg block is heavier, the true center of mass sits closer to it, at 4 m, not 6 m.

6. a) $x_{cm}=\dfrac{(2)(5)+(3)(15)+(5)(35)}{2+3+5}=\dfrac{10+45+175}{10}=\dfrac{230}{10}=23\text{ m}$. b) The classmate took the plain average of the three positions ($\frac{5+15+35}{3}\approx18.3\text{ m}$) and forgot to weight by mass. Drone C is both the farthest out and the heaviest (5 kg), so it pulls the true center of mass well past the naive midpoint estimate, to 23 m.

7. a) Closer to the 10 kg sphere — the more massive object always pulls the center of mass toward itself. b) $x_{cm}=\dfrac{(10)(0)+(2)(3.0)}{10+2}=\dfrac{6}{12}=0.5\text{ m}$ from the 10 kg sphere's center.

8. $x_{cm}=\dfrac{(40)(0)+(20)(60)}{60}=\dfrac{1200}{60}=20\text{ cm}$ — (A).