Question

A 70 kg man's arm, including the hand, can be modeled as BIO 75-cm-long uniform rod with a mass of 3.5 kg. When the man raises both his arms, from hanging down to straight up, by how much does he raise his center of mass?

Answer 1

Answer:

  Δy = 7.1 cm

Explanation:

The center of mass of a body is defined

            $y_{cm}$ = 1 /M ∑$m_{i} y_{i}$i

Where M is the total mass of the body, m mass of each part and ‘y’ height

Let's apply this equation to our case

We locate the reference system on the shoulders

The height of the arms is at its midpoint

            y = -75/2 = 37.5 cm

With arms down

            $y_{cm}$ = 1/70 (63 y₀ - 3.5 37.5 - 3.5 37.5)

            $y_{cm}$ = 1/70 (63 y)₀ - 7 37.5)

With arms up

          $y_{cm}$’= 1/70 (63 y₀ + 3.5 y + 3.5 y)

          $y_{cm}$’= 1/70 (63y₀ + 7 35.5)

let's subtract the two equations

        $y_{cm}$’ - $y_{cm}$ = 1/70 2 (7 35.5)

         Δy = $y_{cm}$’ - $y_{cm}$ = 2 7 35.5 / 70

         ΔY = 7.1 cm