Question

A 68 kg

man's arm, including the hand, can be modeled as a 79-cm
-long uniform cylinder with a mass of 3.3 kg. In raising both his arms, from hanging down to straight up, by how much does he raise his center of gravity?

Answer 1

Step-by-step explanation:

We can calculate the change in the man's center of gravity by considering the initial and final positions of the center of gravity of his arms.

Assuming the man's arms are initially hanging down by his sides, the center of gravity of his arms is located at the midpoint of the cylinder, which is at a distance of L/2 = 79/2 = 39.5 cm from the shoulder joint.

When the man raises his arms straight up, the center of gravity of his arms is located at the top of the cylinder, which is at a distance of L = 79 cm from the shoulder joint.

The change in the man's center of gravity is therefore:

Δh = h_final - h_initial

= L - L/2

= 79 cm - 39.5 cm

= 39.5 cm

Therefore, the man raises his center of gravity by 39.5 cm when he raises both his arms from hanging down to straight up.