A 66 kg man's arm, including the hand, can be modeled as a 78-cm-long uniform cylinder with a mass of 3.6 kg . In raising both his arms, from hanging down to straight up, by how much does he raise his center of gravity?

Answer 1

First, we need to find the center of gravity of each arm. For a uniform cylinder, the center of gravity is in the middle. Thus, the center of gravity of each arm is at a distance of 78/2 = 39 cm.

When the man raises his arms, the center of gravity of each arm moves from 39 cm below shoulder height to 39 cm above shoulder height. So each arm's center of gravity moves by 2 * 39 = 78 cm.

Now we need to find the combined center of gravity of the man and his arms when they are raised. First, let's find the total mass: 66 kg (man) + 2 * 3.6 kg (arms) = 66 + 7.2 = 73.2 kg.

Let x be the distance from the man's original center of gravity to the new combined center of gravity. We can write an equation for the total upward movement of the center of gravity (mass times distance):

66 kg * x + 2 * 3.6 kg * 78 cm = 73.2 kg * combined upward movement

Now solve for the combined upward movement:

combined upward movement = (66x + 2 * 3.6 * 78) / 73.2

We need to find x. Since the man's body doesn't change except for raising his arms, we can assume that the center of gravity of the man without his arms remains at the same height, and therefore x=0.

combined upward movement = (66 * 0 + 2 * 3.6 * 78) / 73.2 = (2 * 3.6 * 78) / 73.2 = 10.08 cm

So, the man raises his center of gravity by 10.08 cm when he raises his arms from hanging down to straight up.