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The Kinetic Energy of Walking. If a person of mass \(M\) simply moved forward with speed \(V\), his kinetic energy would be \(\frac{1}{2} M V^{2}\).However, in addition to possessing a forward motion, various parts of his body (such as the arms and legs) undergo rotation. Therefore, his total kinetic energy is the sum of the energy from his forward motion plus the rotational kinetic energy of his arms and legs. The purpose of this problem is to see how much this rotational motion contributes to the person's kinetic energy. Biomedical measurements show that the arms and hands together typically make up 13\(\%\) of a person's mass, while the legs and feet together account for 37\(\%\) . For a rough (but reasonable) calculation, we can model the arms and legs as thin uniform bars pivoting about the shoulder and hip, respectively. In a brisk walk, the arms and legs each move through an angle of about \(\pm 30^{\circ}\) (a total of \(60^{\circ} )\) from the vertical in approximately 1 second. We shall assume that they are held straight, rather than being bent, which is not quite true. Let us consider a 75 -kg person walking at 5.0 \(\mathrm{km} / \mathrm{h}\) , having arms 70 \(\mathrm{cm}\) long and legs 90 \(\mathrm{cm}\) long. (a) What is the average angular velocity of his arms and legs? (b) Using the average angular velocity from part (a), calculate the amount of rotational kinetic energy in this person's arms and legs as he walks. (c) What is the total kinetic energy due to both his forward motion and his rotation? (d) What percentage of his kinetic energy is due to the rotation of his legs and arms?

Step by step solution

01

Convert speed to m/s

Convert the walking speed from km/h to m/s. Given speed = 5.0 km/h. Use the conversion factor: \[1 \text{ km} = 1000 \text{ m} \quad \text{and} \quad 1 \text{ h} = 3600 \text{ s}\]So, the speed in m/s is:\[V = \frac{5.0 \times 1000}{3600} \approx 1.39 \, \text{m/s}\]

02

Calculate average angular velocity

The arms and legs move through an angle of 60° back and forth in 1 second. Convert this to radians per second.Since 1 revolution = 360° = 2π radians, The angular motion is \(60° = \frac{60 \times \pi}{180} = \frac{\pi}{3} \text{ radians}\).Thus, angular velocity \( \omega = \frac{\text{Total angle}}{\text{Time}} = \frac{\pi/3}{1} \text{ rad/s} = \frac{\pi}{3} \text{ rad/s}\).

03

Find rotational inertia

Model the arms and legs as thin rods rotated about one end. The moment of inertia \(I\) for a rod about one end is given by:\[ I = \frac{1}{3} m L^2 \] The mass of the arms \( m_a = 0.13M \) and their length \( L_a = 0.7 \text{ m} \).The mass of the legs \( m_l = 0.37M \) and their length \( L_l = 0.9 \text{ m} \).

04

Calculate rotational kinetic energy

The rotational kinetic energy \( KE_{r} \) of a rotating body is:\[ KE_{r} = \frac{1}{2} I \omega^2 \]For arms: \[ I_a = \frac{1}{3} m_a L_a^2 = \frac{1}{3} \times 0.13M \times (0.7)^2 \]\[ KE_{r_a} = \frac{1}{2} I_a \left(\frac{\pi}{3}\right)^2 \]For legs: \[ I_l = \frac{1}{3} m_l L_l^2 = \frac{1}{3} \times 0.37M \times (0.9)^2 \]\[ KE_{r_l} = \frac{1}{2} I_l \left(\frac{\pi}{3}\right)^2 \]Calculate both to find the total rotational kinetic energy \( KE_r = KE_{r_a} + KE_{r_l} \).

05

Calculate translational kinetic energy

The translational kinetic energy \( KE_t \) is calculated using:\[ KE_t = \frac{1}{2} M V^2 \]Where \( M = 75 \text{ kg} \) and \( V = 1.39 \text{ m/s} \).\[ KE_t = \frac{1}{2} \times 75 \times (1.39)^2 \]

06

Calculate total kinetic energy

The total kinetic energy \( KE_{total} \) is the sum of translational and rotational kinetic energy:\[ KE_{total} = KE_t + KE_r \]

07

Find percentage of kinetic energy due to rotation

To find the percentage of kinetic energy due to rotation, use:\[ \text{Percentage} = \left(\frac{KE_r}{KE_{total}}\right) \times 100\% \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Kinetic Energy

When we talk about rotational kinetic energy, we are considering the energy that is due to the rotation of an object. This kind of energy is similar to translational kinetic energy, but it involves rotation instead of straightforward motion. It’s important to know that every body part in motion contributes to this energy when walking.

Rotational kinetic energy is calculated using the formula:

• \( KE_{r} = \frac{1}{2} I \omega^2 \)
• Where:
• \( KE_{r} \) is the rotational kinetic energy.
• \( I \) is the moment of inertia.
• \( \omega \) is the angular velocity.

The moment of inertia here depends on how mass is distributed with respect to the axis of rotation. In the case of a walking person, the arms and legs rotating about the shoulder and hip are modeled like thin rods pivoting about one end.

The arms and legs increase their angular velocity and thus their kinetic energy when they rotate. Having an understanding of this concept helps in appreciating how rotation plays a significant role in everyday human movement, like walking.

Translational Kinetic Energy

Translational kinetic energy is all about the straightforward motion of an object's mass. When a person walks, apart from the rotations, there is also translational motion because the person moves in space from one point to another.

This type of kinetic energy can be calculated using the formula:

• \( KE_t = \frac{1}{2} M V^2 \)
• Where:
• \( KE_t \) is the translational kinetic energy.
• \( M \) is the total mass of the person.
• \( V \) is the velocity of movement.

For our exercise, we converted the speed from kilometers per hour to meters per second to get \( V \approx 1.39 \text{ m/s} \). This is crucial since kinetic energy calculations require consistent units.

Translational kinetic energy gives us a vivid picture of how much energy is involved in just moving from one place to another without considering rotational effects. It showcases the basic energy requirements of moving forward.

Moment of Inertia

Moment of inertia is an integral part of studying rotations. It can be thought of as the rotational analog to mass in linear motion. It's a measure of an object's resistance to changes in its rotation.

For a thin rod rotating about one end, the moment of inertia is given by:

• \( I = \frac{1}{3} m L^2 \)
• Where:
• \( I \) is the moment of inertia.
• \( m \) is the mass of the object, for example, the arms or legs.
• \( L \) is the length of the rod (or limb in this case).

Knowing this helps with calculating the rotational kinetic energy as this value \( I \) is integral to the calculations. For instance, using appropriate values for the arms and legs, you can build an accurate model to understand the energy contributions from the rotations.

The moment of inertia heavily depends on how mass is distributed relative to the pivot point, highlighting why objects of the same mass but different shapes can rotate differently. Understanding this concept supports deeper insight into physical movement dynamics and engineering applications.