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 $\pm 30^{\circ}$ about (a total of $60^{\circ}$) from the vertical in approximately 1 second. Assume that they are held straight, rather than being bent, which is not quite true. Consider a 75-kg person walking at 5.0 km/h, having arms 70 cm long and legs 90 cm long. What is the total kinetic energy due to both his forward motion and his rotation?

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 $\pm 30^{\circ}$ about (a total of $60^{\circ}$) from the vertical in approximately 1 second. Assume that they are held straight, rather than being bent, which is not quite true. Consider a 75-kg person walking at 5.0 km/h, having arms 70 cm long and legs 90 cm long. What percentage of his kinetic energy is due to the rotation of his legs and arms?

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 $\pm 30^{\circ}$ about (a total of $60^{\circ}$) from the vertical in approximately 1 second. Assume that they are held straight, rather than being bent, which is not quite true. Consider a 75-kg person walking at 5.0 km/h, having arms 70 cm long and legs 90 cm long. 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.

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 1/2 second. We shall assume that they are held straight, rather than being bent, which is not quite true. Let us consider a $75-\mathrm{kg}$ person running at $12.0 \mathrm{~km} / \mathrm{h}$, having arms $70 \mathrm{~cm}$ long and legs $90 \mathrm{~cm}$ long. What percentage of his kinetic energy is due to the rotation of his legs and arms?

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the string is held in place and the hoop is released from rest. After the hoop has descended 75.0 cm, calculate the speed of its center.

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 1/2 second. We shall assume that they are held straight, rather than being bent, which is not quite true. Let us consider a $75-\mathrm{kg}$ person running at $12.0 \mathrm{~km} / \mathrm{h}$, having arms $70 \mathrm{~cm}$ long and legs $90 \mathrm{~cm}$ long. What is the total kinetic energy due to both his forward motion and his rotation ?

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the string is held in place and the hoop is released from rest. After the hoop has descended 75.0 cm, calculate the angular speed of the rotating hoop.

A light string is wrapped around the outer rim of a solid uniform cylinder of diameter $75.0 \mathrm{~cm}$ that can rotate without friction about an axle through its center. A $3.00 \mathrm{~kg}$ stone is tied to the free end of the string When the system is released from rest, you determine that the stone reaches a speed of after having fallen 2.50 m. What is the mass of the cylinder? 3.50 m/s

A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of $0.600 \mathrm { rad } / \mathrm { s } ^ { 2 }$. Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through $120.0 ^ { \circ }$.

A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of $0.600 \mathrm { rad } / \mathrm { s } ^ { 2 }$. Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim at the start.

Four small spheres, each of which you can regard as a point of mass $0.200 \mathrm{~kg}$, are arranged in a square $0.400 \mathrm{~m}$ on a side and connected by extremely light rods . Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane (an axis through point $O$ in the figure);

Four small spheres, each of which you can regard as a point of mass $0.200 \mathrm{~kg}$, are arranged in a square $0.400 \mathrm{~m}$ on a side and connected by extremely light rods . Find the moment of inertia of the system about an axis that passes through the centers of the upper left and lower right spheres and through point $O$.

Two ladybugs sit on a rotating disk, as shown in the figure (the ladybugs are at rest with respect to the surface of the disk and do not slip). Ladybug 1 is located at a radial distance of R and Ladybug 2 is located at a radial distance of 2R.

a)What is the relationship between the angular speed of ladybug 1 and ladybug 2?

b)What is the relationship between the linear speed of ladybug 1 and ladybug 2?

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceleration α. The flywheel is assumed to be at rest at time t = 0 in this problem. Find the time t_1 it takes to accelerate the flywheel to ω_1 if the angular acceleration is α. Express your answer in terms of ω_1 and α.