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In a long jump, an athlete leaves the ground with an initial angular momentum that tends to rotate her body forward, threatening to ruin her landing. To counter this tendency, she rotates her outstretched arms to "take up" the angular momentum (Fig. 11- 18). In \(0.700 \mathrm{~s}\), one arm sweeps through \(0.500\) rev and the other arm sweeps through \(1.000\) rev. Treat each arm as a thin rod of mass \(4.0 \mathrm{~kg}\) and length \(0.60 \mathrm{~m}\), rotating around one end. In the athlete's reference frame, what is the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders?

Short Answer

Expert verified
The total angular momentum is approximately 6.44 kg m²/s.

Step by step solution

01

Understand the Angular Motion

The angular motion involves the movement of two arms each sweeping through different angles in the same time interval of 0.700 s. Each arm rotates around one end.
02

Convert Revolutions to Radians

Convert the angles from revolutions to radians. Since 1 revolution is equal to \(2\pi\) radians, we have: for one arm, \(0.500\, \text{rev} = 0.500 \times 2\pi = \pi\) radians and for the other arm, \(1.000\, \text{rev} = 1.000 \times 2\pi = 2\pi\) radians.
03

Calculate Angular Velocity

The angular velocity for each arm can be calculated using \(\omega = \frac{\theta}{t}\). For the first arm: \(\omega_1 = \frac{\pi}{0.700}\, \text{rad/s}\) and for the second arm: \(\omega_2 = \frac{2\pi}{0.700}\, \text{rad/s}\).
04

Understand Moment of Inertia for Each Arm

The moment of inertia for a thin rod rotating around one end is given by \( I = \frac{1}{3}mL^2 \). For each arm \((m = 4.0 \text{ kg}, L = 0.60 \text{ m})\), calculate \( I = \frac{1}{3} \times 4.0 \times (0.60)^2 = 0.48 \text{ kg} \, \text{m}^2\).
05

Calculate Angular Momentum for Each Arm

The angular momentum \(L\) is calculated using \(L = I\omega\). For the first arm: \(L_1 = 0.48 \times \frac{\pi}{0.700} \approx 2.15 \text{ kg} \, \text{m}^2/\text{s}\) and for the second arm: \(L_2 = 0.48 \times \frac{2\pi}{0.700} \approx 4.29 \text{ kg} \, \text{m}^2/\text{s}\).
06

Calculate Total Angular Momentum

The total angular momentum of the arms is the sum of the individual angular momenta: \(L_{\text{total}} = L_1 + L_2 \approx 2.15 + 4.29 = 6.44 \text{ kg} \, \text{m}^2/\text{s}\).

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Key Concepts

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

Moment of Inertia
When dealing with rotational motion, the moment of inertia is a key concept that characterizes how mass is distributed relative to an axis of rotation. It describes an object's resistance to changes in its rotational motion. To understand it better, think of how difficult it is to spin objects with different mass distributions, like a thin rod versus a figure skater with arms extended.

For a thin rod rotating about one end, the moment of inertia is calculated using the formula: \[ I = \frac{1}{3} m L^2 \]where \(m\) is the mass of the rod and \(L\) is its length. In our exercise, each arm of the athlete acts as a rod with a mass of \(4.0 \, \text{kg}\) and a length of \(0.60 \, \text{m}\). Plugging in these values, the moment of inertia for each arm was found to be \(0.48 \, \text{kg} \cdot \text{m}^2\).

Remember, the larger the moment of inertia, the harder it is to change the object's rotational speed. This is crucial for the athlete balancing rotational forces during a jump.
Angular Velocity
Angular velocity is the rate at which an object rotates or revolves around an axis. It tells us how fast an angle is changing, usually measured in radians per second. In the context of our problem, angular velocity helps to determine how swiftly each arm of the athlete spins during the rotation.

To find the angular velocity, we use the formula: \[ \omega = \frac{\theta}{t} \]where \(\theta\) is the angular displacement in radians and \(t\) is the time in seconds.
  • The first arm of the athlete spins through \(\pi\) radians in \(0.700 \, \text{s}\), giving an angular velocity of \(\omega_1 = \frac{\pi}{0.700} \, \text{rad/s}\).
  • The second arm spins through \(2\pi\) radians over the same duration, resulting in an angular velocity of \(\omega_2 = \frac{2\pi}{0.700} \, \text{rad/s}\).
Understanding angular velocity helps explain how the athlete manages the arms' motion to counteract the body's planned rotation.
Rotational Motion
Rotational motion involves objects rotating around an axis. It is similar to linear motion, but instead of distance, we consider angles, and instead of speed, we consider angular velocity. Imagine spinning a wheel or twirling a baton - these are examples of rotational motion.

To better understand rotational motion, focus on how forces are balanced. In our athlete’s jump, the arms are strategically rotated to alter the rotation of her body. This is an excellent demonstration of the conservation of angular momentum, ensuring that the rotation of her arms changes the rotation of her body favorably.

The total angular momentum of the system, comprising both arms, equals the vector sum of the individual angular momenta. Calculated as:\[ L_{\text{total}} = L_1 + L_2 \]where \(L_1\) and \(L_2\) are the angular momenta of the individual arms. This rotational adjustment lets the athlete maintain control during her flight and land effectively. Rotational motion is everywhere, from galaxies spinning to kids playing on a merry-go-round, making it a fascinating and vital concept in physics.

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Most popular questions from this chapter

At the instant the displacement of a \(2.00 \mathrm{~kg}\) object relative to the origin is \(\vec{d}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}+(4.00 \mathrm{~m}) \hat{\mathrm{j}}-(3.00 \mathrm{~m}) \hat{\mathrm{k}}\), its velocity is \(\vec{v}=-(6.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}+(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\) and it is subject to a force \(\vec{F}=(6.00 \mathrm{~N}) \hat{\mathrm{i}}-(8.00 \mathrm{~N}) \hat{\mathrm{j}}+(4.00 \mathrm{~N}) \hat{\mathrm{k}}\). Find (a) the acceleration of the object, (b) the angular momentum of the object about the origin, (c) the torque about the origin acting on the object, and (d) the angle between the velocity of the object and the force acting on the object.

A top spins at 30 rev/s about an axis that makes an angle of \(30^{\circ}\) with the vertical. The mass of the top is \(0.50 \mathrm{~kg}\), its rotational inertia about its central axis is \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and its center of mass is \(4.0 \mathrm{~cm}\) from the pivot point. If the spin is clockwise from an overhead view, what are the (a) precession rate and (b) direction of the precession as viewed from overhead?

A sanding disk with rotational inertia \(1.2 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) is attached to an electric drill whose motor delivers a torque of magnitude \(16 \mathrm{~N} \cdot \mathrm{m}\) about the central axis of the disk. About that axis and with the torque applied for \(33 \mathrm{~ms}\), what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

Force \(\vec{F}=(2.0 \mathrm{~N}) \hat{\mathrm{i}}-(3.0 \mathrm{~N}) \hat{\mathrm{k}}\) acts on a pebble with position vector \(\vec{r}=(0.50 \mathrm{~m}) \hat{\mathrm{j}}-(2.0 \mathrm{~m}) \hat{\mathrm{k}}\) relative to the origin. In unit-vector notation, what is the resulting torque on the pebble about (a) the origin and (b) the point \((2.0 \mathrm{~m}, 0,-3.0 \mathrm{~m})\) ?

The rotational inertia of a collapsing spinning star drops to \(\frac{1}{3}\) its initial value. What is the ratio of the new rotational kinetic energy to the initial rotational kinetic energy?

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