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A woman with mass 50 \(\mathrm{kg}\) is standing on the rim of a large disk that is rotating at 0.50 rev \(/ \mathrm{s}\) about an axis through its center. The disk has mass 110 \(\mathrm{kg}\) and radius 4.0 \(\mathrm{m}\) . Calculate the magnitade of the total angular momentum of the woman-plus-disk system. (Assume that you can treat the woman as a point)

Short Answer

Expert verified
The total angular momentum of the system is approximately 5275.33 kg·m²/s.

Step by step solution

01

Convert Rotational Speed

First, convert the rotational speed from revolutions per second to radians per second. Since one revolution equals \( 2\pi \) radians, the angular speed \( \omega \) in radians per second is given by: \[\omega = 0.50\, \text{rev/s} \times 2\pi\, \text{rad/rev} = \pi\, \text{rad/s}.\]
02

Calculate Moment of Inertia for the Disk

The moment of inertia \( I \) for a solid disk rotating about an axis through its center is given by: \[I_{\text{disk}} = \frac{1}{2} M r^2,\] where \( M = 110\, \text{kg} \) is the mass of the disk and \( r = 4.0\, \text{m} \) is the radius. Substituting these values, we get:\[I_{\text{disk}} = \frac{1}{2} \times 110\, \text{kg} \times (4.0\, \text{m})^2 = 880\, \text{kg}\cdot\text{m}^2.\]
03

Calculate Moment of Inertia for the Woman

Since the woman is treated as a point mass at a distance from the axis, the moment of inertia \( I \) is calculated as: \[I_{\text{woman}} = m r^2,\]where \( m = 50\, \text{kg} \) is the mass of the woman and \( r = 4.0\, \text{m} \) is the radius. Thus:\[I_{\text{woman}} = 50\, \text{kg} \times (4.0\, \text{m})^2 = 800\, \text{kg}\cdot\text{m}^2.\]
04

Sum Moments of Inertia

Calculate the total moment of inertia for the system by adding the individual moments of inertia:\[I_{\text{total}} = I_{\text{disk}} + I_{\text{woman}} = 880\, \text{kg}\cdot\text{m}^2 + 800\, \text{kg}\cdot\text{m}^2 = 1680\, \text{kg}\cdot\text{m}^2.\]
05

Calculate Angular Momentum

The angular momentum \( L \) of the system is calculated using the formula:\[L = I_{\text{total}} \cdot \omega,\]where \( \omega = \pi\, \text{rad/s} \) is the angular speed. Thus, \[L = 1680\, \text{kg}\cdot\text{m}^2 \times \pi\, \text{rad/s} = 1680\pi\, \text{kg}\cdot\text{m}^2/s.\]
06

Determine the Magnitude

Finally, calculate the numeric value of the magnitude of the angular momentum:\[L \approx 1680 \times 3.1416\, \text{kg}\cdot\text{m}^2/s = 5275.33\, \text{kg}\cdot\text{m}^2/s.\]

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Key Concepts

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

Rotational Dynamics
Rotational Dynamics is a fascinating area of physics that studies the motion of bodies that rotate. Unlike linear dynamics, where objects move in a straight line, rotational dynamics considers how objects move along a circular path. Central to this study is the angular velocity, which is how fast an object rotates – similar to speed in linear motion but for turning.
In our solution, we converted revolutions per second into radians per second since calculus with angles use radians for precision. Ensuring accurate units are always used is crucial to solving rotational problems effectively.
  • Angular velocity measures how fast something spins and is often initially expressed in revolutions per second (rev/s).
  • To convert this into radians, a full revolution is equivalent to a turn around the circle, or 2Ï€ radians.
The focus on angular variables highlights the unique aspects of rotational dynamics – using rotational equivalents for every linear motion concept, such as velocity, acceleration, and momentum.
Moment of Inertia
Moment of Inertia is a key concept in rotational motion, akin to mass in linear dynamics though for rotational scenarios. It quantifies an object's resistance to changes in its rotation. The greater the moment of inertia, the harder it is to change the object's rotational speed.
This concept depends not just on the mass of the object but also on how the mass is distributed concerning the axis of rotation.
  • For a disk, the formula involves both mass and the square of its radius, divided by two: \[I_{\text{disk}} = \frac{1}{2} M r^2\]
  • For point masses, like the woman standing at the edge of the disk, the formula is different – it takes into account the entire radius because every part of the mass is effectively revolving around the axis: \[I_{\text{woman}} = m r^2\]
Understanding this concept helps in calculating the overall dynamics of a system, which in our problem is vital for finding the total angular momentum.
Rigid Body Motion
Rigid Body Motion in physics describes an ideal scenario where an object does not deform as it moves. That means all the particles in the body maintain the same relative position to each other during movement.
This assumption simplifies calculations in many problems, especially in rotational dynamics, because it means the objects rotate as one unified whole without any flexing or bending.
  • This concept allows us to treat complex systems as simpler models, like considering the woman as a point mass, which makes it easier to calculate moments of inertia.
  • In many calculations, rigid bodies will assume pure rotational motion about a fixed axis, as is seen in the disk scenario.
Rigid body assumptions are fundamental in mechanical engineering and physics because they allow for straightforward analytical modeling of real-world objects, granting better insights into how they behave under rotational forces.

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Most popular questions from this chapter

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3\() .\) When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_{x}\) and \(\alpha_{z}\) are approximately zero and \(v_{x}\) and \(\omega_{z}\) are approximately constant. Rolling without slipping means \(v_{x}=r \omega_{z}\) and \(a_{x}=r \alpha_{x} .\) If an object is set in motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established, A solid cylinder with mass \(M\) and radius \(R\) , rotating with angular speed \(\omega_{0}\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_{k}\) , (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_{x}\) of the center of mass and \(\alpha_{z}\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_{z}=\omega_{0}\) but \(v_{x}=0 .\) Rolling without slipping sets in when \(v_{x}=R \omega_{z}\) . Calculate the distance the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

A 5.00-kg ball is dropped from a height of 12.0 \(\mathrm{m}\) above one end of a uniform bar that pivots at its center. The bar has mass 8.00 \(\mathrm{kg}\) and is 4.00 \(\mathrm{m}\) in length. At the other end of the bar sits another \(5.00-\mathrm{kg}\) ball, unattached to the bar. The dropped ball sticks to the bar after the collision. How high will the other ball go after the collision?

Two metal disks, one with radius \(R_{1}=2.50 \mathrm{cm}\) and mass \(M_{1}=0.80 \mathrm{kg}\) and the other with radius \(R_{2}=5.00 \mathrm{cm}\) and mass \(M_{2}=1.60 \mathrm{kg}\) , are welded together and mounted on a frictionless axis through their common center, as in Problem \(9.89 .\) (a) A light string is wrapped around the edge of the smaller disk, and a 1.50 \(\mathrm{kg}\) block is suspended from the free end of the string. What is the magnitude of the downward acceleration of the block after it is released? (b) Repeat the calculation of part (a), this time with the string wrapped around the edge of the larger disk. In which case is the acceleration of the block greater? Does your answer make sense?

A horizontal plywood disk with mass 7.00 \(\mathrm{kg}\) and diameter 1.00 \(\mathrm{m}\) pivots on frictionless bearings about a vertical axis through its center. You attach a circular model-railroad track of negligible mass and average diameter 0.95 \(\mathrm{m}\) to the disk. A \(1.20-\mathrm{kg}\) , battery-driven model train rests on the tracks. To demonstrate conservation of angular momentum, you switch on the train's engine. The train moves counterclockwise, soon attaining a constant speed of 0.600 \(\mathrm{m} / \mathrm{s}\) relative to the tracks. Find the magnitude and direction of the angular velocity of the disk relative to the earth.

A cord is wrapped around the rim of a solid uniform wheel 0.250 \(\mathrm{m}\) in radius and of mass 9.20 \(\mathrm{kg}\) . A steady horizontal pull of 40.0 \(\mathrm{N}\) to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axe exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?

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