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A rod rotates in the vertical plane around a horizontal axle. A wheel is free to rotate on the rod, as shown in Figure 11.74. A vertical stripe is painted on the wheel remains vertical. Is the translational angular momentum of the wheel relative to location A zero or non-zero? If non-zero, what is its direction? Is the rotational angular momentum of the wheel zero or non-zero? If non-zero, what is its direction? Consider a similar system, but with the wheel welded to the rod (not free to turn). As the rod rotates clock wise, does the stripe on the wheel remain vertical? Is the translational angular momentum of the wheel relatives to location A zero or non-zero? If non-zero, what is its direction? Is the rotational angular momentum of the wheel zero or non-zero? If non-zero, what is its direction?

Short Answer

Expert verified

The direction of the rotational angular momentum and the direction of translational angular momentum is into the page, when the motion of the wheel is in clockwise.

Step by step solution

01

Definition of Angular Momentum.

The rotating inertia of an object or system of objects in motion about an axis that may or may not pass through the object or system is described by angular momentum.

The rotating analogue of linear momentum is angular momentum (also known as moment of momentum or rotational momentum). A closed system's total angular momentum remains constant.

To find the direction of a positive moving charge's magnetic force, point your right thumb in the direction of the velocity (v), your index finger in the direction of the magnetic field (B), and your middle finger in the direction of the resulting magnetic force.

02

Figure shows a rod rotating in the vertical plane.

Below figures shows a rod rotating in the vertical plane with a wheel attached.

03

About the concept of term Translational angular momentum.

The term translation angular momentum is associated with the motion of the center of mass of the system. The magnitude and direction of this differ for different choices of the locations. The expression for a translational angular momentum of a wheel with respect to given locationis,

Ltrans=rCM×Ptot

Here,Ltrans,Ais the translation angular momentum,rCMis the radius vector of the center of mass, andPtotis total translational momentum of the center of mass.

For a given system, the position vectors of center of mass of a wheelrCMwith respect to location and total momentum of the wheelPtotboth possesses non-zero values. Moreover, the directions of these vectors are perpendicular to each other. Hence, the contribution of cross product of the translational angular momentum expressing yields non-zero value.

The right-hand rule can be used to determine the direction of this translational angular momentum. Curl the figures of your right hand in the direction of the plane's rotational motion, and extend your thumb to indicate the unit vector direction of the translational angular momentum, according to this rule. The unit vector (your right thumb) is pointing into the page if you rotate clockwise.

04

About the concept of term Rotational angular Momentum.

The term rotational angular momentum is associated with rotation around the center of mass of the system. In general, rotational angular momentum doesn’t need a subscript because it is calculated relative to the center of mass, and this calculation is unaffected by our choice of the pointA ,and by motion of the center of mass.

For a given system, the position vectors of center of mass of a wheel,rCMwith respect to location center of mass is zero,rA=0(here reference point itself is the center of mass point, note not a pointA ).This yields, zero rotational angular momentum of the wheel. Here, the direction of the rotational angular momentum can’t be assessing. So, it has no direction.

05

When the wheel welded to the rod, the direction of the translational angular momentum is-

The rod was riveted to the wheel. A vertical stripe runs across the wheel. The wheel's stripe turns clockwise as the rod revolves clockwise. As a result, the wheel's stripe does not remain vertical.

The translational angular momentum of the wheel relative to location Ais non-zero quantity, because the center of mass of the wheel changes. Corresponding translational angular momentum of the wheel has some value.

Curl the fingers of our right hand in the direction of rotation in the plane, and the direction of translational angular momentum is in the extended thumb, using the right-hand rule. Because the wheel rotates clockwise, the translation angular momentum is directed into the page.

06

When the wheel welded to the rod, the direction of the rotational angular momentum is-

The wheel's particles move in a circular manner. The rotational angular momentum of the wheel is a non-zero quantity since it is the sum of the angular momentums of the particles.

Curl the fingers of our right hand in the direction of rotational motion in the plane, and the direction of rotational angular momentum is in the extended thumb, using the right hand rule. Because the wheel rotates clockwise, the rotating angular momentum is directed into the page.

When the wheel is turning clockwise, the direction of rotational angular momentum and translational angular momentum is both into the page.

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Most popular questions from this chapter

The Bohr model currently predicts the main energy levels not only for atomic hydrogen but also for other “one-electron” atoms where all but one of the atomic electrons has been removed, such as in He+ (one electron removed) or (two electrons removed) Li++. (a) Predict the energy levels in for a system consisting of a nucleus containing protons and just one electron. You need no recapitulate the entire derivation for the Bohr model, but do explain the changes you have to make to take into account the factor . (b) The negative muon(μ-) behaves like a heavy electron, with the same charge as the electron but with a mass 207 times as large as the electron mass. As a moving μ- comes to rest in matter, it tends to knock electrons out of atoms and settle down onto a nucleus to form a “one-muon” atom. For a system consisting of a lead nucleus ( Pb208has 82 protons and 126 neutrons) and just one negative muon, predict the energy in of a photon emitted in a transition from the first excited state to the ground state. The high-energy photons emitted by transitions between energy levels in such “muonic atoms” are easily observed in experiments with muons. (c) Calculate the radius of the smallest Bohr orbit for a μ- bound to a lead nucleus ( Pb208has 82 protons and 126 neutrons). Compare with the approximate radius of the lead nucleus (remember that the radius of a proton or neutron is about 1×10-15m, and the nucleons are packed closely together in the nucleus).

Comments: This analysis in terms of the simple Bohr model hints at the results of a full quantum-mechanical analysis, which shows that in the ground state of the lead-muon system there is a rather high probability for finding the muon inside the lead nucleus. Nothing in quantum mechanics forbids this penetration, especially since the muon does not participate in the strong intersection. Electrons in an atom can also be found inside the nucleus, but the probability is very low, because on average the electrons are very far from the nucleus, unlike the muon.

The eventual fate of the μ- in a muonic atom is that it either decays into an electron, neutrino, and antineutrino, or it reacts through the weak interaction with a proton in the nucleus to produce a neutron and a neutrino. This “muon capture” reaction is more likely if the probability is high for the muon to be found inside the nucleus, as is the case with heavy nuclei such as lead.

A solid wood top spins at high speed on the floor, with a spin direction shown in figure 11.112

a. Using appropriately labeled diagrams, explain the direction of motion of the top (you do not need to explain the magnitude).

b. How would the motion change if the top had a higher spin rate? Explain briefly.

c. If the top were made of solid steel instead of wood, explain how this would affect the motion (for the same spin rate).

A common amusement park ride is a Ferris wheel (see figure, which is not drawn to scale). Riders sit in chairs that are on pivots so they remain level as the wheel turns at a constant rate. A particular Ferris wheel has a radius of 24 meters, and it make one complete revolution around its axle (at location A) in 20sIn all of the following questions, consider location A(at the center of the axle) as the location around which we will calculate the angular momentum. At the instant shown in the diagram, a child of mass40kg, sitting at location F, is traveling with velocity <7.5,0,0>m/s.

(a.) What is the linear momentum of the child? (b) In the definition L=r×p,what is the vector r? (c) what is r? (d) what is the magnitude of the angular momentum of the child about location A? (e) What is the plane defined by randp(that is, the plane containing both of these vectors)? (f) Use the right-hand rule to determine thecomponent of the angular momentum of the child about locationA. (g) You used the right-hand rule to determine the zcomponent of the angular momentum, but as a check, calculate in terms of position and momentum: What isypx? Therefore, what iszthe component of the angular momentum of the child about locationA? (h) The Ferris wheel keeps turning, and at a later time, the same child is at locationEwith coordinates<16.971,-16.971,0>m relative to location A, moving with velocity<5.303,5.303,0>m/s.Now what is the magnitude of the angular momentum of the child about location A?

An ice skater whirls with her arms and one leg stuck out as shown on the left in Figure 11.93, making one complete turn inThen she quickly moves her arms up above her head and pulls her leg in as shown at the right in Figure 11.93.

(a) Estimate how long it now takes for her to make one complete turn. Explain your calculations, and state clearly what approximations and estimates you make. (b) Estimate the minimum amount of chemical energy she must expended to change her configuration.

A thin metal rod of mass1.3 kg and length0.4 m is at rest in outer space, near a space station (Figure 11.99). A tiny meteorite with mass 0.06 kg travelling at a high speed of strikes the rod a distance 0.2 m from the center and bounces off with speed 60 m/s as shown in the diagram. The magnitudes of initial and final angles to thex-axis of the small mass’s velocity are θi=26° and θf=82°.(a). Afterward, what is the velocity of the center of the rod? (b) Afterward, what is the angular velocity ω of the rod? (c) What is the increase in internal energy of the objects?

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