[b] For Example 10.14 (Calculating the Angular Momentum of a Spinning Skater) which is also in Section 10.5, replace "0.800 rev/s" with "Z rev/s" and then solve part (a) of the example. Again, there is no need to solve part (b). Z=39
Angular Momentum
The momentum of an object is given by multiplying its mass and velocity. Momentum is a property of any object that moves with mass. The only difference between angular momentum and linear momentum is that angular momentum deals with moving or spinning objects. A moving particle's linear momentum can be thought of as a measure of its linear motion. The force is proportional to the rate of change of linear momentum. Angular momentum is always directly proportional to mass. In rotational motion, the concept of angular momentum is often used. Since it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics. To understand the concept of angular momentum first we need to understand a rigid body and its movement, a position vector that is used to specify the position of particles in space. A rigid body possesses motion it may be linear or rotational. Rotational motion plays important role in angular momentum.
Moment of a Force
The idea of moments is an important concept in physics. It arises from the fact that distance often plays an important part in the interaction of, or in determining the impact of forces on bodies. Moments are often described by their order [first, second, or higher order] based on the power to which the distance has to be raised to understand the phenomenon. Of particular note are the second-order moment of mass (Moment of Inertia) and moments of force.
![Calculating the Angular Momentum of a Spinning Skater
Suppose an ice skater, such as the one in Figure 10.23, is spinning at 0.800 rev/ s with her arms extended.
She has a moment of inertia of 2.34 kg · m2 with her arms extended and of 0.363 kg m?with her arms
close to her body. (These moments of inertia are based on reasonable assumptions about a 60.0-kg
skater.) (a) What is her angular velocity in revolutions per second after she pulls in her arms? (b) What is
her rotational kinetic energy before and after she does this?
Strategy
In the first part of the problem, we are looking for the skater's angular velocity w'after she has pulled in
her arms. To find this quantity, we use the conservation of angular momentum and note that the moments
of inertia and initial angular velocity are given. To find the initial and final kinetic energies, we use the
definition of rotational kinetic energy given by
KErot =
10.114
Solution for (a)
Because torque is negligible (as discussed above), the conservation of angular momentum given in
Iw = I'w' is applicable. Thus,
L = L'
10.115
or
Iw = I'w'
10.116
Solving for w'and substituting known values into the resulting equation gives
2.34 kg-m?
T"=(0,363 kem2) (0.800 rev/s)
0.363 kg-m²
5.16 rev/s.
w'
10.117
Solution for (b)
Rotational kinetic energy is given by
1
KErot =
Iw
10.118
The initial value is found by substituting known values into the equation and converting the angular
velocity to rad/s:
KErot = (0.5) (2.34 kg - m²) ((0.800 rev/s)(27 rad/rev))?
10.119
= 29.6 J.
=
The final rotational kinetic energy is
KI
10.120
rot
Substituting known values into this equation gives
KErot'
(0.5) (0.363 kg · m²) [(5.16 rev/s)(27 rad/rev)]²
10.121
191 J.
Discussion
In both parts, there is an impressive increase. First, the final angular velocity is large, although most world-
class skaters can achieve spin rates about this great. Second, the final kinetic energy is much greater than
the initial kinetic energy. The increase in rotational kinetic energy comes from work done by the skater in
pulling in her arms. This work is internal work that depletes some of the skater's food energy.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20154b36-3898-46e5-9631-3dc0bd928319%2F538276a0-5f7a-479f-878d-39228e04c187%2F38g36zl_processed.png&w=3840&q=75)
![[b] For Example 10.14 (Calculating the Angular Momentum of a Spinning Skater) which is also
in Section 10.5, replace "0.800 rev/s" with "Z rev/s" and then solve part (a) of the example.
Again, there is no need to solve part (b).
Z=39](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20154b36-3898-46e5-9631-3dc0bd928319%2F538276a0-5f7a-479f-878d-39228e04c187%2Fby7gn6k_processed.jpeg&w=3840&q=75)
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