EXAMPLE 7.2.1 A long, thin rod of length I and mass m hangs from a pivot point about which it is free to swing in a vertical plane like a simple pendulum. Calculate the total angular momen- tum of the rod as a function of its instantaneous angular velocity . Show that the theorem represented by Equation 7.2.14 is true by comparing the angular momentum obtained using that theorem to that obtained by direct calculation. Solution: The rod is shown in Figure 7.2.3a. First we calculate the angular momentum Lem of the center of mass of the rod about the pivot point. Because the velocity Vem of the center of mass is always perpendicular to the radius vector r denoting its location relative to the pivot point, the sine of the angle between those two vectors is unity. Thus, the mag- nitude of Lem is given by 1 Pom = m 1 =m m ² (1/200) = ² m² ²00 22 Figure 7.2.3b depicts the motion of the rod as seen from the perspective of its center of mass. The angular momentum dLe of two small mass elements, each of size dm symmetrically disposed about the center of mass of the rod, is given by dLe = 2rdp=2rv dm = 2r(rw)λ dr where λ is the mass per unit length of the rod. The total relative angular momentum is obtained by integrating this expression from r=0 to r= 1/2. l/2 Lra = 2Xw [!"² r²dr = 1/2 (21)1² w = (1/2 ml² ) w We can see in the preceding equation that the angular momentum of the rod about its center of mass is directly proportional to the angular velocity of the rod. The constant of proportionality ml²/12 is called the moment of inertia Iem of the rod about its center of mass. Moment of inertia plays a role in rotational motion similar to that of inertial mass in translational motion as we shall see in the next chapter. Figure 7.2.3 Rod of mass m and length ! free to swing in a vertical plane about a fixed pivot. I Vcm dm dm dm 8 (c)
EXAMPLE 7.2.1 A long, thin rod of length I and mass m hangs from a pivot point about which it is free to swing in a vertical plane like a simple pendulum. Calculate the total angular momen- tum of the rod as a function of its instantaneous angular velocity . Show that the theorem represented by Equation 7.2.14 is true by comparing the angular momentum obtained using that theorem to that obtained by direct calculation. Solution: The rod is shown in Figure 7.2.3a. First we calculate the angular momentum Lem of the center of mass of the rod about the pivot point. Because the velocity Vem of the center of mass is always perpendicular to the radius vector r denoting its location relative to the pivot point, the sine of the angle between those two vectors is unity. Thus, the mag- nitude of Lem is given by 1 Pom = m 1 =m m ² (1/200) = ² m² ²00 22 Figure 7.2.3b depicts the motion of the rod as seen from the perspective of its center of mass. The angular momentum dLe of two small mass elements, each of size dm symmetrically disposed about the center of mass of the rod, is given by dLe = 2rdp=2rv dm = 2r(rw)λ dr where λ is the mass per unit length of the rod. The total relative angular momentum is obtained by integrating this expression from r=0 to r= 1/2. l/2 Lra = 2Xw [!"² r²dr = 1/2 (21)1² w = (1/2 ml² ) w We can see in the preceding equation that the angular momentum of the rod about its center of mass is directly proportional to the angular velocity of the rod. The constant of proportionality ml²/12 is called the moment of inertia Iem of the rod about its center of mass. Moment of inertia plays a role in rotational motion similar to that of inertial mass in translational motion as we shall see in the next chapter. Figure 7.2.3 Rod of mass m and length ! free to swing in a vertical plane about a fixed pivot. I Vcm dm dm dm 8 (c)
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Transcribed Image Text:EXAMPLE 7.2.1
A long, thin rod of length I and mass m hangs from a pivot point about which it is free
to swing in a vertical plane like a simple pendulum. Calculate the total angular momen-
tum of the rod as a function of its instantaneous angular velocity . Show that the
theorem represented by Equation 7.2.14 is true by comparing the angular momentum
obtained using that theorem to that obtained by direct calculation.
Solution:
The rod is shown in Figure 7.2.3a. First we calculate the angular momentum Lem of the
center of mass of the rod about the pivot point. Because the velocity Vem of the center
of mass is always perpendicular to the radius vector r denoting its location relative to
the pivot point, the sine of the angle between those two vectors is unity. Thus, the mag-
nitude of Lem is given by
1
Pom
= m
1
=m
m ² (1/200) = ² m² ²00
22
Figure 7.2.3b depicts the motion of the rod as seen from the perspective of its center
of mass. The angular momentum dLe of two small mass elements, each of size dm
symmetrically disposed about the center of mass of the rod, is given by
dLe = 2rdp=2rv dm = 2r(rw)λ dr
where λ is the mass per unit length of the rod. The total relative angular momentum is
obtained by integrating this expression from r=0 to r= 1/2.
l/2
Lra = 2Xw [!"² r²dr = 1/2 (21)1² w = (1/2 ml² ) w
We can see in the preceding equation that the angular momentum of the rod about its
center of mass is directly proportional to the angular velocity of the rod. The constant
of proportionality ml²/12 is called the moment of inertia Iem of the rod about its center
of mass. Moment of inertia plays a role in rotational motion similar to that of inertial mass
in translational motion as we shall see in the next chapter.
Figure 7.2.3 Rod of mass m and length
! free to swing in a vertical plane about a
fixed pivot.
I
Vcm
dm
dm
dm
8
(c)
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