Given that the moment of inertia of a rod of length L and mass M about an axis passing through its center of mass and perpendicular to the length of the rod is I = 12 MI² and that the moment of inertia of a disk of mass M and radius R about an axis passing through its center of mass is I = MR². a) Calculate the moment of inertia of a thin rod of mass m axis through the rod at d₁ = 0.300L, as shown below. 1 = 2.60 kg and length L = 1.60 m about an d₁ Id₁ b) We now add a disk of mass md d₁ =0.300L, as shown below. d₁ dz = 1.30 kg and radius R = 0.160 m to the rod with its center at dz b) calculate the new moment of inertia about an axis through the rod at d₁ = 0.300L. 12d1 = b) We now add a second disk of mass md at the far end of the rod, as shown below. d₁ = 1.30 kg and radius R = 0.160 m to the rod with its center dz c) calculate the new moment of inertia about an axis through the rod at d₁ = 0.300L. 13d1 =
Given that the moment of inertia of a rod of length L and mass M about an axis passing through its center of mass and perpendicular to the length of the rod is I = 12 MI² and that the moment of inertia of a disk of mass M and radius R about an axis passing through its center of mass is I = MR². a) Calculate the moment of inertia of a thin rod of mass m axis through the rod at d₁ = 0.300L, as shown below. 1 = 2.60 kg and length L = 1.60 m about an d₁ Id₁ b) We now add a disk of mass md d₁ =0.300L, as shown below. d₁ dz = 1.30 kg and radius R = 0.160 m to the rod with its center at dz b) calculate the new moment of inertia about an axis through the rod at d₁ = 0.300L. 12d1 = b) We now add a second disk of mass md at the far end of the rod, as shown below. d₁ = 1.30 kg and radius R = 0.160 m to the rod with its center dz c) calculate the new moment of inertia about an axis through the rod at d₁ = 0.300L. 13d1 =
University Physics Volume 1
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Author:William Moebs, Samuel J. Ling, Jeff Sanny
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Chapter11: Angular Momentum
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Problem 89AP: A solid cylinder of mass 2.0 kg and radius 20 cm is rotating counterclockwise around a vertical axis...
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Transcribed Image Text:Given that the moment of inertia of a rod of length L and mass M about an axis passing through its center
of mass and perpendicular to the length of the rod is I = 12 MI² and that the moment of inertia of a
disk of mass M and radius R about an axis passing through its center of mass is I = MR².
a) Calculate the moment of inertia of a thin rod of mass m
axis through the rod at d₁ = 0.300L, as shown below.
1
=
2.60 kg and length L
= 1.60 m about an
d₁
Id₁
b) We now add a disk of mass md
d₁ =0.300L, as shown below.
d₁
dz
=
1.30 kg and radius R
=
0.160 m to the rod with its center at
dz
b) calculate the new moment of inertia about an axis through the rod at d₁ = 0.300L.
12d1
=
b) We now add a second disk of mass md
at the far end of the rod, as shown below.
d₁
=
1.30 kg and radius R
= 0.160 m to the rod with its center
dz
c) calculate the new moment of inertia about an axis through the rod at d₁ = 0.300L.
13d1
=

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