Engineering Mechanics: Dynamics (14th Edition)
Engineering Mechanics: Dynamics (14th Edition)
14th Edition
ISBN: 9780133915389
Author: Russell C. Hibbeler
Publisher: PEARSON
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Chapter 19.2, Problem 1PP

Determine the angular momentum of the 100-kg disk or rod about point G and about point O

Chapter 19.2, Problem 1PP, Determine the angular momentum of the 100-kg disk or rod about point G and about point O , example  1

Chapter 19.2, Problem 1PP, Determine the angular momentum of the 100-kg disk or rod about point G and about point O , example  2

Chapter 19.2, Problem 1PP, Determine the angular momentum of the 100-kg disk or rod about point G and about point O , example  3

Chapter 19.2, Problem 1PP, Determine the angular momentum of the 100-kg disk or rod about point G and about point O , example  4

Expert Solution & Answer
Check Mark
To determine

The angular momentum of the disk or rod about point G and point O .

Answer to Problem 1PP

  1. (a) The angular momentum of the disk or rod about point G and point O is 600kgm2/s(cw)and1800kgm2/s(cw) respectively.
  2. (b) The angular momentum of the disk or rod about point G and point O is 300kgm2/s(cw)and1200kgm2/s(cw) respectively.
  3. (c) The angular momentum of the disk or rod about point G and point O is 800kgm2/s(cw)and2400kgm2/s(cw) respectively.
  4. (d) The angular momentum of the disk or rod about point G and point O is 400kgm2/s(ccw)and700kgm2/s(ccw) respectively.

Explanation of Solution

Given:

The mass of the disk, m=100kg .

Conclusion:

(a)

Express the angular momentum of the disk about point G .

HG=[12mr2]ω (I)

Here, radius of disk is r and angular velocity is ω .

Substitute 100kg for m , 2m for r and 3rad/s(cw) for ω in Equation (I).

HG=[12(100)(2)2](3)=600kgm2/s(cw)

Hence, the angular momentum of the disk about point G is 600kgm2/s(cw) .

Express the angular momentum of the disk about point O .

HO=[12mr2+mr2]ω (II)

Substitute 100kg for m , 2m for r and 3rad/s(cw) for ω in Equation (II).

HO=[12(100)(2)2+100(2)2](3)=(200+400)3=1800kgm2/s(cw)

Hence, the angular momentum of the disk about point O is 1800kgm2/s(cw) .

(b)

Express the angular momentum of the rod about point G .

HG=[112ml2]ω (III)

Here, length of the rod is l .

Substitute 100kg for m , 3m for l and 4rad/s(cw) for ω in Equation (III).

HG=[112(100)(3)2](4)=300kgm2/s(cw)

Hence, the angular momentum of the rod about point G is 300kgm2/s(cw) .

Express the angular momentum of the rod about point O .

HO=[112ml2+m(l/2)2]ω (IV)

Substitute 100kg for m , 3m for l and 4rad/s(cw) for ω in Equation (IV).

HO=[112(100)(3)2+100(3/2)2](4)=(75+225)4=1200kgm2/s(cw)

Hence, the angular momentum of the rod about point O is 1200kgm2/s(cw) .

(c)

Express the angular momentum of the disk about point G .

HG=[12mr2]ω (V)

Substitute 100kg for m , 2m for r and 4rad/s(cw) for ω in Equation (V).

HG=[12(100)(2)2](4)=800kgm2/s(cw)

Hence, the angular momentum of the disk about point G is 800kgm2/s(cw) .

Express the angular momentum of the disk about point O .

HO=[12mr2+mr2]ω (VI)

Substitute 100kg for m , 2m for r and 4rad/s(cw) for ω in Equation (VI).

HO=[12(100)(2)2+100(2)2](4)=(200+400)4=2400kgm2/s(cw)

Hence, the angular momentum of the disk about point O is 2400kgm2/s(cw) .

(d)

Express the angular momentum of the rod about point G .

HG=[112ml2]ω (VII)

Substitute 100kg for m , 4m for l and 3rad/s(ccw) for ω in Equation (VII).

HG=[112(100)(4)2](3)=400kgm2/s(ccw)

Hence, the angular momentum of the rod about point G is 400kgm2/s(ccw) .

Express the angular momentum of the rod about point O .

HO=[112ml2+m(lOG)2]ω (VIII)

Here, length between OandG is lOG .

Substitute 100kg for m , 4m for l , 1m for lOG and 3rad/s(ccw) for ω in Equation (VIII).

HO=[112(100)(4)2+100(1)2](3)=(233.333)3=700kgm2/s(ccw)

Hence, the angular momentum of the rod about point O is 700kgm2/s(ccw) .

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Chapter 19 Solutions

Engineering Mechanics: Dynamics (14th Edition)

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