(a) Explanation Given information: The total mass ( M ) of the aluminum sheet is 1.6 kg. The width ( b ) of aluminum sheet is 120 mm. The breadth ( a ) of aluminum sheet is 160 mm. The angular velocity ( ω x ) of the assembly along x axis is 20 rad/s . The angular velocity ( ω y ) of the assembly along y axis is 0 . The angular velocity ( ω z ) of the assembly along y axis is 0 . Calculation: Find the mass ( m ) of the each aluminum sheet using the equation: m = M 2 Substitute 1.6 kg for M . m = 1.6 2 = 0.8 kg Find the moment of inertia ( I x ) of the assembly about x -axis using the equation: I x = 2 ( 1 3 m b 2 ) Substitute 0.8 kg for m and 120 mm for b . I x = 2 ( 1 3 × 0.8 ) ( 120 mm × 1 m 1 , 000 mm ) 2 = 7.68 × 10 − 3 kg ⋅ m 2 Find the product moment of inertia ( I x y ) of the assembly about xy -axis using the equation: I x y = m ( − a 2 ) ( b 2 ) Substitute 0.8 kg for m , 160 mm for a , and 120 mm for b . I x y = ( 0.8 ) ( − 160 mm × 1 m 1 , 000 mm 2 ) ( 120 mm × 1 m 1 , 000 mm 2 ) = − 3.84 × 10 − 3 kg ⋅ m 2 Find the product moment of inertia ( I x z ) of the assembly about xy -axis using the equation: I x z = m ( a 2 ) ( b 2 ) Substitute 0.8 kg for m , 160 mm for a , and 120 mm for b . I x y = ( 0.8 ) ( 160 mm × 1 m 1 , 000 mm 2 ) ( 120 mm × 1 m 1 , 000 mm 2 ) = 3.84 × 10 − 3 kg ⋅ m 2 Find the x -component of the angular momentum ( H G ) x of the assembly about its mass center G . ( H G ) x = I x ω x − I x y ω y − I x z ω z Substitute 7.68 × 10 − 3 kg ⋅ m 2 for I x , 20 rad/s for ω x , 0 for ω y , and 0 for ω z . ( H G ) x = ( 7.68 × 10 − 3 ) ( 20 ) − I x y ( 0 ) − I x z ( 0 ) = 0.1536 − 0 − 0 = 0.1536 kg ⋅ m 2 /s Find the y -component of the angular momentum ( H G ) y of the assembly about its mass center G . ( H G ) y = − I x y ω x + I y ω y − I y z ω z Substitute − 3

12th Edition

ISBN: 9781260265453

Author: BEER

Publisher: MCG

See similar textbooks

Concept explainers

Conservation of Energy

In physics, the rule of the conservation of energy states that energy doesn’t disappear from the atmosphere; instead, it changes its form from one type to another. Energy is one of the most important physical quantities in nature, and it cannot be create…

Videos

Question

Chapter 18.1, Problem 18.20P

(a)

To determine

The angular momentum HG of the assembly about point G.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 18.1, Problem 18.18P

Chapter 18.1, Problem 18.21P

Students have asked these similar questions

2: A rectangular aluminum block is loaded uniformly in three directions. The loadings are as follows:a 50 kN total resulting compressive load in the x-direction, a 200 kPa uniformly distributed tensile load in they-direction, and a 0.03 MN total resulting tensile load in the z-direction. The block has the following dimensions:L = 1 m, b = 20 cm, h = 350 mm. Use E = 70 GPa and ν = 0.25.Determine the strain in the x and y axes respectively. For the strain in the y-direction to be equal to 0, how much uniformly distributed load inthe surface of y-direction should be added? (+ for tensile, - for compressive) Answers: 5 -1.122 x10-5 / 3 decimals 6 5.102 x10-6 / 3 decimals 7 -0.357 MPa / 3 decimals

A spherical balloon with a diameter of 9 m is filled with water vapour at 200˚C and 200 kPa. Determine the mass of water in the balloon. The R value for water is 0.4615 kJ/kg∙K.

Correct answers are written below. Detailed and correct solution only with fbd. I will upvote. 1: A 3 m alloy shaft fixed at one end has a torsional shearing stress capacity of 55 MPa. Due to improper fabrication, its cross-sectionalarea has become irregularly shaped. Its effective polar moment of inertia has become 2 x10-7 m4, and the maximum torque stress acts at 7.5 cm fromthe center of the shaft.[1]: If the shaft is to be replaced by a properly manufactured solid circular shaft that has a maximumshearing stress capacity of 70 MN/m2, what is the minimum diameter required so it can withstand the sameload? [2]: Calculate the thickness of a hollow circular shaft with the same outside diameter calculated initem [1] that can carry the same load. Limit the maximum shearing stress of the hollow circular shaft to0.09 GPa.Determine the angle of twist on the free end of the shaft. Use G = 150 x103 GPa. [3]: Use the solidcircular shaft from [1] and use the hollow circular shaft from [2].…

Chapter 18 Solutions

VECTOR MECH...,STAT.+DYN.(LL)-W/ACCESS

Ch. 18.1 - A thin, homogeneous disk of mass m and radius r...Ch. 18.1 - Prob. 18.2P Ch. 18.1 - Prob. 18.3P Ch. 18.1 - A homogeneous disk of weight W = 6 lb rotates at...Ch. 18.1 - A homogeneous disk of mass m = 8 kg rotates at the...Ch. 18.1 - A solid rectangular parallelepiped of mass m has a...Ch. 18.1 - Prob. 18.8P Ch. 18.1 - Determine the angular momentum HD of the disk of...Ch. 18.1 - Prob. 18.10P Ch. 18.1 - Determine the angular momentum HO of the disk of...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The angular momentum H G of the assembly about point G .

Solution Summary: The author calculates the mass (m) of each aluminum sheet using the equation: m=M2.

Concept explainers

Videos

The angular momentum HG of the assembly about point G.

Want to see the full answer?

Chapter 18 Solutions