Explanation Given information: The radius ( r ) of quarter circle is 15 in . . The specific weight of γ aluminum is 170 lb / ft 3 . The uniform thickness ( t ) of aluminum plate is 0.05 in . Calculation: Divided the component into three geometric portions 1, 2, and 3. Find the mass of portion 1 in the x − y plane as shown below: m = ρ V = ρ [ 1 4 π r 2 ] t (1) Here, ρ is density of aluminum. Substitute 15 in . for r and 0.05 in . for t . m = ρ [ 1 4 π × ( 15 in . ( 1 ft 12 in . ) ) 2 × 0.05 in . ( 1 ft 12 in . ) ] = ρ [ 1 4 π × 1.5625 × 4.1666 × 10 − 3 ] = 5.1133 × 10 − 3 ρ Find the moment of inertia about x and z axis of portion 1 in the x − y plane as shown below: I x = I z = 1 4 m r 2 (2) Substitute 5.1133 × 10 − 3 ρ for m and 15 in . for r in Equation (2). I x = I z = 1 4 ( 5.1133 × 10 − 3 ρ ) ( 15 in . ( 1 ft 12 in . ) 2 ) = 1 4 ( 5.1133 × 10 − 3 ρ ) 1.5625 = 1.99738 × 10 − 3 ρ Find the moment of inertia about y axis of portion 1 in the x − y plane as shown below: I y = 1 4 m r 2 (3) Substitute 5.1133 × 10 − 3 ρ for m and 15 in . for r in Equation (3). I y = 1 4 ( 5.1133 × 10 − 3 ρ ) ( 15 in . ( 1 ft 12 in . ) 2 ) = 1 4 ( 5.1133 × 10 − 3 ρ ) 1.5625 = 1.99738 × 10 − 3 ρ Find the moment of inertia about x and z axis of portion 2 in the y − z plane as shown below: Here, mass is same for portions 2. I x = I z = 1 2 m r 2 (4) Substitute 5.1133 × 10 − 3 ρ for m and 15 in . for r in Equation (4). I x = I z = 1 2 ( 5.1133 × 10 − 3 ρ ) ( 15 in . ( 1 ft 12 in . ) 2 ) = 1 2 ( 5.1133 × 10 − 3 ρ ) 1.5625 = 3.9948 × 10 − 3 ρ Find the moment of inertia about y axis of portion 2 in the x − y plane as shown below: I y = 1 2 m r 2 (5) Substitute 5.1133 × 10 − 3 ρ for m and 15 in . for r in Equation (5). I y = 1 2 ( 5.1133 × 10 − 3 ρ ) ( 15 in . ( 1 ft 12 in . ) 2 ) = 1 2 ( 5.1133 × 10 − 3 ρ ) 1.5625 = 3.9948 × 10 − 3 ρ Find the mass of portion 3 in the x − y plane as shown below: m = ρ V = ρ [ 1 4 r 2 ] t (6) Here, ρ is density of aluminum. Substitute 15 in . for r and 0.05 in . for t in Equation (6). m = ρ [ 1 2 × ( 15 in . ( 1 ft 12 in . ) ) 2 × 0.05 in . ( 1 ft 12 in . ) ] = ρ [ 1 2 × 1.5625 × 4.1666 × 10 − 3 ] = 3.2552 × 10 − 3 ρ Find the mass moment of inertia ( I mass ) as shown below: ( I x ,mass ) = ρ t I area = ρ t ( 1 12 r 4 ) = ρ ( 1 2 r 2 t ) ( 1 6 r 2 ) = 1 6 m r 2

12th Edition

ISBN: 9781259977121

Author: BEER

Publisher: MCG

See similar textbooks

Concept explainers

Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provi…

Videos

Question

Chapter 9.5, Problem 9.139P

To determine

Find the mass moment of inertia with respect to x,y,z coordinate axes.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 9.5, Problem 9.138P

Chapter 9.5, Problem 9.140P

Students have asked these similar questions

20 in. PROBLEM 15.206 Rod AB is connected by ball-and-socket joints to collar A and to the 16-in.-diameter disk C. Knowing that disk C rotates counterclockwise at the constant rate ₁ =3 rad/s in the zx plane, determine the velocity of collar A for the position shown. 25 in. B 8 in. Answer: -30 in/s =

B Z 001 2.5 ft PROBLEM 15.236 The arm AB of length 16 ft is used to provide an elevated platform for construction workers. In the position shown, arm AB is being raised at the constant rate de/dt = 0.25 rad/s; simultaneously, the unit is being rotated about the Y axis at the constant rate ₁ =0.15 rad/s. Knowing that 20°, determine the velocity and acceleration of Point B. Answers: 1.371 +3.76)+1.88k ft/s a=1.22 -0.342)-0.410k ft/s² X

F1 3 5 4 P F2 F2 Ꮎ Ꮎ b P 3 4 5 F1 The electric pole is subject to the forces shown. Force F1 245 N and force F2 = 310 N with an angle = 20.2°. Determine the moment about point P of all forces. Take counterclockwise moments to be positive. = Values for dimensions on the figure are given in the following table. Note the figure may not be to scale. Variable Value a 2.50 m b 11.3 m C 13.0 m The moment about point P is 3,414 m. × N- If the moment about point P sums up to be zero. Determine the distance c while all other values remained the same. 1.26 m.

Chapter 9 Solutions

VECTOR MECHANICS FOR ENGINEERS: STATICS

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.

Find the mass moment of inertia with respect to x , y , z coordinate axes.

Solution Summary: The author calculates the mass moment of inertia with respect to x,y,z coordinate axes.

Concept explainers

Videos

Find the mass moment of inertia with respect to x,y,z coordinate axes.

Want to see the full answer?

Chapter 9 Solutions