Explanation Given information: The weight of the rod ( W ) is 3 lb. The angular velocity ( ω ) is ( 12 rad/s ) i . The angular acceleration ( α ) is − ( 96rad/s 2 ) i . The distance between DF is 9 inch. The distance between AF is 3 inch. Calculation: Show the free body diagram of the system as in Figure (1). The principal axes for the rod CDE are x ′ and y ′ . Calculate angle ( θ ) in triangle C D F using the relation: cos θ = D F C D Substitute 9 in. for DF and 12 in. for CD . cos θ = 9 in . 12 in . θ = cos − 1 ( 0.75 ) θ = 41.41 ° Calculate angular velocity in the x ′ axis ( ω x ′ ) using the relation: ω x ′ = ω cos θ Here, ω is the angular velocity of the system. Substitute 12 rad / s for ω and 41.41 ° for θ . ω x ′ = ( 12 rad / s ) cos ( 41.41 ° ) = 12 × 0.75 = 9 rad / s Calculate angular velocity in the y ′ axis ω y ′ using the relation: ω y ′ = ω sin θ Substitute 12 rad / s for ω and 41.41 ° for θ . ω y ′ = ( 12 rad / s ) sin ( 41.41 ° ) = 12 × 0.66 = 7.9373 rad / s The angular velocity in the z ′ direction is zero. The mass moment of inertia in x ′ direction is zero, because the torque around the principle axis only affects the acceleration. The value of I ¯ x ′ is zero. Write the expression for the mass moment of inertia in the y ′ direction ( I ¯ y ′ ) as follows: I ¯ y ′ = 1 12 m l 2 (1) Here, m is the mass and l is the length of the rod. Calculate the mass of the rod ( m ) using the formula: m = W g Here, g is the acceleration due to gravity. Substitute 3 lb for W and 32.2 ft / s 2 for g . m = 3 lb 32.2 ft / s 2 = 0.09317 lb ⋅ s 2 / ft Find the moment of inertia about y axis I y ′ : Substitute 0.09317 lb ⋅ s 2 / ft for m and 2 ft for l in equation (1). I ¯ y ′ = 1 12 ( 0.09317 lb ⋅ s 2 / ft ) ( 2 ft ) 2 = 0.007764 × 4 = 0.03106 lb ⋅ s 2 ⋅ ft The moment of inertia for I x ′ and I y ′ are same. Calculate the angular momentum at the point D ( H D ) using the formula: H D = I ¯ x ′ ω x ′ i ′ + I ¯ y ′ ω y ′ j ′ + I ¯ z ′ ω z ′ k ′ Substitute 0 for I ¯ x ′ , 9 rad / s for ω x ′ , 0.03106 lb ⋅ s 2 ⋅ ft for I ¯ y ′ , 7.9373 rad / s for ω y ′ , 0 for I ¯ z ′ , and 0 for ω z ′ . H D = ( 0 ) ( 9 rad / s ) i ′ + ( 0.03106 lb ⋅ s 2 ⋅ ft ) ( 7.9373 rad / s ) j ′ + ( 0 ) ( 0 ) k ′ = 0.03106 × 7.9373 = ( 0.24649 lb ⋅ s ⋅ ft ) j ′ Write the expression for the rate of change of angular momentum at point D ( H ˙ D ) as follows: H ˙ D = ( H ˙ D ) D x ′ y ′ z ′ + Ω × H D (2) Here, ( H ˙ D ) D x ′ y ′ z ′ is the rate of change angular momentum with respect to rotating frame D x ′ y ′ z ′ and Ω is the angular velocity of the frame D x ′ y ′ z ′ . Calculate the angular acceleration in the x ′ direction ( α x ′ ) using the relation: α x ′ = α cos θ Here, α is the angular acceleration. Substitute − 96 rad / s 2 for α and 41.41 for θ . α x ′ = ( − 96 rad / s 2 ) cos 41.41 ° = − 96 × 0

12th Edition

ISBN: 9781264095032

Author: BEER

Publisher: MCGRAW-HILL HIGHER EDUCATION

See similar textbooks

Concept explainers

Power Transmission Elements

It is defined as the energy transfer from one location to another location. The energy will start moving from the energy generation source to the energy applied position to obtain the work output. From this method, a large amount of electric energy moves…

Videos

Question

Chapter 18.2, Problem 18.61P

To determine

The rate of change (H˙D) of the angular momentum (HD) of the rod.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 18.2, Problem 18.60P

Chapter 18.2, Problem 18.62P

Students have asked these similar questions

Plot the displacement diagram for a cam with roller follower of diameter 10 mm. The required motion is as follows; 1- Rising 60 mm in 135° with uniform acceleration and retardation motion. 2- Dwell 90° 3- Falling 60 mm for 135° with Uniform acceleration-retardation motion. Then design the cam profile to give the above displacement diagram if the minimum circle diameter of the cam is 50 mm.

Q1/ A vertical, circular gate with water on one side as shown. Determine the total resultant force acting on the gate and the location of the center of pressure, use water specific weight 9.81 kN/m³ 1 m 4 m

I need handwritten solution with sketches for each

Chapter 18 Solutions

VECTOR MECHANIC

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 rate of change ( H ˙ D ) of the angular momentum ( H D ) of the rod.

Solution Summary: The author shows the free body diagram of the system as in Figure (1).

Concept explainers

Videos

The rate of change (H˙D) of the angular momentum (HD) of the rod.

Want to see the full answer?

Chapter 18 Solutions