Explanation Given information: The mass ( m A B ) of the rod AB is 800 g. The mass ( m 0 ) of the disk is 1.2 kg. The spring constant ( k ) is 12 N/m . The length ( l ) of the rod is 600 mm. The radius ( r ) of the disk is 250 mm. Assume the acceleration due to gravity ( g ) as 9.81 m/s 2 . Calculation: Show the position 1 and position 2 of the rod AB with disk as in Figure (1). For position 1: Calculate the mass moment of inertia of disk ( I G ) d i s k : ( I G ) d i s k = 1 2 m 0 r 2 Substitute 1.2 kg for m 0 and 250 mm for r . ( I G ) d i s k = 1 2 × 1.2 × ( 250 mm × 1 m 1000 mm ) 2 = 0.0375 kg ⋅ m 2 Calculate the mass moment of inertia of rod ( I G ) A B : ( I G ) A B = 1 12 m A B l 2 Substitute 800 g for m A B and 600 mm for l . ( I G ) A B = 1 12 × ( 800 g × 1 kg 1000 g ) × ( 600 mm × 1 m 1000 mm ) 2 = 0.024 kg ⋅ m 2 Calculate maximum velocity ( v m ) A B of the rod AB : ( v m ) A B = ( l 2 − r ) θ m Substitute 600 mm for l and 250 mm for r . ( v m ) A B = ( ( 600 mm × 1 m 1000 mm ) 2 − ( 250 mm × 1 m 1000 mm ) ) θ m = 0.05 θ m Calculate maximum velocity ( v m ) 0 of the disk: ( v m ) 0 = r θ m Substitute 250 mm for r . ( v m ) 0 = ( 250 mm × 1 m 1000 mm ) θ m = 0.25 θ m Express the term ( θ ˙ m ) : θ ˙ m = ω n θ m Write the expression for the kinetic energy ( T 1 ) : T 1 = 1 2 ( I G ) A B θ ˙ m 2 + 1 2 m A B ( v m ) A B 2 + 1 2 ( I G ) disk θ ˙ m 2 + 1 2 m 0 ( v m ) 0 2 Substitute 800 g for m A B , 1.2 kg for m 0 , 0.0375 kg ⋅ m 2 for ( I G ) d i s k , 0.024 kg ⋅ m 2 for ( I G ) A B , 0.05 θ m for ( v m ) A B , and 0.25 θ m for ( v m ) 0 . T 1 = { 1 2 × 0.0375 × θ ˙ m 2 + 1 2 × ( 800 g × 1 kg 1000 g ) × ( 0.05 θ ˙ m ) 2 + 1 2 ( 0.0375 ) θ ˙ m 2 + 1 2 × 1.2 × ( 0.25 θ m ) 2 } = 1 2 [ 0.024 θ ˙ m 2 + 0

11th Edition

ISBN: 9781259633133

Author: BEER

Publisher: MCG

See similar textbooks

Concept explainers

Forced undamped vibrations

In mechanical engineering, vibration is expressed as the term that occurs due to an object's backward and forward movement about a point of equilibrium. Generally, there are three types of vibrations which are,

Videos

Question

Chapter 19.3, Problem 19.83P

To determine

The period (τn) of small oscillation of the system.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 19.3, Problem 19.82P

Chapter 19.3, Problem 19.84P

Students have asked these similar questions

my ID# is 016948724 please solve this problem step by step

MY ID#016948724 please solve the problem step by spet

1 8 4 For the table with 4×4 rows and columns as shown Add numbers so that the sum of any row or column equals .30 Use only these numbers: .1.2.3.4.5.6.10.11.12.12.13.14.14

Chapter 19 Solutions

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

Ch. 19.1 - A particle moves in simple harmonic motion....Ch. 19.1 - Prob. 19.2P Ch. 19.1 - Prob. 19.3P Ch. 19.1 - Prob. 19.4P Ch. 19.1 - Prob. 19.5P Ch. 19.1 - Prob. 19.6P Ch. 19.1 - Prob. 19.7P Ch. 19.1 - Prob. 19.8P Ch. 19.1 - Prob. 19.9P Ch. 19.1 - Prob. 19.10P

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 period ( τ n ) of small oscillation of the system.

Solution Summary: The author shows the position 1 and position 2 of the rod AB with disk as in Figure (1).

Concept explainers

Videos

The period (τn) of small oscillation of the system.

Want to see the full answer?

Chapter 19 Solutions