Loose Leaf For Design Of Machinery (mcgraw-hill Series In Mechanical Engineering)
Loose Leaf For Design Of Machinery (mcgraw-hill Series In Mechanical Engineering)
6th Edition
ISBN: 9781260431308
Author: Robert L. Norton
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 7, Problem 7.1P

A point at a 6.5-in radius is on a body that is in pure rotation with ω = 100 rad/sec and a constant a = 500 rad/sec 2 at point A . The rotation center is at the origin of a coordinate system. When the point is at position A , its position vector makes a 45 ° angle with the X axis. It takes 0.01 sec to reach point B . Draw this system to some convenient scale, calculate the θ and ω of position B , and:

  1. Write an expression for the particle's acceleration vector in position A using complex number notation, in both polar and cartesian forms.
  2. Write an expression for the particle's acceleration vector in position B using complex number notation, in both polar and cartesian forms.
  3. Write a vector equation for the acceleration difference between points B and A . Substitute the complex number notation for the vectors in this equation and solve for the acceleration difference numerically.
  4. Check the result of part c with a graphical method.

(a)

Expert Solution
Check Mark
To determine

An expression for the particle’s acceleration vector in position A.

Answer to Problem 7.1P

Acceleration in polar form is AA=32500jejπ465000ejπ4 .

Acceleration in cartesian form is AA=4366448260j .

Explanation of Solution

Calculation:

Compute the angular velocity at point B.

  ωB=ωA+αt

Here, ωB is the angular velocity at point B, ωA is the initial rotation speed, is the acceleration of rotation, and t is the time to reach point A.

Substitute 500rad/sec2 for α , 0.01 sec for t and 100 rad/sec for ωA .

  ωB=500×0.01+100=95rad/sec

Calculate the angular position at point B.

  θBθA=12αt2+ωAt

Here, θB is the angular position at point B and θA is the angular position at point A.

Substitute 500rad/sec2 for α , 0.01 sec for t and 100 rad/sec for ωA and 45o for θA .

  θBθA=12500×0.012+100×0.01=0.975rad180°πrad=55.863°θB=55.863°+45°=100.863°

Compute the magnitude of the normal component of acceleration at point A.

  aAn=RωA2

Here, aAn is the magnitude of the normal component of acceleration at point A. Substitute the value of R and ωA .

  aAn=6.51002=65000in/sec2

Compute the direction of the normal component of acceleration at point A.

  θAn=θA+180°=45°+180°=225°

Compute the magnitude of the tangential component of acceleration at point A.

  aAt=Rα

Here, aAt is the magnitude of the tangential component of acceleration at point A. Substitute the value of R and α .

  aAt=6.55002=3250in/sec2

Compute the direction of the tangential component of acceleration at point A.

  θAt=θA90°=45°90°=45°

Compute the magnitude of the normal component of acceleration at point B.

  aBn=RωB2

Here, aBn is the magnitude of the normal component of acceleration at point B. Substitute the value of R and ωB .

  aBn=6.5952=58663in/sec2

Compute the direction of the normal component of acceleration at point B.

  θBn=θB+180°=100.863°+180°=280.863°

Compute the magnitude of the tangential component of acceleration at point B.

  aBt=Rα

Here, aBt is the magnitude of the tangential component of acceleration at point B. Substitute the value of R and α .

  aAt=6.55002=3250in/sec2

Compute the direction of the tangential component of acceleration at point A.

  θAt=θA90°=100.863°90°=10.863°

Draw the position vector diagram.

  Loose Leaf For Design Of Machinery (mcgraw-hill Series In Mechanical Engineering), Chapter 7, Problem 7.1P , additional homework tip  1

Computer polar form by using the formula.

  RA=R·ejθA=6.5×ejπ4

Calculate the velocity at point A.

  VA=RjωAejθA=6.5×j×100×ejπ4=650jejπ4

Compute the acceleration vector in position A by the following formula.

  AA=RjαejθARjωA2ejθA=6.5×j×500×ejπ46.5×1002×ejπ4AA=32500jejπ465000ejπ4

Compute the acceleration vector of position A in cartesian form.

  AA=RαsinθA+jcosθARωA2cosθA+jsinθAAA=6.5×500sin45°+jcos45°6.5×1002cos45°+jsin45°AA=4366448260j=436642+482602=65081.27in/sec2

Therefore, the acceleration of point A is 65081.27in/sec2 .

(b)

Expert Solution
Check Mark
To determine

An expression for the particle’s acceleration vector in position B.

Answer to Problem 7.1P

Acceleration in polar form is AB=32500jej0.56035π58662.5ej0.56035π .

Acceleration in cartesian form is AB=1424856999j .

Explanation of Solution

Calculation:

Compute the angular velocity at point B.

  ωB=ωA+αt

Here, ωB is the angular velocity at point B, ωA is the initial rotation speed, is the acceleration of rotation, and t is the time to reach point A.

Substitute 500rad/sec2 for α , 0.01 sec for t and 100 rad/sec for ωA .

  ωB=500×0.01+100=95rad/sec

Calculate the angular position at point B.

  θBθA=12αt2+ωAt

Here, θB is the angular position at point B and θA is the angular position at point A.

Substitute 500rad/sec2 for α , 0.01 sec for t and 100 rad/sec for ωA and 45o for θA .

  θBθA=12500×0.012+100×0.01=0.975rad180°πrad=55.863°θB=55.863°+45°=100.863°

Compute the magnitude of the normal component of acceleration at point A.

  aAn=RωA2

Here, aAn is the magnitude of the normal component of acceleration at point A. Substitute the value of R and ωA .

  aAn=6.51002=65000in/sec2

Compute the direction of the normal component of acceleration at point A.

  θAn=θA+180°=45°+180°=225°

Compute the magnitude of the tangential component of acceleration at point A.

  aAt=Rα

Here, aAt is the magnitude of the tangential component of acceleration at point A. Substitute the value of R and α .

  aAt=6.55002=3250in/sec2

Compute the direction of the tangential component of acceleration at point A.

  θAt=θA90°=45°90°=45°

Compute the magnitude of the normal component of acceleration at point B.

  aBn=RωB2

Here, aBn is the magnitude of the normal component of acceleration at point B. Substitute the value of R and ωB .

  aBn=6.5952=58663in/sec2

Compute the direction of the normal component of acceleration at point B.

  θBn=θB+180°=100.863°+180°=280.863°

Compute the magnitude of the tangential component of acceleration at point B.

  aBt=Rα

Here, aBt is the magnitude of the tangential component of acceleration at point B. Substitute the value of R and α .

  aAt=6.55002=3250in/sec2

Compute the direction of the tangential component of acceleration at point A.

  θAt=θA90°=100.863°90°=10.863°

Draw the position vector diagram.

  Loose Leaf For Design Of Machinery (mcgraw-hill Series In Mechanical Engineering), Chapter 7, Problem 7.1P , additional homework tip  2

Compute the position vector of point B.

  RB=RejθB=6.5ej100.863×π180°

Compute the polar form.

  VB=6.5×j×95×ej100.863×π180°=617.5×j×ej100.863×π180°

Compute the acceleration vector in position B by the following formula.

  AB=RjαejθBRjωB2ejθB=6.5×j×500×ej100.863×π1806.5×952×ej100.863×π180AB=32500jej0.56035π58662.5ej0.56035π

Compute the acceleration vector of position B in cartesian form.

  AB=RαsinθB+jcosθBRωA2cosθB+jsinθBAB=6.5×500sin100.863°+jcos100.863°6.5×952cos100.863°+jsin100.863°AB=1424856999j=142482+569992=58752.79in/sec2

(c)

Expert Solution
Check Mark
To determine

A vector equation for the acceleration difference between points B and A.

Explanation of Solution

Calculation

Calculate the difference in acceleration between the given points.

  ABA=ABAA=1424856999j4366448260j=579128739j

Therefore, the position difference is 579128739j .

(d)

Expert Solution
Check Mark
To determine

The answer from the graphical method.

Explanation of Solution

Calculation Solve the equation ABt+ABn=AAt+AAn+ABA , use velocity scale of 10000in/sec2 .

Construct the acceleration polygon following the below steps.

  1. Locate point OA.
  2. Draw 0.325 in line from the point OA. This represents the velocity of the point AAt in a 45-degree direction.
  3. From the tip of the above line, draw a line of 6.508 in. This represents the acceleration of the point AAn in the indicated direction.
  4. From the point OA, draw a 0.325 line which represents the velocity of the point ABt in 100.863o direction.
  5. From the tip of the point ABt draw a perpendicular line of 5.875 which represents the acceleration of point ABn in the indicated direction.
  6. From the tip of point AAn draw a perpendicular line of 5.875 which represents the acceleration of point ABA in the indicated direction.

  Loose Leaf For Design Of Machinery (mcgraw-hill Series In Mechanical Engineering), Chapter 7, Problem 7.1P , additional homework tip  3

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Please solve this problem as soon as possible My ID# 016948724
The gears shown in the figure have a diametral pitch of 2 teeth per inch and a 20° pressure angle. The pinion rotates at 1800 rev/min clockwise and transmits 200 hp through the idler pair to gear 5 on shaft c. What forces do gears 3 and 4 transmit to the idler shaft? TS I y 18T 32T This a 12 x 18T C 48T 5
Question 1. Draw 3 teeth for the following pinion and gear respectively. The teeth should be drawn near the pressure line so that the teeth from the pinion should mesh those of the gear. Drawing scale (1:1). Either a precise hand drawing or CAD drawing is acceptable. Draw all the trajectories of the involute lines and the circles. Specification: 18tooth pinion and 30tooth gear. Diameter pitch=P=6 teeth /inch. Pressure angle:20°, 1/P for addendum (a) and 1.25/P for dedendum (b). For fillet, c=b-a.

Chapter 7 Solutions

Loose Leaf For Design Of Machinery (mcgraw-hill Series In Mechanical Engineering)

Knowledge Booster
Background pattern image
Mechanical Engineering
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.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
International Edition---engineering Mechanics: St...
Mechanical Engineering
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:CENGAGE L
Text book image
Precision Machining Technology (MindTap Course Li...
Mechanical Engineering
ISBN:9781285444543
Author:Peter J. Hoffman, Eric S. Hopewell, Brian Janes
Publisher:Cengage Learning
Text book image
Automotive Technology: A Systems Approach (MindTa...
Mechanical Engineering
ISBN:9781133612315
Author:Jack Erjavec, Rob Thompson
Publisher:Cengage Learning
Text book image
Principles of Heat Transfer (Activate Learning wi...
Mechanical Engineering
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Cengage Learning
Text book image
Electrical Transformers and Rotating Machines
Mechanical Engineering
ISBN:9781305494817
Author:Stephen L. Herman
Publisher:Cengage Learning
Text book image
Refrigeration and Air Conditioning Technology (Mi...
Mechanical Engineering
ISBN:9781305578296
Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:Cengage Learning
Dynamics - Lesson 1: Introduction and Constant Acceleration Equations; Author: Jeff Hanson;https://www.youtube.com/watch?v=7aMiZ3b0Ieg;License: Standard YouTube License, CC-BY