Determine the Velocity and Acceleration of the Rotating Rod

Click For Summary
SUMMARY

The discussion focuses on determining the instantaneous velocity and acceleration of point A on a rotating rod, given angular velocity (ω = 4.6 rad/s) and angular acceleration (α = 4.4 rad/s²). The correct formulas used are v = ω × r and a = α × r + ω × (ω × r). The user initially miscalculated the radius, using 0.23m instead of the correct diameter of 0.46m, leading to incorrect results. The final correct answers for the velocity and acceleration of point A are V_A = 1.24i + 2.869j and a_A = -12.01i + 8.449j, respectively.

PREREQUISITES
  • Understanding of angular motion and kinematics
  • Familiarity with vector operations in physics
  • Knowledge of the cross product in three-dimensional space
  • Proficiency in using the right-hand rule for rotational dynamics
NEXT STEPS
  • Review the principles of angular velocity and angular acceleration
  • Study vector cross product calculations in physics
  • Learn about the dynamics of rotating bodies and their kinematic equations
  • Explore applications of rotational motion in engineering contexts
USEFUL FOR

Students studying physics, particularly those focusing on rotational dynamics, as well as educators and professionals seeking to reinforce their understanding of angular motion concepts.

Northbysouth
Messages
241
Reaction score
2

Homework Statement


The body is formed of slender rod and rotates about a fixed axis through point O with the indicated angular properties. If ω = 4.6 rad/s and α = 4.4 rad/s2, determine the instantaneous velocity and acceleration of point A.

I have attached an image of the question

Homework Equations



v = wXr

a = αXr + wX(wXr)

The Attempt at a Solution



r = <0.5sin(24)i + 0.23cos(24)i +0.5cos(24)j - 0.23sin(24)j>

r = <0.4135i + 0.3632j + 0k>

w = 4.6k

va = <0i + 0j + 4.6k> X <0.4135i + 0.3632j + 0k>

VA = 1.6707i + 1.9021j

But it says this is wrong. What am I missing?

I know for the second part that I need to use:

aA = αXr + wX(wXr) but I can't seem to get wXr right.

Any advice would be greatly appreciated.
 

Attachments

  • dyn 5.003.png
    dyn 5.003.png
    17.1 KB · Views: 1,492
Physics news on Phys.org
well I see that you are using 0.23m as the distance from the straight part of the rod to the point A. However, I think 0.23m is the radius of the cirular part of the rod, so then the distance you should be using would be the diameter 0.46m.
 
You are correct. Thank you. My final correct answers are:

V_A = 1.24i + 2.869j

a_A = -12.01i + 8.449j
 

Similar threads

Replies
10
Views
3K
  • · Replies 7 ·
Replies
7
Views
5K
Replies
7
Views
8K
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 19 ·
Replies
19
Views
4K
Replies
15
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
5K