Determine the Velocity and Acceleration of the Rotating Rod

AI Thread Summary
The discussion focuses on determining the instantaneous velocity and acceleration of point A on a rotating rod, given its angular velocity (ω = 4.6 rad/s) and angular acceleration (α = 4.4 rad/s²). The initial calculations for velocity and acceleration were incorrect due to using the wrong distance from the axis of rotation; the correct distance is the diameter (0.46m) rather than the radius (0.23m). After correcting this, the final answers for velocity and acceleration were found to be V_A = 1.24i + 2.869j and a_A = -12.01i + 8.449j, respectively. The discussion highlights the importance of accurately identifying the distance in rotational motion problems. This ensures correct application of the relevant equations for velocity and acceleration.
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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.
 

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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
 
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