In a problem, I have the expression of acceleration and velocity in Cartesian coordinates , and it ask me to calculate the tangential and normal acceleration, so we don't know how I can do that, can any one help me?
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$\begingroup$ Do you understand that the velocity is in the tangential direction? $\endgroup$– GhosterCommented May 1, 2023 at 20:26
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$\begingroup$ Yes, so I try to write the expression of Ut using velocity and it's module and substituting in the expression of at but I don't see the key yet? $\endgroup$– AliaCommented May 1, 2023 at 20:39
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$\begingroup$ Have you learned how to project one vector along another vector? $\endgroup$– GhosterCommented May 1, 2023 at 20:49
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$\begingroup$ No, I just know how to do this work with unit vector, did you mean change the base , rather than use Ut I must use at? And I am sorry if my questions seems stupid. $\endgroup$– AliaCommented May 1, 2023 at 21:03
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2$\begingroup$ It's been a long time since I've done this, but can you not turn the velocity vector into a unit vector, and take the dot product of that with the acceleration vector? That is half of your answer. Then you can use vector subtraction for the other half. $\endgroup$– RC_23Commented May 2, 2023 at 1:11
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1 Answer
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The acceleation vector can be split into two components:
- The tangential component $a_\parallel$ is along with (parallel to) the motion, so parallel to the velocity vector.
- The normal component is $a_\perp$ is perpendicular to the motion, so perpendicular to the velocity vector.
Note that these two components form a right-angled triangle with the original acceleration vector. Specifically, the tangential and normal components of the acceleration vector are the cathetuses in this right-angled triangle.
We have tools for finding the cathetuses in a right-angled triangle. All we need to know is the angle. If you can figure out the angle between e.g. the cathetus that represents the tangential component with the original acceleration vector, then just use cosine and sine with this angle to find the tangential and normal components (the adjacent and opposing sides in the right-angled triangle). If this is unclear, then start by drawing a clear sketch.
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$\begingroup$ Thank you, but the problem is that there is nothing to draw it , In the exercise it give me only the velocity and acceleration in Cartesian coordinates $\endgroup$– AliaCommented May 1, 2023 at 21:11
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$\begingroup$ @wissal You have the coordinates? Then draw them in a coordinate system. $\endgroup$– SteevenCommented May 1, 2023 at 21:55
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1$\begingroup$ My teacher send me the solution now , and I find that the trick is just know that we find the tangential acceleration by derivation of velocity's module ! It's new information for me , for drawing I use it to verify teacher's result and I find the same result , thank a lot for helping me $\endgroup$– AliaCommented May 1, 2023 at 22:06