Calculating Normal Vectors for Particle Motion

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To calculate the normal vector for particle motion, the tangential vector T is derived from the velocity vector v, resulting in T={2/3, 2/3, -1/3}. The normal vector N is determined to be perpendicular to T and lies in the plane of motion, calculated as N={0, 4/5, 3/5}. The normal component of acceleration an can be computed using two different equations, leading to discrepancies in values, with one method yielding an=(10/3)(√2) and another giving an=5. It's clarified that T and N are indeed orthogonal, and the normal vector N should represent the component of the acceleration vector that is perpendicular to T. Understanding these relationships is crucial for accurately solving problems involving particle motion.
BennyT
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Homework Statement

[/B]
This problem is from Jon Rogawski's Calculus-Early Transcendentals
At a certain moment, a moving particle has velocity v={2,2,-1} and a={0,4,3}. Find T, N and the decomposition of a into tangential and normal components.

Homework Equations


ANYTHING IN [ ] REPRESENTS A LENGTH OF A QUANTITY and . stands for dot product
T[/B](t)=v(t)/[v(t)]
N(t)=T '(t)/[T '(t)]
an=√([a]^2-at^2)
at=a.T
an=a.N=√([a]^2-at^2)

The Attempt at a Solution


So I'm trying to find an, the normal component of acceleration, but by using the two definitions and equations found in the book I am coming to different values. So I first calculated T={2/3,2/3,-1/3}, and then N={0,4/5,3/5}. Then I calculated at=5/3 and then I calculated an by the first equation (an=√([a]^2-at^2)) to equal an=(10/3)(√2), but using the second equation I get an=5. What am I doing wrong? And also, do T and N always have to be orthogonal? This isn't a graded assignment or anything, but it's causing me frustration. Thanks for your time.
 
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You appear to be assuming that the acceleration vector is itself normal to the tangent vector. That is not necessarily true because you are not told that the acceleration is constant.
 
Ok, so is this reply in response to whether T and N are always orthogonal? Also, for a problem such as the one above, what is the best way to calculate N? Thank you for your response.
 
BennyT said:
Ok, so is this reply in response to whether T and N are always orthogonal? Also, for a problem such as the one above, what is the best way to calculate N? Thank you for your response.
I'm sure it wasn't about whether they are orthogonal. They definitely are.

Vector T is in the direction of motion, you did that correctly.

Vector N is perpendicular to vector T, and is in the plane of the particle's motion. Basically it's in direction of the component of acceleration vector, a, that's perpendicular to vector T .
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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