Example about tangential and normal unit vectors

1. Apr 22, 2014

mech-eng

Here is a example 1.3 from analytical dynamics of Haim Baruh.
a particle moves on a path on the xy plane defined by the curve y=3*x^2 , where x varies with the relation x= sin(a). find the radius of curvature of the path and unit vectors in the normal and tangential directions when a=pi/6. This is example with solution but I tried another way to find unit vectors.I have find the tangential vector true but normal vector was wrong but I dont know where I am wrong. Can you help me please.

First I have tried to find tangential vector. r=sin(a)i + 3(sin(a))^2j as in the solution. Then I thought unit tangential unit vector must be in the direction of tangent to the path which is the same as derivative of position vector r. its derivative I found is cos(a) i + 6sin(a)cos(a)j. then I calculated derivative vector at a=pi/6 as sqrt(3)/2 i + 3/2 sqrt(3) j, which is the same as in the book. Then I have calculated the magnitude of derivative of position vector and then calculated its unit vector as the tangential vector. This was true result. But when calculating normal unit vector something is wrong. While calculating normal unit vector first I thought that it is orthogonal to the tangential unit vector and if I take derivative of tangential unit vector must be in the direction of normal unit vector. I thought derivative of tangential unit vector must be in the same direction with second derivative of position vector which is -sin(a)i + -6sin(a)cos(a)j. Then I have calculated second derivative vector at pi/6 as -1/2i + -2.59808 j. Then I calculate the magnitude as sqrt(7) and unit vector in the normal direction as (-1/2)/sqrt(7)i + -2.59808/sqrt(7) which is -0.18898i+ -0.98198 j. But solution for normal unit vector in the book is -0.9487i+0.3162j. I can't understand what is wrong for my solution.

2. Apr 22, 2014

dauto

The second derivative of the position gives you the derivative of the velocity. The velocity has the same direction as the tangential unit vector but different magnitude. Notably, the magnitude of the velocity may change over time. if v is the velocity vector, v its magnitude, and ut the tangential unit vector than we have

v = v ut, and

d/dt v = v d/dt ut + (dv/dt) ut

The second term in the last equation has the wrong direction and will skew your final reasult

3. Apr 22, 2014

Staff: Mentor

You don't need to take the derivative of the unit tangent vector to get the unit normal vector, although that will certainly give you the unit normal. Don't forget that the unit normal is perpendicular to the unit tangent, so their dot product must be zero. So, if the unit tangent has components (a,b), the unit normal must have components (-b,a) or (b,-a).

Chet

4. Apr 22, 2014

mech-eng

Your approach is really good and very practical. First derivative is velocity and in the tangential direction but should not second derivative which is also derivative of velocity be orthogonal to first derivative. We should also
consider the fact that derivative of a unit vector is orthogonal to itself.

Last edited: Apr 22, 2014
5. Apr 22, 2014

mech-eng

But how about unit vector of first derivative in your last equation? I think it is normal vector. Yes it is magnitude
may change over time but we are dealing with when a=pi/6 and note that second derivative in my first post is
wrongly calculated, sorry for it. but my question still continues from derivative perspective.

Last edited: Apr 22, 2014
6. Apr 22, 2014

Staff: Mentor

If the magnitude of the velocity vector is changing, the derivative of the velocity vector has two components, one in the direction parallel to the velocity vector and the other in the direction normal to the velocity vector. This is what dauto's post was saying. However, if you first evaluated the velocity vector, and then divided it by its own magnitude, you would have a unit vector pointing in the same direction as the velocity vector. The derivative of this unit vector would be a unit vector normal to the velocity vector.

Chet

7. Apr 22, 2014

mech-eng

But aren't first taking derivative then finding unit vector and first finding unit vector then taking
derivative the same things?

8. Apr 22, 2014

Staff: Mentor

No they aren't. Suppose the particle is accelerating in a straight line. What is the time derivative of the velocity vector? Does it have a component tangent to the velocity vector?

Chet

9. Apr 25, 2014

mech-eng

Thank you for this very instructive example. My another question is why do we calculate radius of curvature and torsion of the curve?

10. Apr 25, 2014

Staff: Mentor

These are used to determine the components of the net forces acting on the particle, in conjunction with writing the force balance equations.

Chet