How do i find the radius of this path

[devanlevin]

given
$$\vec{R}$$=(30t-t$$^{3}$$)$$\hat{x}$$+(22t-4t$$^{2}$$)$$\hat{y}$$

find, for time t=2s
1)the acceleration (a)
2)the angular acceleration($$\alpha$$)
3)the radius of the curve of the path it takes, at t=2s

as far as i can see, i can integrate $$\vec{R}$$ twice to get the acceleration, but i need the radius to find out ($$\alpha$$),...
the only thing i could think of to find the radius is using the equation of a circle
(X-Xo)$$^{2}$$+(Y-Yo)$$^{2}$$=r$$^{2}$$
then plugging in the values of $$\vec{R}$$(t=2s) in the X and Y, problem is i don't know the Xo, Yo (centre point) of the circle. thought maybe to plug in X,Y at t=0 for Xo Yo but that just seems wrong.
any ideas

 

[Redbelly98]

... as far as i can see, i can integrate $$\vec{R}$$ twice to get the acceleration ...

Well, actually you don't integrate R. Hopefully you know that, and just typed in the wrong thing.

... problem is i don't know the Xo, Yo (centre point) of the circle.

True, to get angular information there has to be a reference point for the rotation. Since none was given, I'm guessing we use the origin.

 

[devanlevin]

sorry, meant take the derivative of r twice, but that's not really my problem, how do i find the radius?

 

[Redbelly98]

Do you have a calculus book lying around? Curvature of a path or function is usually covered in the 1st semester.

 

[devanlevin]

like i said, only thing i can think of is r^2=(x-x0)^2+(y-y0)^2
dont know where the centre(x0,y0) is?
any ideas

 

[alphysicist]

Redbelly98,

True, to get angular information there has to be a reference point for the rotation. Since none was given, I'm guessing we use the origin.

If I'm reading the question correctly I don't believe that is correct. The path itself defines the center of rotation at any particular time.

As an alternative approach to find the radius of curvature, you could examine the tangent and normal components of the acceleration.

 

[Redbelly98]

If I'm reading the question correctly I don't believe that is correct. The path itself defines the center of rotation at any particular time.

As an alternative approach to find the radius of curvature, you could examine the tangent and normal components of the acceleration.

I'm not convinced. I'd like to know if this is a basic Phys 101 class that is studying introductory rotational dynamics. If so, I'd expect a straightforward use of the origin as reference:
(x0, y0) = (0,0)
r is simply the vector ( x(t), y(t) )
Angle value is taken between r and +x axis.

You're interpretation could be right, but I would only expect that in the case of an advanced honors track course -- which is of course possible. Presumably there is an example or discussion in devanlevin's textbook that would clear this up.

 

[alphysicist]

I'm not convinced. I'd like to know if this is a basic Phys 101 class that is studying introductory rotational dynamics. If so, I'd expect a straightforward use of the origin as reference:
(x0, y0) = (0,0)
r is simply the vector ( x(t), y(t) )
Angle value is taken between r and +x axis.

You mean find $$\frac{d^2\theta}{dt^2}$$ for the coordinate $\theta$? That's possible, but to me doesn't seem to fit the rest of the question.

You're interpretation could be right, but I would only expect that in the case of an advanced honors track course -- which is of course possible.

If they are expected to find the radius of curvature, I think finding the tangential and normal components is something they would be expected to do. Since the velocity, speed, and acceleration were already found in part a, a dot product and using the pythagorean theorem is all that is required to find the two acceleration components. Then one component gives the radius of curvature, and the other gives (alpha) = a_T/R.

(So my point is that these ideas all go together, and the above work takes only a few lines so I think it is well within reason.)

Please correct me if I misunderstand what you were saying, but I think this is much more straightforward than calculating $d^2 \theta/dt^2$ (which seems to involve taking the second derivative of the arctangent of y(t)/x(t) which is a lot of tedium).

Presumably there is an example or discussion in devanlevin's textbook that would clear this up.

I definitely agree with that; just asking for the "angular acceleration" is rather ambiguous since it could go either way.

 

[devanlevin]

sorry about the terminology, I am not taking it from an english textbook and english isn't the language I am studying in,,, what I am looking for when i say "angular acceleration" is the amount of radians/s^2
how do i find the tangent and normal components of the acceleration,

 

[alphysicist]

sorry about the terminology, I am not taking it from an english textbook and english isn't the language I am studying in,,, what I am looking for when i say "angular acceleration" is the amount of radians/s^2

Unfortunately that does not clear up the ambiguity. The question is whether the word "angular" in angular acceleration here refers to the angle that the position vector makes with the x axis, or to the angle it makes with the center of rotation at that particular time.

how do i find the tangent and normal components of the acceleration,

As I was using the terms, the tangential component is parallel to the velocity, and the normal component is perpendicular to the velocity. (Tangential and normal to the path.)

Since you already have the forms for the velocity, speed, and acceleration, you can use the dot product to get the component of the acceleration parallel to the velocity.

How can you then use that to find the normal component?

 

[devanlevin]

the angle it makes with the centre of the circle

 

[devanlevin]

but how would i get the radius, or the "angular acceleration" from the component parallel or normal to the acceleration and velocity?

 

[alphysicist]

but how would i get the radius, or the "angular acceleration" from the component parallel or normal to the acceleration and velocity?

The angular acceleration relative to the center of the path is given by:

$$a_T= \alpha\ R$$
(similar to simple circular motion) where a_T is the tangential component (component parallel to the velocity) and R is the radius of curvature, and alpha is related to the angle of the instantaneous center of rotation.

The normal component is similar to the formula for the centripetal acceleration:

$$a_N = \frac{v^2}{R}$$
where a_N is the normal component (perpendicular to the velocity).

Since you have the speed, once you find a_N and a_T you can find alpha and R.

 

[devanlevin]

thanks a lot,
i found this site with equations for curvature, will i get the same results??
http://mathworld.wolfram.com/RadiusofCurvature.html

 

[alphysicist]

thanks a lot,
i found this site with equations for curvature, will i get the same results??
http://mathworld.wolfram.com/RadiusofCurvature.html

Yes.