Acceleration and radius of curvature of a particle

[ehilge]

Homework Statement

The motion of a particle is defined as:

x=[(t-4)³/6]+t²

y=t³/6-(t-1)²/4

Find the acceleration and radius of curvature at t=2

Homework Equations

a=(dv/dt)(e_t)+(v²/$$\rho$$)(e_n)

where e_t and e_n are the tangential and normal unit vecotrs to the curve and $$\rho$$ is the radius of curvature.

The Attempt at a Solution

So I was able to get the acceleration by differentiating both of the equations twice and plugging in 2. The x component becomes 0 and the y is +1.5m/s. I'm stumped on the radius though. I differentiated to find dv/dt and v and tried solving for rho but it didn't work. That would have been essentially ignoring e_t and e_n since they are unit vectors, so the length is on it they shouldn't affect the outcome, right? Any ideas on how to solve the problem. Also, I think the way the equation is written with the unit vectors in there somewhat confuses me too so if there's anything that could help clear up any confusion with the equation that'd be great too.
Thanks!

 

[Redbelly98]

You do need to account for those unit vectors, at least the e_n one since that is where the radius of curvature comes into play.

Think of a as having a normal and tangential component (instead of x and y components). The normal component may be equated to v²/ρ, according to your equation.

If you can come up with an expression for e_n, that may help get you going.

 

[ehilge]

the equations I have in my notes are

e_n = -sin$$\vartheta$$i + cos$$\vartheta$$j

e_t = cos$$\vartheta$$i + sin$$\vartheta$$j

where $$\vartheta$$ would be the angle between the tangential vector and the x axis

I worked out $$\vartheta$$ = 14 by taking the inverse tan (dy/dt)/(dx/dt).
And then in the original equation, v is the length of <dx/dt, dy/dt> at t=2, and then I'm not exactly sure what dv/dt is. I was thinking that would be the final acceleration, but that would be what is on the left side of the equation so that doesn't make sense. Either way, I have feeling there is an easier way to do all this, any thoughts?

 

[Redbelly98]

I'm not sure of an easier way here.

I agree with the θ = 14° result, which can be used to determine e_n. Can you now find the component of a in the direction of e_n? That result will be equivalent to v²/ρ, so you could then solve for ρ.

dv/dt will not be needed.

 

[ehilge]

ok, I was able to talk with my classmates today and figure out the problem without having to use theta. e_n can be found by taking the unit vector of the vector perpendicular to e_t. e_t is found by differentiating the equation and finding v_x and v_y. Then a = <0,1.5> is dotted with e_n to get a_n. Finally, set a_n equal to v²/rho and solve.
Thanks for your help!

 

[Redbelly98]

You're welcome, glad it worked out.