# Homework Help: Normal component of the acceleration

1. Aug 1, 2010

### Sheen91

1. The problem statement, all variables and given/known data

Given y$$^{2}$$ = 8x$$^{3}$$ + 8x

where x and y are in metres, and y is positive

What is the normal component of the acceleration

When x = 4m, x = 7m/s, x = 2m/s$$^{2}$$

Note : Wasn't sure if it was Calc or Physics, but I was leaning toward Calc, as it is basically Algebra to solve for the answer.

2. Relevant equations

Not sure

3. The attempt at a solution

I understand that "x = 4m, x = 7m/s, x = 2m/s$$^{2}$$" is just "y$$^{2}$$ = 8x$$^{3}$$ + 8x" getting differentiated for each part, first of which doesn't need it.

y$$^{2}$$ = 8x$$^{3}$$ + 8x => y = ($$\sqrt{8x^3 + 8x}$$)

I am just a little lost at "What is the normal component of the acceleration"

Any and all help is greatly appreciated.

Thanks

2. Aug 2, 2010

### Staff: Mentor

These should probably be "... when x = 4m, dx/dt = 7 m/sec, and d^2x/dt^2 = 2m/sec^2. You can't have x being equal to three different quantities with three different units.
Are you sure that you are giving us the exact wording of this problem? It doesn't make much sense to me as you have written it.

3. Aug 2, 2010

### HallsofIvy

If this is not a trajectory in the plane, then the problem makes no sense- there is no "normal component". If it is a trajectory then you need to find the acceleration as the vector $(d^2x/dt^2}\vec{i}+ (d^2y/dt^2)\vec{j}$. Of course, to do that, you have to know the dependence of x and y on t.

Knowing that $y^2= 8x^3+ 8x$ tells you that $2y(dy/dt)= (24x^2+ 8)(dx/dt)$ but uyou still need to know or be able to calculate either dy/dt or dx/dt.

I agree with Mark44- please give the exact wording of the problem. "When x = 4m, x = 7m/s, x = 2m/s2" is impossible.

4. Aug 4, 2010

### Frangecoil

I came across the same question. Sheen has typed the entire question. Only mistake being the repeated x. Which was meant to be x, x' and x''.
I derived the equation twice and subbed in the given x values as i derived to get the corresponding y values. i.e. i solved for y, y' and y". Is this right?
If this is right i still get stuck on how to find the normal component.

5. Aug 4, 2010

### vela

Staff Emeritus
Hint: Since the motion is confined to a plane, the acceleration can be resolved into two components, the tangential component parallel to the velocity and the normal component perpendicular to the velocity.

6. Aug 5, 2010

### Sheen91

***Post removed***

Last edited: Aug 5, 2010
7. Aug 5, 2010

### Sheen91

y² = 8x³ + 8x

2y dy/dt = (24x² + 8) dx/dt

Velocity = sqrt ( (dx/dt)² + (dy/dt)² )

Acceleration = sqrt ( (d²x/dt²)² + (d²y/dt²)² ) = ???? :S (not sure)

2y dy/dt = (24x² + 8) dx/dt

2 d²y/dt² = 48x d²x/dt²

d²y/dt² = 24x d²x/dt²

Therefore:

Acceleration = sqrt ( (d²x/dt²)² + (d²y/dt²)² ) = sqrt( (d²x/dt²)² + (24x . d²x/dt²)² ) = sqrt ( 2² + (24 . 4 . 2)²) = sqrt ( 36868 ) = 192.01 m/s/s

Which doesn't seem right. :S

Can someone look through and see if I did anything wrong. Thank you

PS: I need to find the normal component of the acceleration, so I have yet to start that

Last edited: Aug 5, 2010
8. Aug 5, 2010

### vela

Staff Emeritus
This is good. Come up with numerical answers for what y and y' equal given the information you have about x and x'.

9. Aug 5, 2010

### vela

Staff Emeritus
That's not right. You have to use the product rule.

10. Aug 5, 2010

### Sheen91

Sorry I don't understand how, do I use the product rule in that??

Thanks for all the help so far vela, I really appreciate it

EDIT:
uhmm ohh wait

dy/dx = (dy/dt) / (dx/dt)

2y dy/dt = (24x² + 8) dx/dt

dy/dx = (dy/dt) / (dx/dt) = (24x² + 8) / 2y

d²y/dx² = ( (48x . 2y) - 2(24x² + 8)dy/dx ) / 4y²

I am a little lost.

Last edited: Aug 5, 2010
11. Aug 5, 2010

### Sheen91

y = sqrt (8x³ + 8x) = sqrt (8.4³ + 8.4) = sqrt (544) = 23.324

y' = (24x² + 8) x' / 2y = 58.82

now what?

12. Aug 5, 2010

### Sheen91

I am unsure on how to find the "Normal component of the Acceleration"

Vela can you let me know what to do step by step, rather than just one step at a time. I would really apreciate that.

I also got the numerical values of y and y', and am once again stuck.

Also how do I use the product rule on : 2y dy/dt = (24x² + 8) dx/dt

Have never been taught.

13. Aug 5, 2010

### Sheen91

I think I got it:

Product rule

2y dy/dt = (24x² + 8) dx/dt

(2 dy/dt) . dy/dt + d²y/dt² = (48x . dx/dt) + ( (24x² + 8) . d²x/dt²)

so that means:

d²y/dt² = (48x . dx/dt) + ( (24x² + 8) . d²x/dt²) - (2 dy/dt) . dy/dt

d²y/dt² = -4791.58

That seems very wrong.

Also after you get d²y/dt² and d²x/dt², what do you do? How do you find the normal component?

Last edited: Aug 5, 2010
14. Aug 5, 2010

### vela

Staff Emeritus
You forgot the 2y in front of y''. The RHS looks okay.
You have the two vectors a=axx+ayy and v=vxx+vyy. What you want to do is resolve a into components parallel and perpendicular to v. So how do you find the component of a parallel to v? Hint: Use the dot product.

15. Aug 5, 2010

### Frangecoil

Drawing a vector diagram for this question helps ALOT. It helped me solve it =]