Problem with Circular Motion: Solution Attempt and Equations | Homework Help

[TyErd]

Homework Statement

I've attached the question.

Homework Equations

normal acceleration=(v^2)/r where radius i think is going to be 3-1.2=1.8?
|a|= sqrt(a_t^2 + a_n^2)

The Attempt at a Solution

the equation given x^2/9 + y^2/16 = 1 is distance i believe, so when you differentiate it v=2x/9 + y/8 = 0, and differentiate again would mean a=2/9+1/8=25/72. And these three equations are tangential.
not sure what to do from here...

 

[PeterO]

Homework Statement

I've attached the question.

Homework Equations

normal acceleration=(v^2)/r where radius i think is going to be 3-1.2=1.8?
|a|= sqrt(a_t^2 + a_n^2)

The Attempt at a Solution

the equation given x^2/9 + y^2/16 = 1 is distance i believe, so when you differentiate it v=2x/9 + y/8 = 0, and differentiate again would mean a=2/9+1/8=25/72. And these three equations are tangential.
not sure what to do from here...

On what did you base the idea that radius is 1.8?

 

[emailanmol]

Hey,
See the problem in a different way.

What is the minimum condition for the man to just complete the ellipse?(in terms of velocity at top most point.Why??)

Use Energy equation rather than force.


(Also ask yourself Can we conserve energy of just the man? External force of Normal reaction acts on him.)

 

[TyErd]

On what did you base the idea that radius is 1.8?

well because the radius of the loop is 3m and the centre of gravity of the guy is 1.2m. So when he get to the top, he's only 1.8m away from the centre.

 

[TyErd]

Hey,
See the problem in a different way.

What is the minimum condition for the man to just complete the ellipse?(in terms of velocity at top most point.Why??)

Use Energy equation rather than force.


(Also ask yourself Can we conserve energy of just the man? External force of Normal reaction acts on him.)

the minimum condition is weight force=normal force?

 

[TyErd]

(v^2)/r= mg??

 

[PeterO]

(v^2)/r= mg??

Close!

Either m(v^2)/r= mg or (v^2)/r= g

 

[PeterO]

well because the radius of the loop is 3m and the centre of gravity of the guy is 1.2m. So when he get to the top, he's only 1.8m away from the centre.

3m is the length of one semi-axis of the ellipse - the other is 4m. Not convinced either of them is the radius of curvature at the apex - but it might be? but I certainly do not expect R=3 at the top.

 

[emailanmol]

Correct. v^2/r = g.

What is the radius at this point?
How will you apply conservation of energy?

 

[emailanmol]

Hey, I read your last post now.

See its not necessary radius of curvature will be equal to semi major or minor axis.
Infact during motion radius of curvature is changing.

The way you calculate radius of curvature at any point is by writing (assume m is theta .The parametric angle of ellipse.i can't type theta as Am online via cell)
X=acos(m) Y=bsin(m) .

At lowest point m has value 270 degrees.
(since y is negative , x is 0)
Now the way we calculate radius of curvature at any anypoint
Is by writing
r=v^2 at that point/centripetal acceleration at that point.

You may be surprised as we were supppsed to calculate velocity at that point with this equation.
How is one equation enabling us to soLve for two variables r and v.

However, in essence we are actually using two equations the other being equation of ellipse.
As you will see radius of curvature just depends on a,b and m and is independent of v and a.(as it should be logically.it should depend on path of particle and not on how the path is travelled)


You will get r as
[(a^2)(sin^2m) +(b^2)(cos^m)]/sqrootof[a^2cos^2(m) +(b^2)(sin^(m))]

(How??)(you can write general equation of velocity by differentiating and of centripetal component of acceleration and divide)

Now at topmost point
M has value 90 degrees.
(y is positive x is 0)


Plug in r and use energy equation

 

[TyErd]

so v^2/r=g ; v^2/4=9.81 thus v= 6.26418.
then conservation of energy principle:
0.5mv^2+mgh = 0.5mv^2 + mgh
0.5*85*v^2 + 85(9.81)1.2= 0.5*85*6.26418^2 + 85*9.81*(8-1.2) thus v=12.21 which is incorrect

 

[emailanmol]

R is not 4

 

[TyErd]

im still not quite sure why r isn't 4.

 

[emailanmol]

I just posted the big formula.

At top mist point r is a^2/b.
Plug in and check if answer is right .
And then we will discuss why r isn't 4 :-)

 

[TyErd]

im not exactly sure what a and b is in that formula.

 

[emailanmol]

Oops.my mistake.
Actually general equation of ellipse is take as x^2/a^2 + y^2/b^2 = 1
so
a=3, b=4

 

[TyErd]

so now i have to sub in a=3 b=4 and m=90 into that equation that you gave me?

 

[emailanmol]

Yeah.
You will obtain r=a^2/b for top most point.

Then substitue this new value of v into the energy equation(you have written it correctly).
And find velocity at bottom

 

[TyErd]

okay i did that and i got a velocity of 11.48 m/s. the answer is 11.15 m/s

 

[emailanmol]

Switch value of g from 9.8 m/s^2 to 10 m/s^2 or vice versa (whichever you used) and check once again if the answer matches exactly.

 

[TyErd]

substituting 10 gives me v=11.596

 

[TyErd]

I think my equation is wrong.

 

[emailanmol]

You have usef the equation correctly.
Maybe answer is approximated.

 

[TyErd]

so i was right to include potential energy at the ground where the height would be 1.2m and then when calculating potential energy at the top i have to reduce 1.2m from the height, making it 8-1.2m=6.8m?

 

[emailanmol]

Yes.Change in potential emergy is due to net displacement of COM.

I am prettu sure I have derived the radius of curvature correctly


See, the answer should be 11.48 m/s according to me.

 

[emailanmol]

Wait I got the mistake.

a is not 3 but its 3-1.2
Similarly b is 4-1.2

Since we are intereseted in ellipse of COM.

That gives you the answer in the book.
This is an epic question.which book had it?

 

[TyErd]

It was on a worksheet my lecturer gave us. Can I also ask what this equation is coz we've never used it: [(a^2)(sin^2m) +(b^2)(cos^m)]/sqrootof[a^2cos^2(m) +(b^2)(sin^(m))]

is there a more simpler eqn i cud use, coz i might not get marks...

 

[emailanmol]

No this is the general equation.

You can derive it in less than 2 mins by differentiating the path equations, finding magnitude of v and centripetal acceleration :-)

I don't see any other way of deriving(though simpler geometric ways may exist).
Infact, the simplest way to derive is by differeentiating the path equations as it works always

 

[TyErd]

so differentiate x^2/9 + y^2/16 = 1 ?

 

[emailanmol]

Yes.I ll give u a hint

Its better to write the equation as

X=acos m Y=b sin m .(it is same as writing x^2/a^2+ y^2/b^2=1

Differentiating will give you Vx and Vy from which you can derive magnitude of v at any point.

Double differentiating x will give you Ax and Ay, out of which one component will ne centripetal acceleration.

 

[TyErd]

sorry what is X and Y? are they the vertical and horizontal displacements?

 

[emailanmol]

Yes.:-)

 

[emailanmol]

Also you said,

the equation given x^2/9 + y^2/16 = 1 is distance i believe, so when you differentiate it v=2x/9 + y/8 = 0, and differentiate again would mean a=2/9+1/8=25/72. And these three equations are tangential.


This is not how you differentiate.

X^2/9+ Y^2/16=1

Differentiating
2xdx/9dt +2ydy/16dt =0

dx/dt is horizontal velocity and dy/dt is vertical velocity.

Using equation x=asin m is easier .So differentiate it.

dx/dt or Vx=acos m (dm/dt)

dm/dt is like omega
Ax= -asinm(dm/dt)^2 + a cosm(d2m/dt2)

Here -asin m(dm/dt) is x component of centripetal acceleration.

Magnitude of velocity is sqroot [(dx/dt)^2 +(dy/dt)^2]

 

[TyErd]

i still don't get the radius. i get everything else just not the radius. I don't get how differentiating that somehow gets me a radius. before the mistake the radius was 9/4 now what is it?

 

[TyErd]

so can i actually find out the velocity at the top without using a radius?