Hopefully last question on circular motion (on a cone)

[jiboom]

a smooth hollow circular cone of semi-angle y is fixed with its axis vertical and its vertex A downwards. A particle P,of mass m,moving with constant speed V describes a horizontal circle on the inner surface of the cone in a plane which is at a distance b above A.

show V^2=gb...done

b) if P is attached to one of a light elastic string PQ of natural length a and modulus of elasticity mg,find V^2 if

i)Q is attached to A
ii)Q is passed through a small hole at A and is attached to a particle of mass m hanging freely in equilibrium

for this how do i find the extension of the string? i am missing soemthing here as i need this extension to find the tension but is a<b or a>b?or does it not matter?

 

[cupid.callin]

for this how do i find the extension of the string? i am missing soemthing here as i need this extension to find the tension but is a<b or a>b?or does it not matter?

well I don't remember anything about modulus of elasticity (its been a long time since i studies phy) but of by tension you mean tension in string then notice this:

Q is passed through a small hole at A and is attached to a particle of mass m hanging freely in equilibrium

 

[jiboom]

if I am thinking correctly about this question to find the tension in the string i use

modulus of elasticity=(tension x natural length)/extension

but i can't see how to find the extension without knowing if a<b or a>b as i need to know by how much,if any, the string is stretched passed its length a.

 

[jiboom]

anyone able to help with this? I am thinking the length of the string is set by the motion of the particle,but have no idea how to decipher it?!

 

[JHamm]

Might be easier to think of the string instead as a spring, the modulus of elasticity is the "k" value of the spring, the "x" is the difference between the length a and the distance from the point A to the point Q.

 

[jiboom]

Might be easier to think of the string instead as a spring, the modulus of elasticity is the "k" value of the spring, the "x" is the difference between the length a and the distance from the point A to the point Q.

but this what I am asking hoiw to find. i don't see how to find the extension,i only know the length AQ in terms of alpha and b. i don't know how a relates to b, or is this contained within the motion and i can't unlock it from the physics?

 

[cupid.callin]

b) if P is attached to one of a light elastic string PQ of natural length a and modulus of elasticity mg,find V^2 if

i)Q is attached to A

I guess there is a flaw in question,
suppose all the conditions in first question are given ... and just a string is attached where b=a*cos(y)

Now the string will experience no external force and thus will just keep on spinning along with particle. and because string can be assumed light (obvious) its own weight will have no effect on motion

 

[JHamm]

I don't think the question states what the value of 'a' is, in that case the square of the extension or compression is $(a-\frac{b}{\cos y})^2$. From this you can find the force from the string.

 

[cupid.callin]

And what if $a = \frac{b}{\cos y}$. Which is quite possible

Well i guess you should answer in 2 cases,

I when $a = \frac{b}{\cos y}$
II when $a ≠ \frac{b}{\cos y}$

 

[JHamm]

Whether it is or not makes little difference, the answer will be algebraic anyway, a note at the end on what happens if indeed $a = \frac{b}{\cos y}$ might be nice but isn't necessary.

 

[cupid.callin]

and what about $a > \Large{\frac{b}{\cos y}}$ ? there won't be any tension then.

 

[jiboom]

im glad other someone else is seeing my problem. if a>b/cos y then i would have no extension,i don't have a spring so compression is no good!

this is what makes me think this is not allowed because of the motioin being described but i don't have the "physics" knowledege to deduce this.

if there is no tension then would not the v^2 be the same when there was no string? so it is pointless using a>b/cos y?

 

[JHamm]

and what about $a > \Large{\frac{b}{\cos y}}$ ? there won't be any tension then.

Can't believe I missed that, in that case I guess you'd just have to take the two cases
$$a < \frac{b}{\cos y}$$
and
$$a \geq \frac{b}{\cos y}$$