 #1
 63
 14
Homework Statement:
 Determination of velocity of particle in hollow cone
Relevant Equations:
 F=mv^2/r
Hi, please could I ask for help with the following question:
A smooth hollow circular cone of semiangle α, is fixed with its axis vertical and its vertex A downwards. A particle P, of mass m, moving with constant speed V, decribes a horizontal circle on the inner surface of the cone in a plane which is at a distance b above A. If P is attached to one end of a light elastic string PQ of natural length a and of modulus of elasticity mg, find V^2 if:
1) Q is attached to A.
2) Q is passed through a small hole at A and is attached to a particle of mass m hanging freely in equilibrium.
I've done part 1. It's part 2 that has me stuck.
Here's my diagram and answer for part 1: (I have indicated items relating to the elastic string in red)
Here's my reasoning:
We are told that the modulus of elasticity of the elastic string is mg, this means that for a natural length a and an extension x that the tension in the elastic string is T = mgx/a (Equation 1)
Also, r = b tan(α)
Resolving Horizontally:
R cos(α) + T sin(α) = (mV^2) / (b tan(α)) (Equation 2)
Resolving Vertically:
R sin(α) = T cos(α) + mg
Substituting in this equation for T (from Equation 1) we get:
R = ( mgx cos(α) ) / (a sin(α)) + (mg) / sin(α) (Equation 3)
Using Equation 1 and Equation 3 to substitute for R and T in Equation 2 we obtain:
( mgx (cos(α))^2 ) / (a sin(α)) + mg cos(α) / sin(α) + ( mgx sin(α) ) / a = (mV^2) / (b tan(α))
This rearranges and simplifies to give:
V^2 = gb ( 1 + x / (a * cos(α)))
Finally we substitute in for x using:
cos(α) = b / (a + x) which gives x = (b / (cos(α) )  a to obtain
V^2 = gb( 1  sec(α) + (b/a) (sec(α))^2 ) which is the answer in the book
But now for part 2. Here's my diagram:
I reasoned that the only thing that has changed form part 1 is that we no longer have:
T = mgx/a because clearly the extension is now > x, it is x + k.
k is the extra extension caused by hanging the mass m to the setup in part 1.
Is k is the extension produced in the elastic string by a force mg which would mean that k = a
OR
Is k the extension produced in the elastic string by a force mg cos(α) which would mean that k = a cos(α) ?
I beleive these equations still hold:
R cos(α) + T sin(α) = (mV^2) / (b tan(α))
R sin(α) = T cos(α) + mg
x = (b / (cos(α) )  a
With either value of k I can't match the book answer of:
V^2 = gb (1 + sec(α) )
Thanks for any help,
Mitch.
A smooth hollow circular cone of semiangle α, is fixed with its axis vertical and its vertex A downwards. A particle P, of mass m, moving with constant speed V, decribes a horizontal circle on the inner surface of the cone in a plane which is at a distance b above A. If P is attached to one end of a light elastic string PQ of natural length a and of modulus of elasticity mg, find V^2 if:
1) Q is attached to A.
2) Q is passed through a small hole at A and is attached to a particle of mass m hanging freely in equilibrium.
I've done part 1. It's part 2 that has me stuck.
Here's my diagram and answer for part 1: (I have indicated items relating to the elastic string in red)
Here's my reasoning:
We are told that the modulus of elasticity of the elastic string is mg, this means that for a natural length a and an extension x that the tension in the elastic string is T = mgx/a (Equation 1)
Also, r = b tan(α)
Resolving Horizontally:
R cos(α) + T sin(α) = (mV^2) / (b tan(α)) (Equation 2)
Resolving Vertically:
R sin(α) = T cos(α) + mg
Substituting in this equation for T (from Equation 1) we get:
R = ( mgx cos(α) ) / (a sin(α)) + (mg) / sin(α) (Equation 3)
Using Equation 1 and Equation 3 to substitute for R and T in Equation 2 we obtain:
( mgx (cos(α))^2 ) / (a sin(α)) + mg cos(α) / sin(α) + ( mgx sin(α) ) / a = (mV^2) / (b tan(α))
This rearranges and simplifies to give:
V^2 = gb ( 1 + x / (a * cos(α)))
Finally we substitute in for x using:
cos(α) = b / (a + x) which gives x = (b / (cos(α) )  a to obtain
V^2 = gb( 1  sec(α) + (b/a) (sec(α))^2 ) which is the answer in the book
But now for part 2. Here's my diagram:
I reasoned that the only thing that has changed form part 1 is that we no longer have:
T = mgx/a because clearly the extension is now > x, it is x + k.
k is the extra extension caused by hanging the mass m to the setup in part 1.
Is k is the extension produced in the elastic string by a force mg which would mean that k = a
OR
Is k the extension produced in the elastic string by a force mg cos(α) which would mean that k = a cos(α) ?
I beleive these equations still hold:
R cos(α) + T sin(α) = (mV^2) / (b tan(α))
R sin(α) = T cos(α) + mg
x = (b / (cos(α) )  a
With either value of k I can't match the book answer of:
V^2 = gb (1 + sec(α) )
Thanks for any help,
Mitch.
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