Determination of the velocity of a particle in a hollow cone

• gnits
In summary, the conversation discusses a problem involving a smooth hollow circular cone and a particle moving in a circle on its inner surface. The first part of the question is solved by using equations related to the modulus of elasticity of an elastic string. In the second part, the only change is the addition of a particle hanging freely on the string, leading to a simple solution of tension being equal to mg. The summary concludes with a thank you for the help.
gnits
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 semi-angle 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 believe 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.

Attachments

• cone.png
3.9 KB · Views: 220
In part 2, isn't the tension just mg?

Yes! I made that way too complicated. I see it now. Thank you.

1. How is the velocity of a particle in a hollow cone determined?

The velocity of a particle in a hollow cone is determined by calculating the change in position of the particle over time. This can be done using the formula v = ∆r/∆t, where v is the velocity, ∆r is the change in position, and ∆t is the change in time.

2. What factors affect the velocity of a particle in a hollow cone?

The velocity of a particle in a hollow cone is affected by various factors such as the angle and speed at which the particle is launched, the shape and size of the cone, and any external forces acting on the particle.

3. Can the velocity of a particle in a hollow cone change over time?

Yes, the velocity of a particle in a hollow cone can change over time. This can be due to external forces acting on the particle, changes in the shape or size of the cone, or changes in the angle or speed at which the particle is launched.

4. How does the velocity of a particle in a hollow cone differ from the velocity of a particle in a solid cone?

The velocity of a particle in a hollow cone differs from the velocity of a particle in a solid cone because the hollow cone has a larger surface area, which results in more air resistance and a slower velocity. Additionally, a particle in a hollow cone may also experience changes in velocity due to the changing shape and size of the cone.

5. What are the applications of determining the velocity of a particle in a hollow cone?

The determination of the velocity of a particle in a hollow cone has various applications in fields such as physics, engineering, and aerodynamics. It can be used to study the effects of air resistance, understand projectile motion, and design efficient structures and machines. It also has applications in sports, such as determining the trajectory of a ball in sports like baseball or golf.

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