Chronology |
Current Month |
Current Thread |
Current Date |

[Year List] [Month List (current year)] | [Date Index] [Thread Index] | [Thread Prev] [Thread Next] | [Date Prev] [Date Next] |

*From*: David Bowman <David_Bowman@georgetowncollege.edu>*Date*: Sat, 6 Mar 2021 21:33:13 +0000

I tried my hand at explicitly solving for the motion of the falling object hanging from a spool by a flexible inextensible cord with appreciable *non-negligible*mass per unit length of cord. I successfully came up with an expression for the elapsed time from when the system was released in terms of the distance fallen by the object. Unfortunately, that expression involved a complicated integral that I don't know how to do in closed form (short of appealing to some sort of approximation method such as a power series expansion or over simplification of the model's relevant functions). But in case anyone is actually interested in the explicit formula of the solution, here is the motion's undoable 2nd integral in terms of its quite doable first integral.

Let M_0 = the *initial* dangling mass *including* the mass of any initially unwound cord.

Let σ = the linear mass density (mass per unit length) of the cord.

Let I_0 = the *initial* moment of inertia of the spool *including* that due to the initially wound up portion of the cord.

Let R_0 = the initial mean radius of the contact point on the spool from which the cord begins to come off.

Let δ = the mean thickness of a monolayer of cord on the spool, i.e. the effective radius of the spool decreases (assumed smoothly) by δ with each complete rotation of the spool.

Let t = elapsed time since release of the system from *rest*.

Let x = length of cord unwrapped since release, i.e. net descent distance of the hanging object from its initial height.

Let g = gravitational field strength.

With these definitions the expression for t in terms of x is:

t = Integral{0, x| du sqrt((1 + (I_0 - σ*R_0^2*u/2)/D)/(2*g*u))}

where D is defined for mathematical expression simplification purposes as

D = (M_0 + σ*u/2)*(sqrt((R_0 + δ/2)^2 - δ*u/ Π ) - δ/2)^2 , and

u = integration variable taken between the lower limit, 0, and upper limit, x.

Fortunately is not too hard to see that the initial behavior of the system before any significant amount of cord has unwrapped is

t = sqrt(2*x*(1 + I_0/(M_0*R_0^2))/g) or

x = g*t^2/(2*(1 + I_0/(M_0*R_0^2)))

as is expected , initially.

David Bowman

**References**:**[Phys-L] Rotational Motion Pulley Question***From:*Michael Barr <zubarsky@gmail.com>

**Re: [Phys-L] Rotational Motion Pulley Question***From:*John Denker <jsd@av8n.com>

**Re: [Phys-L] Rotational Motion Pulley Question***From:*Brian Whatcott <betwys1@sbcglobal.net>

**Re: [Phys-L] Rotational Motion Pulley Question***From:*Jeffrey Schnick <JSchnick@Anselm.Edu>

- Prev by Date:
**Re: [Phys-L] Rotational Motion Pulley Question** - Next by Date:
**[Phys-L] Science Fairs and a classical example of the "Scientific Method"** - Previous by thread:
**Re: [Phys-L] Rotational Motion Pulley Question** - Next by thread:
**[Phys-L] Science Fairs and a classical example of the "Scientific Method"** - Index(es):