# Bicycle Type Chain [Mechanics] Problem

1. Jun 2, 2004

### maverick280857

Hi guys..

I am a new member of physicsforums.com (I am a science student at the pre-undergrad level). I am unable to solve a problem involving motion of a chain along a circular track, in classical mechanics. The problem is from an old book on mechanics. Please read the problem (below) and offer your advice...

The problem:

"A flexible bicycle-type chain of length (1/2)*Π*r (0.5 * pie * radius) has a mass RHO per unit length and is released after being held by its upper end in an initial rest condition in the smooth circular channel. Determine the acceleration 'a' which all links experience just after release. Also find the expression for the tension T in the chain as a function of θ (theta)¸ for the condition immediately following release. Isolate a differential element of the chain as a free body and apply the appropriate motion equation."

The Diagram (drawn in Paint: the red dots represent the chain, the diag is otherwise vague and not as good as the original book version) is on the link,

I have gone as far as getting T (The tension) as a function of θ¸ but the function also includes 'a' which has to be eliminated.

Any ideas/solutions/approaches are appreciated...

Cheers
Vivek

---------------------------------------------
"Experiment without theory is
lame. Theory without
experiment is blind."

2. Jun 6, 2004

### maverick280857

For the Moderator

(FOR THE MODERATOR)

HI

I think I posted this in a wrong place...can you please shift it to the homework forum?

Cheers
Vivek

3. Jun 10, 2004

### maverick280857

Hi

I am including my solution so that you can help me better...

$$dm = \rho r d\phi$$

Forces

$$\sum dF_{\phi} = T(\phi + d\phi) + g dmsin\phi - T(\phi) = dma_{t}$$

which gives,

$$dT = dm(a_{t} - gsin\phi) = \rho r (a_{t} - gsin\phi)d\phi$$

Assuming

$$T(\phi = 0) = T(\phi = \frac{\pi}{2}) = 0$$

(I think there could be a mistake in the choice of these boundary values.)

$$\int_{0} ^{T(\phi)} dT = \rho r a_{t} \int_{0}^\phi d\phi - g\int_{0}^\phi sin\phi d\phi$$

or

$$T(\phi) = \rho r a_{t} \phi - \rho r g(1-cos\phi)$$

Also

$$T(\phi = \frac{\pi}{2}) = 0 \Rightarrow \rho r a_{t}\frac{\pi}{2} = \rho r g \Rightarrow a_{t} = \frac{2g}{\pi}$$

So substituting for the acceleration in the expression for tension, we have

$$T(\phi) = \rho r g(\frac{2\phi}{\pi} - (1 - cos\phi))$$

$$T(\phi) = \rho r g(\frac{2\phi}{\pi} - sin\phi)$$