# Introductory mechanics problem

1. Dec 30, 2005

### Benny

Hi, can someone help me with the following question?

Q. A particle is suspended between two cylinders each of radius p (rho) by means of a massless rope of length l. It's instantaneous position can be determined by the angular displacement a (alpha). Motion of the particle is restricted to the vertical plane XY of an inertial system of coordinates.

To visualise the situation picture two circles which are tangent to each other. 'Squeezed' in between these two circles at the point of tangency is the rope of length l. The particle is attached to the 'bottom' of the rope. The positive X axis is vertically downwards and the positive Y axis is horizontally to the right.

As for the part about the particle's displacement being determined by the angle alpha; place the usual 2D cartesian coordinate system (I am purposely not denoting the axes by pronumerals so as to avoid ambiguity) on the circle on the 'right' such that the origin of that coordinate system coincides with the centre of that circle. The angle between the 'negative' horizontal axis and a straight line through the origin which is negative to the left of the origin, is alpha.

So basically the rope is wrapped around one of the circles (the one on the right) and the extent of the 'wrap' is indicated by the angle alpha. In other words, the length of the rope which is wrapped around the circle is pa (rho multiplied by alpha) - which is the usual formula for arc length of a circle.

Sorry about the lack of a diagram.

Anyway the problem is to determine the components of the absolute velocity and acceleration of the particle along the interial system of coordinates as a function of the angular displacement alpha.

I need help with determining the X and Y components of the displacement. Once I have that I can just differentiate to get the velocity and acceleration.

For a quick check I will include the answer for the velocity components.

$$v_X = \rho \mathop \alpha \limits^. \cos \left( \alpha \right) - l\mathop \alpha \limits^. \sin \left( \alpha \right) + \rho \alpha \mathop \alpha \limits^. \sin \left( \alpha \right)$$

$$v_Y = - \rho \mathop \alpha \limits^. \sin \left( \alpha \right) + l\mathop \alpha \limits^. \cos \left( \alpha \right) - \rho \alpha \mathop \alpha \limits^. \cos \left( \alpha \right)$$

I can't make substantial progress in determining the X and Y components of position of the particle. Initially the angle alpha is zero so that the rope is vertically downwards. I was thinking that for the X component would be the difference between the length of the rope L and some function of arc length.

But not much is making sense at the moment since when I look at this problem, I don't see any right angled triangles so the basic trig relations don't seem to work. Can someone please help me get started? Any help would be good thanks.

Edit: I am familiar with the use of intrinsic (tangential, normal and binormal components) coordinates. So I don't necessarily need to work out the displacement components if is it easier to avoid doing so.

Last edited: Dec 30, 2005