# Slope rotating around a vertical axis.

#### peripatein

This is NOT a HW question. I'd appreciate an explanation of the following:
I would like to determine the forces acting on a mass set between two springs of constant k on a slope (the slope's angle is alpha). The slope revolves around the vertical axis with angular velocity w and the mass could only move along the slope in between the springs. Please see attachment. Suppose I choose my axes so that my x-axis is parallel to the slope. While calculating the centrifugal and coriolis forces acting on the mass, only the components of omega vertical to my x-axis should be taken under consideration (the cross product would otherwise yield zero). However, aren't there two components of omega vertical to the x-axis (projection of omega on z as well as its projection on y)?

#### Attachments

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#### olivermsun

Do you consider the mass to be constrained so that it only moves in the xz-plane, or are the spring attachments free to rotate in any direction?

#### peripatein

Why x-z plane? The rotation is in the x-y plane (or r-theta if you will)! And the mass can only oscillate along the x-axis.

#### olivermsun

Okay, how about if I ask this way: are you calling your vertical axis y? (Your OP mentions the projection of omega on z).

#### peripatein

My coordinate system is simply rotated by an angle alpha counter-clockwise. See attachment.

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#### olivermsun

So your coordinate axes are rotating, with angular velocity ω parallel to the z-axis. The mass has to move in the positive-z direction to move outward (positive-x direction) while staying stuck to the slope, right?

#### peripatein

Has to move in the positive-z direction? Why? It can only move along my x-axis. I am not following.

#### olivermsun

It's a "slope," implying that dz/dx > 0 along the slope. So, if x changes then z must too. Unless I'm misunderstanding the problem, the mass is not allowed to move straight outward (in the positive x-direction) and through the slope surface.

The spring forces are also directed along the slope, by the way, as is the (net) force due to gravity.

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