Forces of NatureDate: 10/9/95 at 13:38:56 From: Anonymous Subject: sliding inclined rod (Physics) Here is an elementary physics question which has been puzzling me. I know virtually no physics, but I do have a good grasp of vector mechanics, so if you could please phrase your answer in terms of mathematical vector operations (instead of using physics terminology) it would be much appreciated. Okay, my understanding is that any motion in a system initially at rest must be a result of Forces present in that system (and these forces in turn can all be expressed as vectors). Now, suppose we have an "almost upright" rod which touches the (frictionless) ground at the single point at its end and makes an angle Theta with the ground. As far as I can see, the only Force involved would be Gravity, which would be directed straight down. (Though there may also be some kind of "tension" Force, internal to the rod holding it together? In any case, these Forces would presumably be parallel to the rod and would not affect my question). Now my intuition tells me that what will happen is that eventually the rod will be flush on the ground and moving with a velocity parallel to the ground. Could you please explain all the forces which resolved here? I don't see how gravity alone (pointing downward) could result in sideways motion. Thanks. Date: 10/9/95 at 14:11:49 From: Doctor Andrew Subject: Re: sliding inclined rod (Physics) There is one more force. It is called the normal force and it pushes against the rod from the ground in the vertical direction (perpendicular to the ground). It keeps the rod from going through the ground. Its magnitude is equal and opposite to the vertical force the rod puts on it. You could construct a diagram which includes the rod, a vector representing gravity acting downward on the center of gravity of the rod, and a vector representing the normal force of the ground pushing on the end of the rod. The y position of the bottom of the rod is fixed and there is no friction, so the x position of the center of gravity of the rod is also fixed. The normal force exerts a torque on the rod causing it to rotate as it falls. The magnitude of the normal force depends on the mass and moment of inertia of the rod. You can set up an equation for the y-displacement and acceleration, and another one for the rotation and the rotational acceleration. The magnitude can be determined because the y-position of the center of the rod and the rotation of the rod are linked. (y = l/2 sin a) where y is the displacement of the rod, l is the length of the rod and a is the angle of the rod with the ground. The linear accelerations and velocities are tied with the rotational ones in the same way. Hope this helps. -Doctor Andrew, The Geometry Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/