Force applied on suspending/falling mass

AI Thread Summary
A falling rod subjected to a horizontal force at one end will experience torque, with the axis of rotation determined by the specific conditions rather than being fixed at the center of mass. For a non-hinged body like a flipping car, the axis of rotation does not always pass through the center of gravity, and identifying the correct axis requires analysis of the forces and moments acting on the body. The net torque about any axis is independent of other axes, allowing for calculations of net rotation over time. The relationship between torque, moment of inertia, and angular acceleration is analogous to linear motion equations. Understanding these principles is crucial for analyzing rotational dynamics effectively.
jakesee
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Hi,

May I ask, if there is a sufficiently long rod falling vertically downwards due to gravity, and then at an instance, a horizontal force is applied at one end of the rod, will this force create a torque on the rod? If so, where is the axis of rotation?

Second question, if a body, not hinged, is experiencing rotation, e.g. a flipping car in "mid air" in a collision, does the axis of rotation ALWAYS go through its center of gravity. If not, how do we find out which axis?.

Third question, say, in the flipping car example, is the "Net Torque" about any axis independant of any other axis? Is there a method to calculate the "net rotation about net axis" so that we can integrate over time to get the resultant orientation? Similar to integrating velocity vector to get position vector kind of maths?


Sorry, that's aquite abit. Hope I can get some help. Thanks for helping.
 
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Hi jakesee! :smile:

Yes, if there's a torque (a moment) about the centre of mass, then the body will rotate.

The position of the axis of rotation depends on the exact figures (it won't usually be the centre of mass).

The total angle of rotation doesn't depend on the axis. You can get the dynamics from torque = moment of inertia times angular acceleration, just like the linear F = ma.

See http://en.wikipedia.org/wiki/Instant_centre_of_rotation" for some details. :wink:
 
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Thanks for the replies, I'll be reading the suggested topics a while before coming back. thanks thanks. =)
 

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