Continuing from the following scenario from my previous question Centripetal force of a rotating rigid body? :
Consider someone pushing a roundabout in a playground. Initially the roundabout is stationary, but when it is pushed, it rotates with increasing rotational speed.
The force of the push is balanced by the reaction force exerted by the support at the centre of the roundabout. The forces are equal in magnitude and opposite in direction, so the roundabout is in translational equilibrium. But they have different lines of action, so there is a resultant torque, causing the playground to rotate and have angular momentum.
Okay, suppose the roundabout's rotational speed now stabilises (ie. the resultant torque becomes zero and the roundabout is in rotational equilibrium). I infer that this happens only when the pushing force is removed (otherwise there would be a resultant torque as described in the yellow box). But if so, what (force) is keeping the roundabout rotating at its constant rotational speed (assuming no friction)? Isn't circular/rotational motion a forced motion??
Answer
Roundabout has angular momentum [as well as kinetic (rotational) energy] , which cannot just disappear. So you need some negative torque (e.g. frictional torque) [or negative work of torque] to change angular momentum [kinetic energy] in order to stop roundabout.
$$\vec{\tau} = \frac{\text{d}\vec{L}}{\text{d}t}$$ (rotational form of 2nd Newton law)
It is an analogous question, why a particle keeps moving. It keeps moving because it has momentum [as well as kinetic (linear) energy]. So you need some negative force (e.g. kinetic friction) [or negative work of force] to change momentum [kinetic energy] in order to force particle to stop.
$$\vec{F} = \frac{\text{d}\vec{p}}{\text{d}t}$$ (linear form of 2nd Newton law)
If differential forms of 2nd Newton law present a problem, you can also use non-diferential:
$$\vec{F} = m \vec{a}, \vec{\tau} = I \vec{\alpha}$$
But beware, the second (rotational form) is valid only for fixed axes. For non-fixed axes you have to use complex Euler's equations.
No comments:
Post a Comment