The discussion centers on the relationship between torque and angular momentum in the context of a cone rolling on its side. The participants clarify that torque, represented as ##\tau = r \times F##, can act in a direction opposite to the change in angular momentum ##\Delta L##, which is directed in the +y direction when the cone's center of mass moves in the -y direction. The confusion arises from the need to consider both the angular momentum about the apex and the center of mass, leading to a deeper understanding of how forces and torques interact in rotational motion.
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No the normal force on the back wheel doesn't increase. The downwards force exerted by the back wheel decreases to zero as a consequence of the 'upwards' torque exerted on the bicycle frame by the brake pads that are locked on the front wheel.Happiness said:Suppose we analyse a braking bicycle such that a sudden brake is applied to its front wheel, making it stop completely. The bicycle will then flip forward. The back wheel will be lifted up and the bicycle will undergo circular motion about the front wheel before it topples. The back wheel is on the outside of this vertical circle (or semi-circle). So do you mean that the normal force on the back wheel, ##N_b## is greater than that on the front wheel, ##N_f##?
mfb said:there is a counter-torque from the normal force (+gravity), which is stronger at the outside compared to a cone that is at rest.
rcgldr said:So I'm wondering if the vertical reaction forces of the flat surface to any imbalance in the vertical forces from the cone cancel any net "outward" torque exerted on the cone related to the radial forces, and if so, if it's mathematically possible to show this is the case.
I'm wondering if the normal force related counter-torque exactly cancels the radial (centripetal acceleration) related torque.mfb said:I don't understand your post.