Torque opposite in direction to change in angular momentum?

[andrewkirk]

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, $N_f$?

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.

I'm not sure that that helps with the cone though.

 

[rcgldr]

there is a counter-torque from the normal force (+gravity), which is stronger at the outside compared to a cone that is at rest.

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 don't understand your post.

I'm wondering if the normal force related counter-torque exactly cancels the radial (centripetal acceleration) related torque.