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  • There doesn't appear to be any force pushing the rider to make a wider turn, this obviously happens on the flat and can also happen on less steeply banked tracks.

    Would R(v) be the force at play?

  • Objects in equilibrium (ie no net forces) continue to go in the direction and speed they were already heading. Forces are only needed to change direction.

  • If the bike/rider was going round the bend but started to drift out/up the track wouldn’t that be a change in direction?

  • On the bend at speed, the rider is not in equilibrium - they are accelerating towards the centre of the turn radius. The centripetal force is R(h) in the diagram, the horizontal component of R.

    @M_V

    There doesn't appear to be any force pushing the rider to make a wider turn

    You could draw a force diagram with respect to the rider's own frame of reference while cornering, and then there would be an apparent centrifugal force. It's cancelled by R(h) if they stay on their line; if it's greater than R(h) they are drifting out/up.

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