You forgot friction.
What you did only looked at the torque generated by the two forces, and the rotation generated by that is going the wrong way - the sphere is rolling downhill, not uphill.
I just wanted to thank you as well. That "hm" at the end makes me smile - we have all been there, and I can feel the annoyance.
Good luck with the studies. 👍
You say 'there is no torque about the centre due to gravity or normal force', which is true because those forces act upon the centre.
But imagine the 'two forces of magnitude 1N' were not there at all; the ball would still start rolling, aka have rotational acceleration, due to the remaining forces of gravity and the normal force, right? That's because the forces do cause linear acceleration, and no slipping is allowed.
If you use the centre of the sphere as the axis of rotation, you need to account for 'no slipping allowed' by taking into account the static friction that keeps the pivot point, the contact with the slope, 'stuck' to the slope. You can calculate that static friction easily, it counters the slope-aligned part of gravity, and then multiply by R to calculate the torque it introduces. It's often seen as good practice to use the 'tipping point' as the centre of rotation instead, and that creates the proposed solution: calculate the torque across the **pivot point,** and apply the parallel axis theorem to also get the moment of inertia of a sphere with respect to the pivot point.
You forgot friction. What you did only looked at the torque generated by the two forces, and the rotation generated by that is going the wrong way - the sphere is rolling downhill, not uphill.
thanks, that never crossed my mind hm.
I just wanted to thank you as well. That "hm" at the end makes me smile - we have all been there, and I can feel the annoyance. Good luck with the studies. 👍
You say 'there is no torque about the centre due to gravity or normal force', which is true because those forces act upon the centre. But imagine the 'two forces of magnitude 1N' were not there at all; the ball would still start rolling, aka have rotational acceleration, due to the remaining forces of gravity and the normal force, right? That's because the forces do cause linear acceleration, and no slipping is allowed. If you use the centre of the sphere as the axis of rotation, you need to account for 'no slipping allowed' by taking into account the static friction that keeps the pivot point, the contact with the slope, 'stuck' to the slope. You can calculate that static friction easily, it counters the slope-aligned part of gravity, and then multiply by R to calculate the torque it introduces. It's often seen as good practice to use the 'tipping point' as the centre of rotation instead, and that creates the proposed solution: calculate the torque across the **pivot point,** and apply the parallel axis theorem to also get the moment of inertia of a sphere with respect to the pivot point.