The area element in plane polar coordinates is . Thus the axial moment of inertia is
For the removed material, the angular width is , and
Its moment of inertia is therefore , leaving
The torque equation gives constant angular acceleration, so the time to reach the prescribed angular speed from rest is
To find the centre of mass after removing material, let point along the bisector of the missing sector in the disc. The original disc has zero first mass moment about its centre. The missing sector has zero transverse first moment by reflection symmetry, while its component along is
because the radial integral is . The remaining first moment is . Dividing by the given mass gives the body-fixed centre of mass
It lies away from the missing sector. In the inertial frame, let be the moment the applied torque stops and let give the orientation of that sector's bisector. Then, with ,
The centre of mass has inward centripetal acceleration of magnitude . The required real force is the constraint reaction exerted by the fixed rod and its support. Its horizontal resultant is ; a zero applied axial torque does not imply a zero resultant force.
For a freely rolling rigid body, decompose the kinetic energy into translation of its centre of mass G and rotation about G. The centre of mass after removing material lies on the side opposite H:
Another use of the parallel axis theorem gives
Thus at the same angular speed,
This compares the rotational component, not the total rolling kinetic energy. The total also contains and need not be smaller than . Describing all rolling motion as instantaneous rotation about its contact point includes that translational contribution; it is not the rotational component intended by this comparison.