Consider a square block kept on a rough horizontal surface (friction is sufficient enough to prevent slipping/sliding). You apply a horizontal force F on the topmost point ,constant in magnitude and direction. The magnitude is adjusted so that the block doesn't gain significant kinetic energy. Due to rotation about the axis passing through the line of contact of the ground and block, the potential energy of the block increases as it loses contact with the surface. How does it gain potential energy? I'm confused which force is responsible for this gain.
Thanks in advance.
[Physics] Toppling of a Rigid body
rotational-dynamics
Best Answer
If I understood your question correctly, what gets the Center of Mass to rise (and hence gain potential energy) is the couple (torque) that F creates:
Just like what happens in an imperfectly balanced wheel (let's say that more mass is in the lower part), where you get enough torque to get the wheel to start rotating on its axis. In this case, similarly, the Center of Mass rises, you just need enough Torque to overcome the torque from the gravitational force (which creates a torque as soon as the wheel rotates because the CM isn't exactly on the axis of rotation).
In a rigid-body, staticity is not only from $\sum\limits_{i=1}^{n} F_i = 0$
but also from the sum of all present torques $\sum\limits_{i=1}^{n} R_i x F_i = 0$ which, in this case, is not happening
$\int\limits_{0}^{\theta_f} M_F(\theta) d\theta = W_{nonconservative} = \Delta E_{mec} = mgh + 1/2 I {\omega}^2$
with $h = lsqrt2 /2 - l/2$ and I from H.S. theorem
The integral actually solves out to constants, as $l\sqrt2 F$ applied along the arc of circumference