Roly-Poly toy

Center of mass for this bodies lies exactly at the body. In what ever direction it is pushed centre of mass position is not changed. Which means centre of mass displacement is 0. This implies all the bodies will come to their initial position.

$x_{cm} = \dfrac{m_1x_1 + m_2x_2 + ...}{m_1 + m_2 + ...}$ Where $x_{cm}$ is the change in position of centre of mass which is 0. $\implies m_1x_1 + m_2x_2 + ... = 0$ This says that the sum of product of masses and change in position of corresponding bodies is equal to $0$ . This implies that all the bodies' have reached their initial positions.

WARNING: THIS IS ONLY BY USING COMMON SENSE.

The mass can't be distributed evenly because it would be impossible for it to rebound. If most of the mass is at the top then it would stay wobbling down and cannot rebound (or bounce back) and thus the answer is it is at the bottom because then it [the top] could just move while the bottom is trying to keep the toy still and the bottom is the reason why it [the toy] stops after a while.

2 points (from Chris Cheek) : - if the mass were all at the top then it would also flip upside down and remain there (point 1) and the reason that it finally comes to rest is due to friction and the resulting loss of momentum that was imparted by the punch (point 2)

It can be shown that the center of mass is the point such that the resulting external force and torque applied to it produces the same effect as the application of all those external forces to the body.

This allows us to consider the force of gravity as a vector pointing towards the ground applied to the center of mass. Let's conduct our analysis choosing the point of contact with the ground as the origin (this allows us to avoid considering the normal force).

In case of point A (in the situation showed), this means that the body will experience a torque that makes it rotate counterclockwise , thus making the situation worse. Point C is the only one that is exerting a finite torque that restores the body to its initial position: point B might do the same job as point A with only a slight fluctuation.

Because of the form of the toy's base, point C (or a point that is a little lower) will always exert a restoring torque.

Note: directions of rotation are given by the right hand rule .

It’s evil; it rises up each time by the strength of its purely malevolent hatred of all humanity.

It's simply some common sense. If all the mass will be distributed at the bottom because of gravity or the greatest force of attraction will only the towards the earth (9.8m/s). Now, if you punch the object, the mass will be at rest at the moment but while it goes down, there are some disturbance to the matter pulling it upwards but gravity uncle will pull the matter back down again no matter what ! If you didn't understood then listen, First, the mass is distributed at the bottom keeping it in stationery motion. Now, if you punch it backwards you are actually trying to distribute the mass at the top while there is more mass at the bottom. So the mass pulled upwards is pulled downwards again which leads the object to get back in straight position.

This is a basic problem. Thank about it. If you have EVER lift one of these things up, which side is heavier? The bottom. You don’t need super math skills to solve this, just basic knowledge.

The centre of mass is at the bottom of the toy this makes the toy more stable. The centre of mass is at the bottom because the lower surface of the toy has a larger surface area than the top, sort of like a cone.

The object is stable (returns to equilibrium), as long as moving away equals center of mass having to rise (beside from moving sideways). This happens when the center of mass lies below the center of the smallest curvature in the contact point. So actually a combination of a low center of mass and a blunt underside are needed. This phenomenon also happens with a classical guitar.

An object is in balance as long as its centre of gravity lies in or above the area of the zone where it touches the ground. This can only happen when (most of) the mass is in the bodom, whereas the centre of gravity is as low as possible.

Think about torque about the contact point with the ground. If the mass is distributed only at the top or evenly distributed, a considerable torque exists due to gravity (because a "considerably" large lever arm exists) that prevents the rebound of the object. Only plausible choice is C.

This sounds so much like a meme $\frac{X}{D}$

For your amusement : https://youtu.be/JEKLN2V5_gI

The toy's centre of mass would have to be in the base, lower down than its widest part. In any position, the toy's weight will effectively act downwards at the centre of mass. When the toy is in contact with the ground, a couple forms consisting of the toy's weight and the normal reaction from the ground. When the centre of mass is suitably placed, this couple will always be oriented in such a direction as to rotate the toy back to the upright position.