Cut at center of mass

For the bat to be able to balance about its center of mass, the net moment due to both the pieces about it should be zero.

Let the centers of mass of the two pieces be at distance $l_1$ and $l_2$ from the center of mass of the whole bat. Looking at the diagram, we can say that $l_1 > l_2$ , since the mass in piece 1 is distributed at a greater distance on average compared to piece 2.

The moment of a force is given by $\text{Force} \times \text{Distance}$ . The moments due to the two pieces about the center of mass are $m_1 g \ l_1$ anti-clockwise, and $m_2 g \ l_2$ clockwise respectively. They cancel each other out, so their magnitudes are equal $m_1g \ l_1 = m_2 g \ l_2$ . This is equivalent to $\dfrac{m_1}{m_2} = \dfrac{l_2}{l_1}$ . Since $l_1 > l_2$ , we have $m_1 < m_2$ .

Hint: Moment = mass x length.

Suppose the length of piece 1 is L1, its mass is M1, and length of piece is L2 and mass is M2.

We know that the bat is cut through centre of mass,

=> Moment 1 = Moment 2

=> M1 x L1 = M2 X L2 ---------------- eq (i)

It is evident from the image above that L1 is greater than L2,

hence for eq (i) to be correct, M2 must be greater than M1.

Moment of mass has to be same for the either side about COM. As can be seen from the diagram the mass distribution on the left side is less denser but the distance from COM is more, thereby giving equal moment of mass.

The cut should be a curve in order to get to equal weight pieces.A straight cut would work if it the center of mass would be located in a cylindrical portion of the bat.

I dont know why other answers are going into moment of force etc. Its not wrong, but answer is really simple here, follows directly from weighted average definition.

The two bodies are now divided into $m_1$ and $m_2$ and have distances $l_1$ and $l_2$ from old centre of mass. For centre of mass to be in old position, we need $m_1l_1 = m_2l_2$ (not taking signs, its obvious here). So piece with more distance from centre of mass has lesser mass.

Since $l_2 < l_1$ so $m_2 > m_1$ .

co-ordinates of C.o.M in the $x$ component is $m_1$ $x_1$ $+$ $m_2$ $x_2$ $/$ $m_1$ $+$ $m_2$ . In the y component it is $m_1$ $y_1$ $+$ $m_2$ $y_2$ $/$ $m_1$ $+$ $m_2$ .Suppose the given point is taken as the origin(taking cartesian reference system).Then when we evaluate (more precisely $integrate$ because the mass distribution is continuous) the above formulas for each point (obviously points to the left be have negative $x$ points to the right will have positive $x$ ) we will get $(0,0)$ . But just a quick glance will tell us that $x_{left}$ $>$ $x_{right}$ so obviously $m_{right}$ $>$ $m_{left}$ .

Merely an analytical solution not very technical.

We could also use the torque/moment of the mass solution as given by other brilliantians but for C.o.M to coincide with C.G. there has to be uniform gravitational field intensity which has not been given in the question.