# 10.6: Translational Motion of the Center of Mass

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The velocity of the center of mass is found by differentiation,

$\overrightarrow{\mathbf{V}}_{\mathrm{cm}}=\frac{1}{m_{\mathrm{sys}}} \sum_{i=1}^{i=N} m_{i} \overrightarrow{\mathbf{v}}_{i}=\frac{\overrightarrow{\mathbf{p}}_{\mathrm{sys}}}{m_{\mathrm{sys}}} \nonumber$

The momentum is then expressed in terms of the velocity of the center of mass by

$\overrightarrow{\mathbf{p}}_{\mathrm{sys}}=m_{\mathrm{sys}} \overrightarrow{\mathbf{V}}_{\mathrm{cm}} \nonumber$

We have already determined that the external force is equal to the change of the momentum of the system (Equation (10.4.9)). If we now substitute Equation (10.6.2) into Equation (10.4.9), and continue with our assumption of constant masses $$m_{i}$$, we have that

$\overrightarrow{\mathbf{F}}^{\mathrm{ext}}=\frac{d \overrightarrow{\mathbf{p}}_{\mathrm{sys}}}{d t}=m_{\mathrm{sys}} \frac{d \overrightarrow{\mathbf{V}}_{\mathrm{cm}}}{d t}=m_{\mathrm{sys}} \overrightarrow{\mathbf{A}}_{\mathrm{cm}} \nonumber$

where $$\overrightarrow{\mathbf{A}}_{\mathrm{cm}}$$ the derivative with respect to time of $$\overrightarrow{\mathbf{V}}_{\mathrm{cm}}$$ is the acceleration of the center of mass. From Equation (10.6.3) we can conclude that in considering the linear motion of the center of mass, the sum of the external forces may be regarded as acting at the center of mass.

Example 10.4 Forces on a Baseball Bat

Suppose you push a baseball bat lying on a nearly frictionless table at the center of mass, position 2, with a force $$\overrightarrow{\mathbf{F}}$$ (Figure 10.7). Will the acceleration of the center of mass be greater than, equal to, or less than if you push the bat with the same force at either end, positions 1 and 3