Question
Starting from the definition of the position vector of the center of mass, show that$$\sum_{k=1}^n m_k\left(\mathbf{r}_k-\mathbf{r}\right)=\mathbf{0}, \quad \sum_{k=1}^n m_k\left(\mathbf{v}_k-\mathbf{v}\right)=\mathbf{0}$$Where were these identities used?
Step 1
For a system of n particles with masses m₁, m₂, ..., mₙ and position vectors r₁, r₂, ..., rₙ, the center of mass position vector r is defined as: $$\mathbf{r} = \frac{\sum_{k=1}^n m_k\mathbf{r}_k}{\sum_{k=1}^n m_k} = \frac{\sum_{k=1}^n Show more…
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