Question

Prove that

$\large\rm { \sum \tau = I \alpha}$​

Answer 1

we know that

$\large\rm { \:\:\:\:\:\:\:\:\:\:\: \sum F = ma}$

$\large\rm { \: }$

$\large\rm { \:\:\:\:\:\:\:\:\:\:\: \sum r \times F = mr \times a}$

$\large\rm { \: }$

$\large\rm { \:\:\:\:\:\:\:\:\:\:\: \sum r \times F = mr \times ( \alpha \times r)}$

$\large\rm { \: }$

$\large\rm { \:\:\:\:\:\:\:\:\:\:\: \sum r \times F = m \alpha ( r \cdot r) - r ( r \cdot \alpha)}$

$\large\rm { \: }$

$\large\rm { \:\:\:\:\:\:\:\:\:\:\: \sum r \times F = m \alpha r^{2} - 0}$

$\large\rm { \: }$

$\large\rm { \:\:\:\:\:\:\:\:\:\:\: \sum r \times F = mr^{2} \alpha}$

$\large\rm { \: }$

$\large\rm { \:\:\:\:\:\:\:\:\:\:\: \boxed{\therefore \sum \tau = I \ \alpha}}$