Establish the relationship between Torque and Moment of Inertia.
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Consider a rigid body rotating about a given axis with a uniform angular acceleration a, under the action of a torque. Let the body consist of particles of masses $m_1, m_2, m_3 ... m_n$ at perpendicular distances $\text{r}_1,\text{r}_2,\text{r}_3\dots\text{r}_{\text{n}}$ respectively from the axis of rotation, (see Fig.). As the body is rigid, angular acceleration a of all the particles of the body is the same. However, their linear accelerations are different because of different distances of the particles from the axis. If $\text{a}_1,\text{a}_2,\text{a}_3\dots\text{a}_{\text{n}}$ are the respective linear accelerations of the particles, then $\text{a}_1=\text{r}_1\alpha,\text{a}_2=\text{r}_2\alpha,\text{a}_3=\text{r}_3\alpha$Force on particle of mass $m_1$ is $\text{f}_1=\text{m}_1\text{a}_1=\text{m}_1\text{r}_1\alpha$
Moment of this force about the axis of rotation $=\text{f}_1\text{r}=(\text{m}_1\text{r}_1\alpha)\times\text{r}_1=\text{m}_1\text{r}^2_1\alpha$ Similarly, moments of forces on other particles about the axis of rotation are $\text{m}_2\text{r}^2_2\alpha,\text{m}_2\text{r}^2_3\alpha\dots\text{m}_{\text{n}}\text{r}_{\text{n}}^2\alpha$ $\therefore$ Torque acting on the body, $\tau$ $=\text{m}_1\text{r}^2_1\alpha+\text{m}_2\text{r}^2_2\alpha+\text{m}_3\text{r}^2_3\alpha+\dots\text{m}_{\text{n}}\text{r}_{\text{n}}\alpha$ $=(\text{m}_1\text{r}^2_1+\text{m}^2\text{r}^2_2+\text{m}_3\text{r}^2_3+\dots\text{m}_{\text{n}}\text{r}^2_{\text{n}})\alpha$ $\tau=\Bigg(\sum\limits_{\text{i}=\text{l}}\limits^{\text{i}=\text{n}}\text{m}_{\text{i}}\text{r}^2_{\text{i}}\Bigg)\alpha$ $\text{or }\tau=\text{I}\alpha\text{ or }\vec{\tau}=\text{I}\vec{\alpha}$
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