Question

A uniform disc of radius $r$ spins with angular velocity $\omega$ and angular acceleration $\alpha$. If the centre of mass of the disc has linear acceleration $a$, find the magnitude and direction of acceleration of the points 1,2 and 3

Answer 1

The acceleration of point $1{\overrightarrow{a}}_C={\overrightarrow{a}}_{1C}+{\overrightarrow{a}}_C$

But ${\overrightarrow{a}}_{1C}={\left({\overrightarrow{a}}_{1C}\right)}_{tangentiak}+{\left({\overrightarrow{a}}_{1C}\right)}_{radial}$

$\Rightarrow {\overrightarrow{a}}_C=\alpha R\hat{i}+{\omega }^{2}R\hat{j};{\overrightarrow{a}}_{1C}=a\hat{i}$

Hence ${\overrightarrow{a}}_1=\left({\omega }^{2}R\right)\hat{i}+\alpha R\hat{j}$

Similarly the acceleration ${\overrightarrow{a}}_2={\overrightarrow{a}}_{2C}+{\overrightarrow{a}}_C$

Here ${\overrightarrow{a}}_{2C}=(\alpha R\hat{i}-{\omega }^{2}R\hat{j});{\overrightarrow{a}}_C=a\hat{i}$

Hence, ${\overrightarrow{a}}_2=(a+\alpha R)\hat{i}-{\omega }^{2}R\hat{j}$

Also acceleration of $3:{\overrightarrow{a}}_3={\overrightarrow{a}}_{3C}+{\overrightarrow{a}}_C$

Here ${\overrightarrow{a}}_{3C}=-{\omega }^{2}R\hat{i}-\alpha R\hat{j};{\overrightarrow{a}}_C=a\hat{i}$

Hence ${\overrightarrow{a}}_3=(a-{\omega }^{2}R)\hat{i}-\alpha R\hat{j}$