Question

A ring of radius $2\text{m}$ performs combined translational and rotational motion on a frictionless horizontal surface with an angular acceleration $4{\text{rad/s}}^2$ and the acceleration of its centre $a=4{\text{m/s}}^2$ as shown in figure. Find the acceleration of point $D$.

• A. $8\hat{i}{\text{m/s}}^2$
• B. $(4\hat{i}+8\hat{j}){\text{m/s}}^2$
• C. $(4\hat{i}-4\hat{j}){\text{m/s}}^2$
• D. $8\hat{j}{\text{m/s}}^2$

Answer 1

The correct option is B $(4\hat{i}+8\hat{j}){\text{m/s}}^2$

Radius of ring, $r=2\text{m}$
From right hand thumb rule, $+vez-$axis comes out perpendicularly outward to $(x-y)$ plane.
$\Rightarrow \overrightarrow{\alpha }=-4\hat{k}{\text{rad/s}}^2$



Acceleration of centre of ring ${\overrightarrow{a}}_O=4\hat{i}{\text{m/s}}^2$
For net acceleration of point $D$,
$\overrightarrow{a_D}=\overrightarrow{a_O}+{\overrightarrow{a}}_{DO}...\left(i\right)$
${\overrightarrow{a}}_{DO}=\overrightarrow{\alpha }\times {\overrightarrow{r}}_{DO}$
Here, ${\overrightarrow{r}}_{DO}=-2\hat{i}\text{m}$
Substituting in Eq $\left(i\right)$ we get,
${\overrightarrow{a}}_D=4\hat{i}+\left[\right(-4\hat{k})\times (-2\hat{i}\left)\right]$
${\overrightarrow{a}}_D=4\hat{i}+8(\hat{k}\times \hat{i})$
$\therefore {\overrightarrow{a}}_D=(4\hat{i}+8\hat{j}){\text{m/s}}^2$