Question

A solid sphere of radius $r=2\text{m}$ is under a combination of rotational and translational motion as shown in figure.


If acceleration of COM $\left(O\right)$ is $a_o=2{\text{m/s}}^2$ and horizontal component of acceleration at point $A$ is $a_A=6{\text{m/s}}^2$, then find angular acceleration $\alpha$.

• A. $4{\text{rad/s}}^2$
• B. $6{\text{rad/s}}^2$
• C. $2{\text{rad/s}}^2$
• D. $1{\text{rad/s}}^2$

Answer 1

The correct option is C $2{\text{rad/s}}^2$

Given:
$a_A=6{\text{m/s}}^2;a_0=2{\text{m/s}}^2$
As we know that,
${\overrightarrow{a}}_A={\overrightarrow{a}}_{COM}+(\overrightarrow{\alpha }\times {\overrightarrow{r}}_{OA})$
$\Rightarrow {\overrightarrow{a}}_A={\overrightarrow{a}}_0+(\overrightarrow{\alpha }\times {\overrightarrow{r}}_{OA})$
$\Rightarrow {\overrightarrow{a}}_A-{\overrightarrow{a}}_0=\overrightarrow{\alpha }\times {\overrightarrow{r}}_{OA}....\left(i\right)$
Here,
${\overrightarrow{a}}_A=6\hat{i}{\text{m/s}}^2$
${\overrightarrow{a}}_0=2\hat{i}{\text{m/s}}^2$
$\overrightarrow{\alpha }\times {\overrightarrow{r}}_{OA}=(-\alpha \hat{k})\times \left(2\hat{j}\right)$
${}^{\prime }{\alpha }^{\prime }$ being the magnitude of angular acceleration.
$\Rightarrow \overrightarrow{\alpha }\times {\overrightarrow{r}}_{OA}=-2\alpha (\hat{k}\times \hat{j})=+2\alpha \hat{i}{\text{rad/s}}^2$
On substituting in Eq.$\left(i\right)$ we get,
$6\hat{i}-2\hat{i}=2\alpha \hat{i}$
On equating the coefficients of $\hat{i}$ both sides:
$\Rightarrow 4=2\alpha$
Or, $\alpha =2{\text{rad/s}}^2$
$\therefore$ option $\left(c\right)$ is correct.

Alternate solution:
We know that acceleration of top most point in rolling motion is
$a_A=a_0+r\alpha$
$\because$ The tangential acceleration $\left(r\alpha \right)$ of point $A$ and acceleration of centre of mass$\left(a_0\right)$, both are along the same direction (i.e $+vex-$ direction).
$\Rightarrow 6=2+2\alpha$
$\therefore \alpha =2{\text{rad/s}}^2$