In a rigid body in plane motion, the point R is accelerating with respect to point P at 10∠180° m/s² . If the instantaneous acceleration of point Q is zero, the acceleration (in m/s² ) of point R is

The correct option is 8∠217°.

1. Coordinate System and Geometry Setup:
By analyzing the diagram in the provided image:

• Point $Q$ is located at the origin of a Cartesian coordinate system, i.e., $Q(0,0)$.
• Point $P$ is along the positive $y$-axis, at a distance of 12 units: $P(0,12)$.
• Point $R$ is along the positive $x$-axis, at a distance of 16 units: $R(16,0)$.
• The hypotenuse connecting point $P$ and point $R$ has a length of 20 units, verifying that the triangle $PQR$ is a right-angled triangle since: ${12}^2+{16}^2=144+256=400={20}^2$.

The relative position vector of point $R$ with respect to point $P$ is:
${\overrightarrow{r}}_{R/P}=16\hat{i}-12\hat{j}$

2. Using Relative Acceleration Formula:
For a rigid body undergoing plane motion, the acceleration of point $R$ relative to point $P$ is given by:
${\overrightarrow{a}}_{R/P}=\overrightarrow{\alpha }\times {\overrightarrow{r}}_{R/P}-{\omega }^2{\overrightarrow{r}}_{R/P}$
where $\overrightarrow{\alpha }=\alpha \hat{k}$ is the angular acceleration and $\omega$ is the angular velocity of the rigid body.

Substituting the cross product:
$\overrightarrow{\alpha }\times {\overrightarrow{r}}_{R/P}=\left(\mathrm{��}\hat{k}\right)\times (16\hat{i}-12\hat{j})=12\alpha \hat{i}+16\alpha \hat{j}$

And the centripetal term:
$-{\omega }^2{\overrightarrow{r}}_{R/P}=-16{\omega }^2\hat{i}+12{\omega }^2\hat{j}$

Combining these gives:
${\overrightarrow{a}}_{R/P}=(12\alpha -16{\omega }^2)\hat{i}+(16\alpha +12{\omega }^2)\hat{j}$

3. Determining Angular Parameters:
We are given that the relative acceleration of point $R$ with respect to point $P$ is $10\angle {180}^{\circ }{\text{m/s}}^2$, which in vector form is:
${\overrightarrow{a}}_{R/P}=-10\hat{i}$

Equating the components:
1) $12\alpha -16{\omega }^2=-10$
2) $16\alpha +12{\omega }^2=0$

From equation (2):
$\alpha =-0.75{\omega }^2$

Substituting this value into equation (1):
$12(-0.75{\omega }^2)-16{\omega }^2=-10$
$-9{\omega }^2-16{\omega }^2=-10$
$-25{\omega }^2=-10\Rightarrow {\omega }^2=0.4{\text{rad}}^2/{\text{s}}^2$

Now, find the angular acceleration $\alpha$:
$\alpha =-0.75\times 0.4=-0.3{\text{rad/s}}^2$

4. Calculating the Absolute Acceleration of Point R:
We are given that the instantaneous acceleration of point $Q$ is zero (${\overrightarrow{a}}_Q=0$). The absolute acceleration of point $R$ is therefore:
${\overrightarrow{a}}_R={\overrightarrow{a}}_{R/Q}=\overrightarrow{\alpha }\times {\overrightarrow{r}}_{R/Q}-{\omega }^2{\overrightarrow{r}}_{R/Q}$

With ${\overrightarrow{r}}_{R/Q}=16\hat{i}$, we have:
${\overrightarrow{a}}_R=\left(\alpha \hat{k}\right)\times \left(16\hat{i}\right)-{\omega }^2\left(16\hat{i}\right)=-16{\omega }^2\hat{i}+16\alpha \hat{j}$

Substitute the values of ${\omega }^2=0.4$ and $\alpha =-0.3$:
${\overrightarrow{a}}_R=-16\left(0.4\right)\hat{i}+16(-0.3)\hat{j}=-6.4\hat{i}-4.8\hat{j}$

5. Computing Magnitude and Direction:
The magnitude of ${\overrightarrow{a}}_R$ is:
$\left|{\overrightarrow{a}}_R\right|=\sqrt{{(-6.4)}^2+{(-4.8)}^2}=\sqrt{40.96+23.04}=\sqrt{64}=8{\text{m/s}}^2$

The direction angle $\theta$ in the Cartesian plane (where both components are negative, putting it in the third quadrant) is:
$\theta ={180}^{\circ }+{\tan }^{-1}\left(\frac{4.8}{6.4}\right)={180}^{\circ }+{36.87}^{\circ }\approx {217}^{\circ }$

Thus, the acceleration of point R is 8∠217° m/s².