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 Setup and Geometry:
From the given image, we can establish a Cartesian coordinate system with the origin at point Q(0, 0).
- The point P is located along the vertical y-axis at a distance of 12 units:
$r_P=12j$
- The point R is located along the horizontal x-axis at a distance of 16 units:
$r_R=16i$
- Therefore, the position vector of R relative to P is:
$r_{R/P}=r_R-r_P=16i-12j$

2. Relative Acceleration Relationship:
For a rigid body in plane motion, the acceleration of point R relative to point P is given by:
$a_{R/P}=\alpha \times r_{R/P}-{\omega }^2{r}_{R/P}$
where:
- $\alpha =\alpha k$ is the angular acceleration vector.
- $\omega$ is the angular velocity of the rigid body.
Let's compute the cross product term:
$\alpha \times r_{R/P}=\left(\alpha k\right)\times (16i-12j)=12\alpha i+16\alpha j$
Now, compute the centripetal acceleration term:
$-{\omega }^2{r}_{R/P}=-{\omega }^2(16i-12j)=-16{\omega }^{2}i+12{\omega }^{2}j$
Combining these gives the relative acceleration vector:
$a_{R/P}=(12\alpha -16{\omega }^2)i+(16\alpha +12{\omega }^2)j$

3. Solving for Angular Motion Parameters:
We are given that the relative acceleration is:
$a_{R/P}=10\angle 180°=-10i$
Equating the components:
$12\alpha -16{\omega }^2=-10$ (Equation 1)
$16\alpha +12{\omega }^2=0$ (Equation 2)
From Equation 2, we can express ${\omega }^2$ in terms of $\alpha$:
${\omega }^2=-\frac{4}{3}\alpha$
Substituting this into Equation 1:
$12\alpha -16(-\frac{4}{3}\alpha )=-10$
$12\alpha +\frac{64}{3}\alpha =-10$
$\frac{100}{3}\alpha =-10\Rightarrow \alpha =-0.3{\text{rad/s}}^2$
Then, solving for ${\omega }^2$:
${\omega }^2=-\frac{4}{3}(-0.3)=0.4{\text{(rad/s)}}^2$

4. Acceleration of Point R:
We are given that the instantaneous acceleration of point Q is zero ($a_Q=0$). Therefore, the acceleration of R is simply:
$a_R=a_Q+\alpha \times r_{R/Q}-{\omega }^2{r}_{R/Q}$
Since $r_{R/Q}=16i$:
$a_R=\left(\alpha k\right)\times \left(16i\right)-{\omega }^2\left(16i\right)=-16{\omega }^{2}i+16\alpha j$
Substituting $\alpha =-0.3$ and ${\omega }^2=0.4$:
$a_R=-16\left(0.4\right)i+16(-0.3)j=-6.4i-4.8j{\text{m/s}}^2$

5. Magnitude and Direction of a_R:
- Magnitude:
$\left|a_R\right|=\sqrt{{(-6.4)}^2+{(-4.8)}^2}=\sqrt{40.96+23.04}=\sqrt{64}=8{\text{m/s}}^2$
- Angle with the positive x-axis (since both components are negative, the vector lies in the third quadrant):
$\theta =180°+\arctan \left(\frac{4.8}{6.4}\right)=180°+36.87°\approx 217°$
Therefore, the acceleration of point R is 8∠217°.