Question

In a rigid body in plane motion, the point R is accelerating with respect to point P at $10\angle {180}^{o}m/s^2$. If the instantaneous acceleration of point Q is zero acceleration (in m\s${}^2$) of point R is

• A. $8\angle {233}^o$
• B. $10\angle {225}^o$
• C. $10\angle {217}^o$
• D. $8\angle {217}^o$

Answer 1

The correct option is D $8\angle {217}^o$


As acceleration of point Q is zero, therefore we can assume this rigid body PQR is hinged at Q at this instant.

${\overrightarrow{a}}_{RP}={\overrightarrow{a}}_R-{\overrightarrow{a}}_P$

It is given as $10m/s^2$ at an angle of ${180}^o$, that means only radial acceleration is there at that instant and reference is PR.

$a_{RP}=\left(RP\right){\omega }^2=10$

$\Rightarrow 20{\omega }^2=10$

$\Rightarrow \omega =\frac{1}{\sqrt{2}}$

Since point R has only radial acceleration, so total acceleration is only radial. therefore tangential acceleration is zero.

${\alpha }_{body}=0$

$a_R=QR\left({\omega }^2\right)=16\times \frac{1}{2}=8m/s^2$

and will be in the horizontal backward direction,by our reference is along PR, so the angle of it from reference is ($180+\theta$) from $\Delta PQR$

$tan\theta =\frac{12}{16}$

$\Rightarrow \theta ={36.8698}^o$

$\text{So},180+36.8698=216.8698\underset{–}{\sim }{270}^o$
So the answer is $8\angle {217}^o$