Question

Consider the $L$ shaped bar shown in the figure. At the instant shown, the bar is rotating at $4rad/s$ and is slowing down at the rate of $2rad/s^2$
(a) Find the acceleration of point $A$
(b) Find the relative acceleration $a_{B/A}$ of point $B$ with respect to point $A$..

Answer 1

Acceleration of A will have 2 parts, tangential$a_t$ and radial$a_r$.

$a_t=\alpha \times r=-2\times \left(2\right)=-4m/s^2$ -ve sign is because it is slowing down

$a_r=r\times {\omega }^2=32m/s^2$

total a$=\sqrt{{32}^2+4^2}=32.25m/s^2$

Similarly acceleration of B will have 2 parts.

$a_t=\alpha \times r^{{}^{\prime }}=-2\times \left(2\sqrt{2}\right)=-4\sqrt{2}m/s^2$

$a_r=r^{{}^{\prime }}\times {\omega }^2=32\sqrt{2}m/s^2$

total a$=\sqrt{({32}^2\times 2)+(4^2\times 2)}=45.61m/s^2$

This is original acceleration of B

But we have to take A as the reference frame which is non inertial.

In this a pseudo acceleration acts in opposite direction.

Hence it is 45.61-35.25= $13.36m/s^2$