Question

Alls Sams
EXERCISE 6.4
Motion of Rigid body, Rotational variables, Relation between linear and angular
variables, Rigid body constraint
Spe. 21. Consider the 'L' shaped bar shown in figure. At the instant shown, the bar is rotating at 4 rad/s and is slowing
rod down at the rate of 2 rad/s2.
y-axis
В
2 m
Motor
X-axis
2m-
a. Find the acceleration of point A.
b. Find the relative acceleration as of point B with respect to point A.​

Answer 1

Answer:

The acceleration of A is 32.25 m/s²

The relative acceleration of B w.r.t. point A is 13.4 m/s²

Explanation:

The question is:

Consider the L shaped bar as shown in the figure. At the instant shown, the bar is rotating at 4 rad/s² and is slowing down at the rate of 2 rad/s².

(a) Find the acceleration of point A

(b) Find the relative acceleration of point B with respect to point A.​

Solution:

The diagram is attached

At point A the acceleration will have two components, the tangential acceleration $a_t$ and teh radial acceleration $a_r$

If the angular acceleration is $\alpha$ then

$\alpha=-2$ rad/s²

Therefore, the tangential acceleration

$a_t=\alpha\times r$

$\implies a_t=-2\times 2$

$\implies a_t=-4$ m/s^2

The radial acceleration will be the centripetal acceleration, therefore

$a_r=\omega^2r$

Given the angular velocity $\omega=4$ rad/s

Thus

$a_r=4^2\times 2$

$\implies a_r=32$ m/s²

Thus total acceleration at A

$a_A=\sqrt{a_t^2+a_r^2}$

$\implies a_A=\sqrt{(-4)^2+(32)^2}$

$\implies a_A=\sqrt{4^2+4^2\times 8^2}$

$\implies a_A=4\sqrt{1+64}$

$\implies a_A=4\sqrt{65}$

$\implies a_A=32.25$ m/s²

Similarly for point B

The radial distance will be $\sqrt{2^2+2^2}$ or $2\sqrt{2}$

Thus the tangetial acceleration at B

$a_t=2\sqrt{2}\times(-2)$

$a_t=-4\sqrt{2}$

The radial acceleration will be

$a_r=4^2\times 2\sqrt{2}$

$\implies a_r=32\sqrt{2}$

Total acceleration at B

$a_B=\sqrt{a_t^2+a_r^2}$

$\implies a_B=\sqrt{(-4\sqrt{2})^2+(32\sqrt{2})^2}$

$\implies a_B=\sqrt{4^2\times 2+4^2\times 8^2\times 2}$

$\implies a_B=4\sqrt{2+128}$

$\implies a_B=4\sqrt{130}$

$\implies a_B=45.65$ m/s²

The relative acceleration of B with respect to A

$a_{B/A}=45.65-32.25$

$\implies a_{B/A}=13.4$ m/s²