Question

A uniform slender bar $AB$ of mass $m$ is suspended as shown from a small cart of the same mass $m$. Neglect the effect of friction. Then,

• A. magnitude of acceleration of point $A$ immediately after a horizontal force $F$ has been applied at $B$ is $\frac{16F}{5m}$
• B. magnitude of acceleration of point $A$ immediately after a horizontal force $F$ has been applied at $B$ is $\frac{2F}{5m}$
• C. magnitude of acceleration of point $B$ immediately after a horizontal force $F$ has been applied at $B$ is $\frac{16F}{5m}$
• D. magnitude of acceleration of point $B$ immediately after a horizontal force $F$ has been applied at $B$ is $\frac{2F}{5m}$

Answer 1

Answer: (B), (C)

The correct options are
B magnitude of acceleration of point $A$ immediately after a horizontal force $F$ has been applied at $B$ is $\frac{2F}{5m}$
C magnitude of acceleration of point $B$ immediately after a horizontal force $F$ has been applied at $B$ is $\frac{16F}{5m}$

Let the acceleration of bar and cart be a and $a^{\prime }$ respectively; also $\alpha$ be the angular acceleration of the bar.
Writing newton's equation of motion
For the bar: $F-F^{\prime }$ = $ma$ (i)
For the cart: $F^{\prime }$ = $m{a}^{\prime }$ (ii)
Torque equation for rod about centre of mass
For the rotation of the rod
$(F+F^{\prime })\frac{L}{2}$ = $\frac{M{L}^2}{12}\alpha$
$F+F^{\prime }$ = $\frac{ML\alpha }{6}$ (III)
The acceleration of point $A$ on the rod will be same as the acceleration of the cart
$\overline{a}={\overline{a}}_{A.C}+{\overline{a}}_C$
$-a^{\prime }=\alpha \frac{L}{2}-a$ or
$a=a-\frac{L}{2}\alpha$ (iv)
On solving equations (i)(II)(III) and (Iv), we get
$a$ = $\frac{7F}{5m}$ and
$\alpha$ = $\frac{18F}{5mL}$
$|a^{\prime }|$ = $|a_A|$ = $a-\frac{L}{2}\alpha$ = $-\frac{2F}{5m}$
$|a_B|$ = $a$ +$\frac{L}{2}\alpha$ = $\frac{16f}{5m}$