Question

A uniform slender bar AB of mass m is suspended as shown from a small cart of the same mass m. Neglecting the effect of friction, determine the accelerations of points A and B immediately after a horizontal force F has been applied at B

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Let the acceleration of bar and cart be a and a' respectively; also α be the angular acceleration of the bar.

Writing newton's equation of motion

For the bar: F - F' =ma ...(i)

For the cart: F' = ma' ...(ii)

Torque equation for rod about center of mass

For rotation of the rod (F + F) $\frac{L}{2}=\frac{M{L}^2}{12}\alpha$

F+F'=$\frac{ML\alpha }{6}$ ...(iii)

The acceleration of point A on the rod will be same as the acceleration of the cart,

-a' $=\alpha \frac{L}{2}$-a or a' $=a-\frac{L}{2}\alpha$ ${\overrightarrow{a}}_A={\overrightarrow{a}}_{AC}+{\overrightarrow{a}}_c$ ...(iv)
On solving eqation (i), (ii), (iii) and (iv), we get, $\alpha =\frac{7F}{5m}and\alpha =\frac{18F}{5mL}$
|a'| $=\left|a_A\right|=a-\frac{L}{2}-\alpha -\frac{2F}{5m}$; $\left|a_B\right|=a+\frac{L}{2}\alpha =\frac{16F}{5m}$