Question

The assembly of four masses connected by rigid mass-less rods is kept on a smooth horizontal floor as shown in the figure. Under the applied force $2F$, the magnitude of angular acceleration of the assembly at the instant shown is:

• A. $\dfrac{F}{mc}$
• B. $\dfrac{F}{2mc}$
• C. $\dfrac{2F}{3mc}$
• D. $\dfrac{F}{3mc}$

Hint:

When masses are connected by massless rods, only point masses contribute to the moment of inertia. Always sum $mr^2$ of individual masses to find $I$.

Answer 1

The Correct Option is B

The assembly consists of four point masses, each of mass $m$, located at a distance $c$ from the central point $O$ along horizontal and vertical rods. Since the rods are massless, the moment of inertia of the system about $O$ is:
\[ I = m c^2 + m c^2 + m c^2 + m c^2 = 4mc^2 \] The external force acting is $2F$ applied horizontally at the top mass. This force produces a torque about point $O$. The moment arm is the perpendicular distance $c$, so the torque is:
\[ \tau = (2F)(c) = 2Fc \] Using rotational dynamics, the angular acceleration is:
\[ \alpha = \frac{\tau}{I} = \frac{2Fc}{4mc^2} = \frac{F}{2mc} \] Thus, the angular acceleration of the assembly is:
\[ \alpha = \frac{F}{2mc} \]