Question

A square made of $4$ uniform rods each of mass $m=1/4kg$ and length $l=1m$ each, is hinged about mid point $O$ of $AB$ as shown and is lying on a smooth horizontal table. Two forces $F=154N$ act on the square in the direction shown. They are always perpendicular to the rods $AB$ and $AD$. Find the angular velocity (in rad/s) of the square by the time it rotates by and angle of $\pi /2$
(Take $\pi =\frac{22}{7}$)

Answer 1

Since this is angular motion we need to talk in terms of angular quantities only.

Torque is the angular form of force. It plays the same role in angular motion as force does in linear motion.

The formula for torque is force$\times$perpendicular distance between the point of application of force and the point of axis of rotation. Remember the word perpendicular.

The first force produces a torque of $154\times OA=154\times 0.5$

The second produces a torque of $154\times AD=154\times 1$ Note that it is AD and not OD $\because$ it should be perpendicular.

Since both the torques produces rotation in the same direction (anticlockwise) they get added up ordinarily.

$total$ $torque=\tau =154\times 1.5$

In order to calculate the angular velocity we need to find out the angular acceleration $\alpha$.

$\alpha =\frac{\tau }{I}$ where $I$ is the moment of inertia of the assembly of rods.

$I$ of rod AB=$I_{AB}=\frac{m{l}^2}{12}$

$I$ of AD about its centre $=I_{{}_{CAB}}=\frac{m{l}^2}{12}$

$\therefore I_{AD}$ about $O$ can be calculated by the theorem of parallel axes. $I=I_C+M{h}^2$

By pythagoras theorem, distance between O and the midpoint of AD=$\frac{1}{\sqrt{2}}$

$\therefore I_{AD}=\frac{m{l}^2}{12}+\frac{m}{2}$

Similarly $I_{BC}=\frac{m{l}^2}{12}+\frac{m}{2}$

Distance between O and midpoint of DC is 1metre.

$\therefore I_{DC}=\frac{m{l}^2}{12}+m$

Total $I$ =$I_T=I_{AB}+I_{DC}+I_{AD}+I_{BC}$

$I_T=\frac{m{l}^2}{3}+2m=\frac{1}{12}+\frac{1}{2}=\frac{7}{12}$ $kg{m}^2$

Kinematic Equations can be applied for rotational motion in a similar way

${\omega }^2=2\alpha \theta$ just as ${{}_V}^2=2as$

$\alpha =\frac{154\times 1.5}{\left(\frac{7}{12}\right)}=11\times 3\times 12$

$\therefore {\omega }^2=2\times 11\times 3\times 12\times \left(\frac{\pi }{2}\right)=2\times 11\times 3\times 12\times \frac{11}{7}$

$\omega =\sqrt{\frac{2}{7}\times {11}^2\times 6^2}$

ANSWER:- $66\times \sqrt{\frac{2}{7}}$ radians.