Rotational Equilibrium

Rotational Equilibrium

For Translational Equilibrium

$\sum \vec{F}=0$

For Rotational Equilibrium

• For rotational equilibrium of the system, the resultant torque acting on it must be zero.

i.e.,
$
\sum \tau=0
$

Various Cases of Equilibrium
$
\text { 1. } \sum \vec{F}=0 \text { and } \sum \vec{\tau}=0
$

Forces are equal and act along the same line.

The body will be in both Translational and Rotational equilibrium.

i.e., It will remain stationary if initially it was at rest.

$\text { 2. } \sum \vec{F}=0 \text { and } \sum \tau \neq 0$

Forces are equal and do not act along the same line.

Rotation of the body will happen i.e. spinning of the body.

$\text { 3. } \sum F \neq 0 \text { and } \sum \vec{\tau}=0$

Forces are unequal and act along the same line.

The body will be in Translational motion.

i.e., slipping of body
$
\text { 4. } \sum F \neq 0 \text { and } \sum \tau \neq 0
$

Forces are unequal and do not act along the same line.

The body will be in both Rotation and translation motion.

i.e. rolling of a body.

Couple Force

A couple is defined as a combination of two equal and oppositely directed forces but not acting along the same line.
$
\text { i.e., } \sum \vec{F}=0 \text { and } \sum \tau \neq 0
$

A torque by a couple is given by
$
\vec{\tau}=\vec{r} \times \vec{F}
$

In the case of a couple both forces are externally applied.

Work done by torque in twisting the wire is given by
$
W=\frac{1}{2} C \cdot \theta^2
$

Recommended Topic Video

Solved Examples Based on Rotational Equilibrium

Example 1: As shown in the diagram a ladder of mass M and length l is placed in equilibrium against a smooth vertical wall and a rough horizontal surface. If $\theta$ be the angle of inclination of the rod with horizontal then what is the normal reaction of the wall on the ladder

1) $\frac{1}{2} m g \cot \theta$
2) $\frac{1}{2} m g \tan \theta$
3) $m g \cos \theta$
4) $\frac{1}{2} m g$

Solution:

$\begin{aligned}
& \text { Equilibrium } \\
& \sum \vec{F}=0 \text { means Translational equilibrium } \\
& \sum \vec{\tau}=0 \text { means Rotational equilibrium }
\end{aligned}$

Draw FBD

condition of translational equilibrium

$\begin{aligned}
& \quad \sum_r F_x=0 \Rightarrow F_r-N=0 \\
& F_r=N \ldots(1) \\
& \text { Similarly } \sum_0 F_y=0 \\
& N_1-m g=0 \\
& N_1=m g \ldots .(2)
\end{aligned}$

Taking torque about the centre of the rod

and using $\sum \tau_c=0$
from (1) and (2)
$
\begin{aligned}
& N_1 \frac{l}{2} \cos \theta-f_r \frac{l}{2} \sin \theta-N \frac{l}{2} \sin \theta=0 \\
& m g \frac{l}{2} \cos \theta-N l \sin \theta=0 \\
& N=\frac{m g}{2 \tan \theta}=\frac{1}{2} m g \cot \theta
\end{aligned}
$

Example 2: An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in the figure. If AB= BC, and the angle made by AB with downward vertical is , then:

1) $\tan \theta=\frac{1}{2 \sqrt{3}}$
2) $\tan \theta=\frac{1}{2}$
3) $\tan \theta=\frac{2}{\sqrt{3}}$
4) $\tan \theta=\frac{1}{3}$

Solution:

This can be calculated by using either $\tau=r_1 F$ or $\tau=r \cdot F_1$ $r_1=$ perpendicular distance from the origin to the line of force.
$F_1=$ component of force perpendicular to the line joining force.

$
m g C 1 X 1=m g \frac{L}{2} \sin \theta
$
here $\mathrm{C} 1 \mathrm{X} 1$ is the perpendicular distance as shown in figure $\mathrm{mg} \mathrm{C}_2 \mathrm{x}_2=\left(\mathrm{S}_1 \mathrm{~S}_2-\mathrm{S}_2 \mathrm{C}_2\right) \mathrm{mg}=\mathrm{mg} \frac{L}{2} \cos \theta-\mathrm{mg} \mathrm{L} \sin \theta$, here $\mathrm{C} 1 \mathrm{X} 2$ is the perpendicular distance as shown in figure
$
\begin{aligned}
& \mathrm{mg} \mathrm{C}_1 \mathrm{x}_1=\mathrm{mg} \mathrm{C}_2 \mathrm{x}_2 \\
& m g \frac{L}{2} \sin \theta=m g \frac{L}{2} \cos \theta-m g L \sin \theta \\
& \Rightarrow \tan \theta=\frac{1}{3}
\end{aligned}
$

Example 3: A body mass $m=10 \mathrm{~kg}$ is attached to one end of a wire of length $0.3 \mathrm{~m}$. The maximum angular speed ( in rad s${ }^{-1}$ ) with which it can be rotated about its other end in space station is (Breaking stress of wire $=4.8 \times 10^7 \mathrm{Nm}^{-2}$ and area of cross-section of the wire $=10^{-2} \mathrm{~cm}^2$ ) is:

1) 4

2) 8

3) 6

4) 1

Solution:

$\begin{aligned}
\frac{F}{A} & =\frac{m v^2}{l A} \\
\Rightarrow v & =\sqrt{\frac{l F}{m}}=\sqrt{\frac{0.3}{10} \times 4.8 \times 10^7 \times 10^{-6}} \\
& =\sqrt{3 \times 48 \times 10^4 \times 10^{-6}}=1.2 \mathrm{~m} / \mathrm{s} \\
\omega & =\frac{v}{l}=\frac{1.2}{0.3}=4 \mathrm{rad} / \mathrm{s}
\end{aligned}$

Hence, the correct option is (1).

Example 4: $\mathrm{A}$ body is said to be in equilibrium if
1) $\sum \vec{F}=0$
2) $\sum \vec{F}=0$
3) both $\sum \vec{F}=0, \sum \vec{T}=0$
4) None of these

Solution:

For the equilibrium of the body, $\sum \vec{F}=0$ and $\sum \vec{T}=0$
Both translational and rotational equilibrium conditions are satisfied.
Hence, the answer is the option 3.

Example 5: Shown in the figure is a rigid and uniform one-meter-long rob AB held in the horizontal position by two strings tried to its ends and attached to the ceiling. The rod is of mass 'm' and has another weight of mass 2 m hung at a distance of 75 cm from A. The tension in the string at A is :

1) 0.5 mg

2) 2 mg

3) 0.75 mg

4) 1 mg

Solution:

As shown in the figure
The tension in the string at $\mathrm{A}$ is $\mathrm{T}$
So
As Net torque about $B=0$
$
\begin{aligned}
& \Rightarrow T \times 100=m g \times 50+2 m g \times 25 \\
& \Rightarrow T=m g
\end{aligned}
$

Summary

In short, rotational equilibrium occurs when the sum of all torques on a system is zero, resulting in constant angular momentum. It's a key concept in rotational motion, essential for understanding the balance in systems where forces and torques are applied, and is crucial for exams like JEE Main and NEET.