Edited By Vishal kumar | Updated on Sep 24, 2024 01:53 PM IST

Torque is something that everyone understands. People may find it difficult to express, but they are aware of it. Still, it's simply understandable if you picture a window or a door. A door must be pushed away from the hinge in order to be opened. This makes it possible to open the door with relatively little effort. More force is needed the closer you go to the hinge.

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In this article, we will cover the concept of Torque. This topic falls under the broader category of Rotational Motion which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and others. Over the last ten years of the JEE Main exam (from 2013 to 2023), more than nineteen questions have been asked on this concept. And for NEET five questions were asked from this concept.

Let's read this entire article to gain an in-depth understanding of Torque.

The vector product of the Force vector and position vector is known as **torque**.

$\vec{\tau}=\underset{r}{\rightarrow} \times \underset{}{\vec{F}}$

Its direction is always perpendicular to the plane containing vector r and F and with the help of the right-hand screw rule, we can find it.

The magnitude of torque is calculated by using either

$\tau=r_1 F \text { or } \tau=r \cdot F_1$

$r_1=$ perpendicular distance from the origin to the line of force.

$F_1$ = component of force perpendicular to line joining force.

- $\tau=r . F . \sin \phi$

Where $\phi$ is the angle between vector $\mathrm{r}$ and $\mathrm{F}$

- $\tau_{\max }=r . F\left(\right.$ when $\left.\phi=90^{\circ}\right)$

Solved Examples Based on Work Done By a Constant Force

$\tau_{\min }=0\left(\text { when } \phi=0^0\right)$

If a pivoted, hinged body tends to rotate due to an applied force, then it is said that torque is acted on the body by force.

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In the rotation of a hinged door when we apply torque with the help of force $F$.

- SI Unit- Newton-metre

- Dimension- $M L^2 T^{-2}$

- If a body is acted upon by more than one force, then we get the resultant torque by doing the vector sum of each torque.

$$

\tau=\tau_1+\tau_2+\tau_3 \ldots \ldots

$$

Just like force is the cause of translatory motion similarly Torque is the cause of rotatory motion.

**Example 1: Let $\vec{F}$ be the force acting on a particle having a position vector $\vec{r}$ and $\vec{T}$ be the torque of this force about the origin. Then**

**1) $\vec{r} \cdot \vec{T}=0$ and $\vec{F} \cdot \vec{T} \neq 0$ 2) $\vec{r} \cdot \vec{T} \neq 0$ and $\vec{F} \cdot \vec{T}=0$ 3) $\vec{r} \cdot \vec{T} \neq 0$ and $\vec{F} \cdot \vec{T} \neq 0$ 4) $\vec{r} \cdot \vec{T}=0$ and $\vec{F} \cdot \vec{T}=0$**

**Solution:**

We know that $\vec{\tau}=\vec{r} \times \vec{f}$

The angle between $\mathrm{\tau}$ and $\mathrm{F}$ is $90^{\circ}$ and between $\mathrm{\tau}$ and $\mathrm{r}$ is 900 for example, if we take the dot product between two vectors as $\theta=90^{\circ} \Longrightarrow \cos \theta=\cos 90^{\circ}=0$

$

\begin{aligned}

& \vec{r} \cdot \vec{\tau}=\vec{r} \cdot(\vec{r} \times \vec{f})=0 \\

& \therefore \vec{f} \cdot \vec{\tau}=\vec{f} \cdot(\vec{r} \times \vec{f})=0

\end{aligned}

$

Hence, the answer is option (4).

**Example 2: A force of $-F \hat{k}$ acts on $\mathrm{O}$, the origin of the coordinate system. The torque about the point $(1,-1)$ is : 1) $-F(\hat{i}-\hat{j})$ 2) $F(\hat{i}-\hat{j})$ 3) $-F(\hat{i}+\hat{j})$ 4) $F(\hat{i}+\hat{j})$**

**Solution:**

Torque -

$

\underset{\tau}{\vec{\tau}}=\underset{r}{\vec{F}} \times \underset{\vec{F}}{\overrightarrow{ }}

$

wherein

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 line joining force.

given $\vec{\tau}=\vec{r} \times \vec{F}$

$

\begin{aligned}

& \vec{F}=-F \hat{k}, \vec{r}=\hat{i}-\hat{j} \\

& \therefore \quad \vec{r} \times \vec{F}=\left|\begin{array}{ccc}

\hat{i} & \hat{j} & \hat{k} \\

1 & -1 & 0 \\

0 & 0 & -F

\end{array}\right| \\

& =\hat{i} F-\hat{j}(-F)=F(\hat{i}+\hat{j})

\end{aligned}

$

Hence, the answer is option (4).

**Example 3: A force of 40 N acts on a point B at the end of an L-shaped object, as shown in the figure. The angle θ that will produce the maximum moment of the force about point A is given by :**

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

**Solution:**

Torque

$

\underset{\tau}{\vec{r}}=\underset{F}{\rightarrow}

$

wherein

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 line joining force.

torque about $\mathrm{A}=\tau_A=(F \cos \theta) \cdot 4+(F \sin \theta) \cdot 2$

for maximum torque

$

\frac{d \tau_A}{d \Theta}=0

$

or $\quad 4 f(-\sin \theta)+2 f \cos \theta=0$

$

2 \sin \theta=\cos \theta \text { or } \tan \theta=\frac{1}{2}

$

Hence, the answer is option (1).

**Example 4: A uniform rod AB is suspended from a point X, at a variable distance x from A, as shown. To make the rod horizontal, a mass m is suspended from its end A. A set of (m, x) values is recorded. The appropriate variables that give a straight line, when plotted, are :**

**1) $m, x$ 2) $m, \frac{1}{x}$ 3) $m, \frac{1}{x^2}$ 4) $m, x^2$**

**Solution:**

Torque -

$

\vec{\tau}=\underset{r}{\rightarrow} \times \underset{F}{\vec{F}}

$

wherein

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 line joining force.

Balancing Torque w.r.t point of suspension

$

\begin{aligned}

& m g x=M g\left(\frac{l}{2}-x\right) \\

& \Rightarrow m x=M \frac{l}{2}-M x \Rightarrow m=\frac{M l}{2} \cdot \frac{1}{x}-M

\end{aligned}

$

This represents a straight line

Hence, the answer is option (2).

**Example 5: The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1N, and the distance of the particle from the origin is 5 m, the angle between the force and the position vector is (in radians) :**

**1) 0.52**

**2) 30**

**3) 10**

**4) 1.04**

**Solution**

$\begin{aligned}

& T=\vec{r} \times \vec{F} \\

& T=|\vec{r}| \cdot|\vec{F}| \cdot \sin \theta \ldots-(1) \\

& T=2.5 \mathrm{Nm} \\

& |\vec{r}|=5 m \\

& |\vec{F}|=1 N \\

& \text {-put in (1) } \\

& T=2.5=1 \times 5 \times \sin \theta \\

& \sin \theta=0.5=\frac{1}{2} \\

& \theta=\frac{\pi}{6} \\

& \theta=0.52 \text { radian } \\

&

\end{aligned}$

Hence, the answer is option (1).

In short, torque can be conceptualised as an object's twist around a certain axis, similar to how a linear force is a push or a pull. Another definition of torque is the product of the force's magnitude and the force's line of action's perpendicular distance from the axis of rotation. It is possible to find the torque immediately given the length of the moment arm.

1. Define torque.

Torque, or moment of force, is the ability of a force to induce rotational motion in a body. The product of the force's magnitude and the perpendicular distance between the force's line of action and the speed of rotation is another definition of torque.

The expression for torque is

=**r**X**F**

Where,

** r** - the position vector of the point

**F** - the force acting on the body

2. What are the different types of torque?

Different types of torque:

Deflecting torque

Controlling torque

Damping torque

3. Define Damping torque.

Damping torque is a physical mechanism for managing the movement of a system by providing motion that opposes the system's natural oscillation.

It acts only while a system is in motion

4. Define couple.

A couple is a concept which includes the forces of equal magnitude but opposite direction that are distanced by a perpendicular distance such that their lines of action do not meet, resulting in a turning effect.

5. Is torque a scalar or vector quantity?

The expression for torque is

=**r**X**F**

Where,

**r **- the position vector of the point

**F** - the force acting on the body

Here, the product of **r **and **F** is called the vector product or cross product.

When two vectors are vectorized, a new vector is created that is perpendicular to both vectors. As a result, torque is a vector quantity.

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**Correct Answer: ** Work and torque

**Solution : ** The correct answer is **Work and torque**.

The units for work and torque are the same. Both work and torque have units of measurement in the International System of Units (SI) called joule (J). The force exerted on an object multiplied by its displacement is what is known as work. The force applied to an object multiplied by the distance between the pivot point and the force application point is known as torque.

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