# Torque and Couple

Let a force $\vec{F}$ is applied on a particle located at a point P. The torque on the particle about a point O is given by \begin{align} \vec{\tau}=\vec{\mathrm{OP}}\times\vec{F}=\vec{r}\times\vec{F}, \nonumber \end{align} where $\vec{r}=\vec{\mathrm{OP}}$ is the position vector from O to P. The torque $\vec{\tau}$ is defined about a specific point. The phrase 'torque about an axis' is the component of $\vec{\tau}$ along that axis. This formula is also valid for torque on a rigid body. A couple is two equal and opposite forces acting at two points of a body. The torque due to a couple is given by $\tau=Fd$, where $d$ is the perpendicular distance between the lines of action of two forces (each of magnitude $F$). The torque due to a couple is independent of the point about which torque is defined.

## Solved Problems on Torque from IIT JEE

### Problem from IIT JEE 2016

The position vector $\vec{r}$ of a particle of mass m = 0.1 kg is given by the equation \begin{align} \vec{r}(t)=\alpha t^3\,\hat\imath+\beta t^2\,\hat\jmath, \nonumber \end{align} where $\alpha=10/3$ m/s3 and $\beta=5$ m/s2. At $t=1$ s, which of the following statement(s) is(are) true about the particle?

1. The velocity $\vec{v}$ is given by $\vec{v}=(10\,\hat\imath+10\,\hat\jmath)$ m/s.
2. The angular momentum $\vec{L}$ with respect to the origin is given by $\vec{L}=-5/3\,\hat{k}$ N-m s.
3. The force $\vec{F}$ is given by $\vec{F}=(\hat\imath+2\hat\jmath)$ N.
4. The torque $\vec{\tau}$ with respect to the origin is given by $\vec{\tau}=-20/3\,\hat{k}$ N-m.

Solution: The expressions for the velocity and the acceleration are given by \begin{align} \label{tzb:eqn:1} \vec{v}(t)&=\mathrm{d} \vec{r}/\mathrm{d}t \\ &=3\alpha t^2 \,\hat\imath+2 \beta t\;\hat\jmath, \end{align} \begin{align} \label{tzb:eqn:2} \vec{a}(t)&=\mathrm{d} \vec{v}/\mathrm{d}t \\ &=6\alpha t\,\hat\imath+2 \beta \;\hat\jmath. \end{align} Substitute $\alpha=10/3$ m/s3, $\beta=5$ m/s2, and $t=1$ s in above equation to get \begin{align} \vec{v}=(10\,\hat\imath+10\,\hat\jmath)\,\mathrm{m/s} \end{align} The position vector at $t=1$ s is \begin{align} \vec{r}=(10/3\,\hat\imath+5\,\hat\jmath)\,\mathrm{m} \end{align} The angular momentum with respect to the origin is given by \begin{align} \vec{L}&=\vec{r}\times(m\vec{v})\nonumber\\ &=(10/3\,\hat\imath+5\,\hat\jmath)\times 0.1(10\,\hat\imath+10\,\hat\jmath)\\ &=-5/3\,\hat{k} \,\mathrm{Nms}. \end{align} The acceleration at $t=1$ s is \begin{align} \vec{a}=(20\,\hat\imath+10\,\hat\jmath)\,\mathrm{m/s^2}. \end{align} Apply Newton's second law to get force on the particle as \begin{align} \vec{F}=m\vec{a}=(2\,\hat\imath+\,\hat\jmath)\,\mathrm{N}. \end{align} The torque with respect to the origin is given by \begin{align} \vec{\tau}&=\vec{r}\times\vec{F} \\ &=(10/3\,\hat\imath+5\,\hat\jmath)\times(2\,\hat\imath+\hat\jmath)\nonumber\\ &=(-20/3\,\hat{k})\,\mathrm{Nm}. \nonumber \end{align} We encourage you to show that $\vec{\tau}=\mathrm{d}\vec{L}/\mathrm{d}t$.

### Problem from JEE Mains 2019

A particle of mass $m$ is moving along a trajectory given by \begin{align} &x=x_0+a\cos\omega_1 t,\\ &y=y_0+b\sin\omega_2 t.\nonumber \end{align} The torque, acting on the particle about the origin, at $t=0$ is

1. $m(-x_0b+y_0a)\omega_1^2\,\hat{k}$
2. $my_0 a \omega_1^2\,\hat{k}$
3. zero
4. $-m(x_0b\omega_2^2+y_0a\omega_1^2)\,\hat{k}$

Solution: The position, velocity, and acceleration of the particle at time $t$ are given by \begin{align} \vec{r}(t) & =(x_0+a\cos\omega_1 t)\,\hat\imath+ (y_0+b\sin\omega_2 t)\,\hat\jmath,\nonumber \\ \vec{v}(t) & =\mathrm{d}\vec{r}/\mathrm{d}t \\ & =-a\omega_1\sin\omega_1 t\,\hat\imath+b \omega_2 \cos\omega_2 t\,\hat\jmath,\nonumber \\ \vec{a}(t) & =\mathrm{d}\vec{v}/\mathrm{d}t \\ &=-a\omega_1^2\cos\omega_1 t\,\hat\imath-b \omega_2^2 \sin\omega_2 t\,\hat\jmath.\nonumber \end{align} Substitute $t=0$ to get position $\vec{r}=(x_0+a)\,\hat\imath+y_0\,\hat\jmath$ and acceleration $\vec{a}=-a\omega_1^2\,\hat\imath$. The force on the particle at this instant is $\vec{F}=m\vec{a}$ and torque about the origin is \begin{align} \vec{\tau}&=\vec{r}\times\vec{F}=m\vec{r}\times\vec{a}\nonumber\\ &=m\left((x_0+a)\,\hat\imath+y_0\,\hat\jmath\right)\times (-a\omega_1^2\,\hat\imath) \\ & =my_0a\omega_1^2\hat{k}.\nonumber \end{align} We encourage you to find angular momentum $\vec{L}(t)$ of the particle about the origin and show that $\vec{\tau}(t)=\mathrm{d}\vec{L}(t)/\mathrm{d}t$. If $\omega_1=\omega_2$ then given trajectory is \begin{align} \frac{(x-x_0)}{a^2}+\frac{(y-y_0)}{b^2}=1, \nonumber \end{align} which is an ellipse centered at $\mathrm{O^\prime} \,(x_0,y_0)$ with semi-axis $a$ and $b$. The particle's motion is similar to the planetary motion. The torque on the particle about $\mathrm{O^\prime}$ is zero i.e., $\vec{\tau}_\mathrm{O^\prime}=\vec{0}$. Thus, angular momentum $\vec{L}_\mathrm{O^\prime}$ of the particle about $\mathrm{O}^\prime$ is conserved. The direction of $\vec{L}_\mathrm{O^\prime}$ is perpendicular to the $x\text{-}y$ plane. You can also show that $v_B=v_A a/b$. We encourage you to prove each of these statements.

### Problem from JEE Mains 2019

A slab is subjected to two forces $\vec{F}_1$ and $\vec{F}_2$ of same magnitude $F$ as shown in the figure. Force $\vec{F}_2$ is in x-y plane while $\vec{F}_1$ acts along z-axis at the point $(2\hat\imath+3\hat\jmath)$. The moment of force about the point O will be 1. $(3\hat\imath-2\hat\jmath+3\hat{k})F$
2. $(3\hat\imath-2\hat\jmath-3\hat{k})F$
3. $(3\hat\imath+2\hat\jmath-3\hat{k})F$
4. $(3\hat\imath+2\hat\jmath+3\hat{k})F$

Solution: The force $\vec{F}_1=F\hat{k}$ acts at the point $\vec{r}_1=(2\hat\imath+3\hat\jmath)$ and the force $\vec{F}_2=(-F\sin30\,\hat\imath-F\cos30\,\hat\jmath)$ acts at the point $\vec{r}_2=6\hat\jmath$. The torque of $\vec{F}_1$ and $\vec{F}_2$ about the origin O is given by \begin{align} \vec\tau_O&=\vec{r}_1\times\vec{F}_1+\vec{r}_2\times\vec{F}_2\nonumber\\ &=(2\hat\imath+3\hat\jmath)\times (F\hat{k}) \nonumber \nonumber\\ &\quad + (6\hat\jmath)\times(-F\sin30\,\hat\imath-F\cos30\,\hat\jmath)\nonumber\\ &=(3\hat\imath-2\hat\jmath+3\hat{k})F.\nonumber \end{align}

### Problem from JEE Mains 2019

To mop-clean a floor, a cleaning machine presses a circular mop of radius $R$ vertically down with a total force $F$ and rotates it with a constant angular speed about its axis. If the force $F$ is distributed uniformly over the mop and if coefficient of friction between the mop and the floor is $\mu$, the torque, applied by the machine on the mop is

1. $\mu F R/3$
2. $\mu F R/6$
3. $\mu F R/2$
4. $2\mu F R/3$

Solution: The angular acceleration of the mop is zero because it is rotating with a constant angular speed. Thus, net torque on the mop is zero. Hence, the torque on the mop due to the machine is equal and opposite to the frictional torque on it due to the ground. The force applied by the machine is balanced by the normal force from the ground i.e., $N=F$ (we assume mop's weight to be very small in comparison to $F$). Since $F$ is distributed uniformally over the mop, we expect the same for the normal force $N$. Thus, normal force per unit area of the mop is $F/\pi R^2$. Consider a circular ring of radius $r$ and thickness $\mathrm{d}r$. The normal force on the ring is $\mathrm{d}N=2Fr\mathrm{d}r/R^2$.

The frictional force on each element of the ring is tangential. The frictional torque on each element of the ring about the centre O adds up. Thus, net frictional frictional torque on the ring is \begin{align} \mathrm{d}\tau=r (\mu\mathrm{d}N)=\frac{2\mu F}{R^2}\, r^2\mathrm{d}r. \nonumber \end{align} The direction of frictional torque on each ring of the mop is same (clockwise). Thus, net frictional torque on the mop is \begin{align} \tau&=\int_0^R \mathrm{d}\tau \\ &=\frac{2\mu F}{R^2}\int_0^R r^2\mathrm{d}r \\ &=\frac{2}{3}\mu F R. \nonumber \end{align}

### Problem from IIT JEE 2018

Consider a body of mass 1.0 kg at rest at the origin at time $t=0$. A force $\vec{F}=(\alpha t\,\hat\imath+\beta\,\hat\jmath)$ is applied on the body, where $\alpha=1.0$ N/s and $\beta=1.0$ N. The torque acting on the body about the origin at time $t=1.0$ second is $\vec{\tau}$. Which of the following statements is (are) true?

1. $\left| \vec{\tau} \right|=\frac{1}{3}$ Nm
2. The torque $\vec{\tau}$ is in the direction of the unit vector $+\hat{k}$
3. The velocity of the body at $t=1$ sec is $\vec{v}=\frac{1}{2}\left(\hat\imath+2\hat\jmath\right)$ m/s
4. The magnitude of displacement of the body at $t=1$ sec is 1/6 m

Solution: Apply Newton's second law to get the acceleration of the body \begin{align} \label{acc:eqn:1} \vec{a}=\vec{F}/m=\alpha t\,\hat\imath+\beta\,\hat\jmath, \end{align} where $m=1$ kg, $\alpha=1$ N/s and $\beta=1$ N. Integrate above equation to get the velocity of the body \begin{align} \label{acc:eqn:2} \vec{v}&=\vec{v}_0+\int_0^t \!\vec{a}\,\mathrm{d}t \\ &=\vec{0}+\int_0^t \!(\alpha t\,\hat\imath+\beta\,\hat\jmath)\,\mathrm{d}t \nonumber\\ &=\int_0^t \!\alpha t \, \mathrm{d}t\;\hat\imath+ \int_0^t \!\beta \,\mathrm{d}t\;\hat\jmath \\ &=\frac{1}{2}\alpha t^2\;\hat\imath+ \beta t\;\hat\jmath, \end{align} where initial velocity $\vec{v}_0=\vec{0}$ because the body starts from rest. Note that the directions of the unit vectors $\hat\imath$ and $\hat\jmath$ do not vary with time. Substitute $t=1$ sec in above equation to get the velocity \begin{align} \vec{v}=\frac{1}{2}(\hat\imath+2\hat\jmath) \mathrm{m/s}. \end{align} Integrate the expression for velocity to get the displacement of the body \begin{align} \label{acc:eqn:3} \vec{r}&=\vec{r}_0+\int_0^t \!\vec{v}\,\mathrm{d}t \\ &=\vec{0}+\int_0^t \!\left(\frac{1}{2}\alpha t^2\,\hat\imath+\beta t\,\hat\jmath\right)\mathrm{d}t \nonumber\\ &=\frac{1}{6}\alpha t^3\;\hat\imath+ \frac{1}{2}\beta t^2\;\hat\jmath, \end{align} where initial displacement $\vec{r}_0=\vec{0}$ because body starts from the origin. Substitute $t=1$ sec in the above equation to get the displacement \begin{align} \vec{r}&=(\frac{1}{6}\hat\imath+\frac{1}{2}\hat\jmath)\,\mathrm{m}\\ |\vec{r}|&=\sqrt{10}/6 \,\mathrm{m}. \end{align} The torque on the body is given by \begin{align} \label{acc:eqn:4} \vec{\tau}&=\vec{r}\times\vec{F} \\ &=\left(\frac{1}{6}\alpha t^3\;\hat\imath+ \frac{1}{2}\beta t^2\;\hat\jmath \right) \times \left(\alpha t\,\hat\imath+\beta\,\hat\jmath\right) \nonumber\\ &=\frac{1}{6}\alpha\beta t^3\,\hat{k}-\frac{1}{2}\alpha\beta t^3\,\hat{k} \\ &=-\frac{1}{3}\alpha\beta t^3\,\hat{k}. \end{align} Substitute $t=1$ sec in the above equation to get \begin{align} \vec{\tau}&=-\frac{1}{3}\hat{k}\\ |\vec{\tau}|&=\frac{1}{3}. \end{align} I encourage you to find the expression for the angular momentum of the body about the origin and show that $\vec{\tau}=\mathrm{d}\vec{L}/\mathrm{d}t$.