7.7 Torque and angular momentum

In this section, we shall acquaint ourselves with two physical quantities (torque and angular momentum) which are defined as vector products of two vectors. These as we shall see, are especially important in the discussion of motion of systems of particles, particularly rigid bodies.

Rotational Dynamics

When one switches a fan on, the centre of the fan remains unmoved while the fan rotates with an angular acceleration. As the centre of mass remains at rest, the external forces acting on the fan must add to zero. This means that one can have angular acceleration even if the resultant external force is zero. But then why do we need to switch on the fan in order to start it? If an angular acceleration may be achieved with zero total external force, why does not a wheel chair start rotating on the floor as soon as one wishes it to do so. Why are we compelled to use our muscles to set it into rotation? In fact, one cannot have angular acceleration without external forces.

What is then the relation between the force and the angular acceleration? We find that even if the resultant external force is zero, we may have angular acceleration. We also find that without applying an external force we cannot have an angular acceleration. What is responsible for producing angular acceleration? The answer is torque which we define below.

Torque (Moment of force)

Torque is responsible for producing angular acceleration. Torque is also referred as couple (moment of force). 

If a force acts on a single particle at a point \(P\) whose position with respect to the origin \(O\) is given by the position vector \(\mathbf{r}\) (Fig. 7.18), the moment of the force acting on the particle with respect to the origin \(\mathrm{O}\) is defined as the vector product
\(
\tau=\mathbf{r} \times \mathbf{F} \dots(7.23)
\)
The moment of force (or torque) is a vector quantity having its direction perpendicular to \(\mathbf{r}\) and \mathbf{F} according to the rule of the cross product. The symbol \(\tau\) stands for the Greek letter tau. The magnitude of \(\tau\) is
\(
\tau=r F \sin \theta \dots(7.24a)
\)
where \(r\) is the magnitude of the position vector \(\mathbf{r}\), i.e. the length \(\mathrm{OP}, F\) is the magnitude of force \(\mathbf{F}\) and \(\theta\) is the angle between \(\mathbf{r}\) and \(\mathbf{F}\) as shown.

Moment of force has dimensions \(\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2}\). Its dimensions are the same as those of work or energy. It is, however, a very different physical quantity than work. Moment of a force is a vector, while work is a scalar. The SI unit of moment of force is Newton metre \((\mathrm{N} \mathrm{m})\). The magnitude of the moment of force may be written
\(
\begin{aligned}
& \tau=(r \sin \theta) F=r_{\perp} F \dots(7.24b)\\
& \text { or } \tau=r F \sin \theta=r F_{\perp} \dots(7.24c)
\end{aligned}
\)
where \(r_{\perp}=r \sin \theta\) is the perpendicular distance of the line of action of \(\mathbf{F}\) from the origin and \(F_{\perp}(=F \sin \theta)\) is the component of \(\mathbf{F}\) in the direction perpendicular to \(\mathbf{r}\). Note that \(\tau=0\) if \(r=0, F=0\) or \(\theta=0^{\circ}\) or \(180^{\circ}\). Thus, the moment of a force vanishes if either the magnitude of the force is zero, or if the line of action of the force passes through the origin.
One may note that since \(\mathbf{r} \times \mathbf{F}\) is a vector product, properties of a vector product of two vectors apply to it. If the direction of \(\mathbf{F}\) is reversed, the direction of the moment of force is reversed. If the directions of both \(\mathbf{r}\) and \(\mathbf{F}\) are reversed, the direction of the moment of force remains the same.

Rigid Body Rotating About a Given Axis of Rotation

Now consider a rigid body rotating about a given axis of rotation \(A B\) (figure aboveLet \(F\) be a force acting on the particle \(P\) of the body. \(F\) may not be in the plane \(A B P\). Take the origin \(O\) somewhere on the axis of rotation.

The torque of \(F\) about \(O\) is \(\vec{\tau}=\vec{r} \times \vec{F}\). Its component along \(O A\) is called the torque of \(\vec{F}\) about \(O A\). To calculate it, we should find the vector \(\vec{r} \times \vec{F}\) and then find out the angle \(\theta\) it makes with \(O A\).
The torque about \(O A\) is then \(|\vec{r} \times \vec{F}| \cos \theta\).
The torque of a force about a line is independent of the choice of the origin as long as it is chosen on the line. This can be shown as given below. Let \(O_1\) be any point on the line \(A B\) (figure above). The torque of \(F\) about \(O_1\) is
\(
\overrightarrow{O_1 P} \times \vec{F}=\left(\overrightarrow{O_1 O}+\overrightarrow{O_P}\right) \times \vec{F}=\overrightarrow{O_1 O} \times \vec{F}+\overrightarrow{O P} \times \vec{F}
\)
As \(\overrightarrow{O_1 O} \times \vec{F} \perp \overrightarrow{O_1 O}\), this term will have no component along \(A B\).
Thus, the component of \(\overrightarrow{O_1 P} \times \vec{F}\) is equal to that of \(\overrightarrow{O P} \times \vec{F}\)

There are some special cases that occur frequently.

Case-I

\(
\vec{F} \| \overrightarrow{A B}
\)
\(\vec{r} \times \vec{F}\) is perpendicular to \(\vec{F}\), but \(\vec{F} \| \overrightarrow{A B}\), hence \(\vec{r} \times \vec{F}\) is perpendicular to \(\overrightarrow{A B}\). The component of \(\vec{r} \times \vec{F}\) along \(\overrightarrow{A B}\) is, therefore, zero.

Case-II

\(
F \text { intersects } A B \text { (say at } \mathrm{O} \text { ) }
\)

Taking the point of intersection as the origin, we see that \(\vec{r}(=\overrightarrow{O P})\) and \(\vec{F}\) are in the same line. The torque about \(O\) is \(\vec{r} \times \vec{F}=0\). Hence the component along \(O A\) is zero.

Case III

\(
\vec{F} \perp \overrightarrow{A B} \text { but } \vec{F} \text { and } A B \text { do not intersect. }
\)
In three dimensions, two lines may be perpendicular without intersecting each other. For example, a vertical line on the surface of a wall of your room and a horizontal line on the opposite wall are mutually perpendicular but they never intersect. Two nonparallel and nonintersecting lines are called skew lines.

The figure above shows the plane through the particle \(P\) that is perpendicular to the axis of rotation \(A B\). Suppose the plane intersects the axis at the point \(O\). The force \(F\) is in this plane. Taking the origin at \(O\),
\(
\vec{\tau}=\vec{r} \times \vec{F}=\overrightarrow{O P} \times \vec{F}
\)
\(
\text { Thus, } \quad \tau=r F \sin \theta=F \text {. }(O S)
\)
where \(O S\) is the perpendicular from \(O\) to the line of action of the force \(\vec{F}\). The line \(O S\) is also perpendicular to the axis of rotation. It is thus the length of the common perpendicular to the force and the axis of rotation.

The direction of \(\vec{\tau}=\overrightarrow{O P} \times \vec{F}\) is along the axis \(A B\) because \(\overrightarrow{A B} \perp \overrightarrow{O P}\) and \(\overrightarrow{A B} \perp \vec{F}\). The torque about \(A B\) is, therefore, equal to the magnitude of \(\vec{\tau}\) that is \(F .(O S)\).

Thus, the torque of \(F\) about \(A B=\) magnitude of the force \(F \times\) length of the common perpendicular to the force and the axis. The common perpendicular \(O S\) is called the lever arm or moment arm of this torque.
The torque may try to rotate the body clockwise or anticlockwise about \(A B\). Depending on the convenience of the problem one may be called positive and the other negative. It is conventional to take the torque positive if the body rotates anticlockwise as viewed through the axis.

Case IV

\(\vec{F}\) and \(\overrightarrow{O A}\) are skew but not perpendicular.
Take components of \(\vec{F}\) parallel and perpendicular to the axis.

The torque of the parallel part is zero from case I and that of the perpendicular part may be found as in case III.

In most of the applications that we shall see, cases I, II or III will apply.
If there are more than one forces \(\vec{F}_1, \vec{F}_2, \vec{F}_3, \ldots\) acting on a body, one can define the total torque acting on the body about a given line.

To obtain the total torque, we have to get separately the torques of the individual forces and then add them.
\(
\vec{\tau}=\overrightarrow{r_1} \times \vec{F}_1+\overrightarrow{r_2} \times \vec{F}_2+\ldots
\)
You may be tempted to add the forces \(\vec{F}_1, \vec{F}_2\), \(\vec{F}_3, \ldots\) vectorially and then obtain the torque of resultant force about the axis. But that won’t always work. Even if \(\vec{F}_1+\vec{F}_2+\ldots=0, \vec{r}_1 \times \vec{F}_1+\vec{r}_2 \times \vec{F}_2+\ldots\) may not. However, if the forces act on the same particle, one can add the forces and then take the torque of the resultant.

Example 7.10: Consider a pulley fixed at its centre of mass by a clamp. A light rope is wound over it and the free end is tied to a block. The tension in the rope is T. (a) Write the forces acting on the pulley. How are they related? (b) Locate the axis of rotation. (c) Find the torque of the forces about the axis of rotation.

Solution:

(a) The forces on the pulley are (figure above)
(i) attraction by the earth, \(M g\) vertically downward,
(ii) tension \(T\) by the rope, along the rope,
(iii) contact force \(N\) by the support at the centre.
\({N}=T+M g\) (centre of mass of the pulley is at rest, so Newton’s 1st law applies).
(b) The axis of rotation is the line through the centre of the pulley and perpendicular to the plane of the pulley.
(c) Let us take the positive direction of the axis towards the reader.
The force \(M g\) passes through the centre of mass and it intersects the axis of rotation. Hence the torque of \(M g\) about the axis is zero (Case II). Similarly, the torque of the contact force \(N\) is also zero.
The tension \(T\) is along the tangent of the rim in the vertically downward direction. The tension and the axis of rotation are perpendicular but never intersect. Case III applies. Join the point where the rope leaves the rim to the centre. This line is the common perpendicular to the tension and the axis. Hence the torque is \(T . r\) (positive, since it will try to rotate the pulley anticlockwise).

Angular momentum of a particle

Like moment of a force (Torque), angular momentum is also a vector product. It could also be referred to as moment of (linear) momentum. Consider a particle of mass \(m\) and linear momentum \(\mathbf{p}\) at a position \(\mathbf{r}\) relative to the origin \(\mathrm{O}\). The angular momentum \(l\) of the particle with respect to the origin \(O\) is defined to be
\(
l=\mathbf{r} \times \mathbf{p} \dots(7.25a)
\)

Like moment of a force, angular momentum is also a vector product. It could also be referred to as moment of (linear) momentum. Consider a particle of mass \(m\) and linear momentum \(\mathbf{p}\) at a position \(\mathbf{r}\) relative to the origin \(\mathrm{O}\). The angular momentum \(l\) of the particle with respect to the origin \(O\) is defined to be
\(
l=\mathbf{r} \times \mathbf{p} \dots(7.25a)
\)
The magnitude of the angular momentum vector is
\(
l=r p \sin \theta \dots(7.26a)
\)
where \(p\) is the magnitude of \(\mathbf{p}\) and \(\theta\) is the angle between \(\mathbf{r}\) and \(\mathbf{p}\). We may write
\(
l=r p_{\perp} \text { or } r_{\perp} p \dots(7.26b)
\)
where \(r_{\perp}(=r \sin \theta)\) is the perpendicular distance of the directional line of \(\mathbf{p}\) from the origin and \(p_{\perp}(=p \sin \theta)\) is the component of \(\mathbf{p}\) in a direction perpendicular to \(\mathbf{r}\). We expect the angular momentum to be zero \((l=0)\), if the linear momentum vanishes \((p=0)\), if the particle is at the origin \((r=0)\), or if the directional line of \(\mathbf{p}\) passes through the origin \(\theta=0^{\circ}\) or \(180^{\circ}\).

Angular momentum of the particle about the axis of the Circle (\(L=I \omega\))

Suppose a particle is going in a circle of radius \(r\) and at some instant, the speed of the particle is \(v\) (figure below). Let’s find the angular momentum of the particle about the axis of the circle.

As the origin may be chosen anywhere on the axis, we choose it at the centre of the circle. Then \(\vec{r}\) is along a radius and \(\vec{v}\) is along the tangent so that \(\overrightarrow{r_{\rightarrow}}\) is perpendicular to \(\vec{v}\) and \(l=|\vec{r} \times \vec{p}|=m v r\). Also \(\vec{r} \times \vec{p}\) is perpendicular to \(\vec{r}\) and \(\vec{p}\) and hence is along the axis. Thus, the component of \(\vec{r} \times \vec{p}\) along the axis is \(mvr \) itself.

Next, consider a rigid body rotating about an axis \(A B\) (figure above). Let the angular velocity of the body be \(\omega\). Consider the ith particle going in a circle of radius \(r_i\) with its plane perpendicular to \(A B\).
The linear velocity of this particle at this instant is \(v_i=r_i \omega\). The angular momentum of this particle about \(A B=m_i v_i r_i\) \(=m_i r_i^2 \omega\). The angular momentum of the whole body about \(A B\) is the sum of these components, i.e.,
\(
L=\sum m_i r_i^2 \omega=I \omega
\)
where \(I\) is the moment of inertia of the body about \(A B\).

Relationship between Torque and Angular momentum

The physical quantities, moment of a force and angular momentum, have an important relation between them. It is the rotational analogue of the relation between force and linear momentum. For deriving the relation in the context of a single particle, we differentiate \(\boldsymbol{l}=\mathbf{r} \times \mathbf{p}\) with respect to time,
\(
\frac{\mathrm{d} \boldsymbol{l}}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{r} \times \mathbf{p})
\)
Applying the product rule for differentiation to the right-hand side,
\(
\frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{r} \times \mathbf{p})=\frac{\mathrm{d} \mathbf{r}}{\mathrm{d} t} \times \mathbf{p}+\mathbf{r} \times \frac{\mathrm{d} \mathbf{p}}{\mathrm{d} t}
\)
Now, the velocity of the particle is \(\mathbf{v}=d \mathbf{r} / d t\) and \(\mathbf{p}=m \mathbf{v}\)
Because of this \(\frac{\mathrm{d} \mathbf{r}}{\mathrm{d} t} \times \mathbf{p}=\mathbf{v} \times m \mathbf{v}=0\),
as the vector product of two parallel vectors vanishes. Further, since \(\mathrm{d} \mathbf{p} / \mathrm{d} t=\mathbf{F}\),
\(
\mathbf{r} \times \frac{\mathrm{d} \mathbf{p}}{\mathrm{d} t}=\mathbf{r} \times \mathbf{F}=\boldsymbol{\tau}
\)
Hence \(\frac{\mathrm{d}}{\mathrm{d} t}(\mathbf{r} \times \mathbf{p})=\boldsymbol{\tau}\)
or \(\frac{\mathrm{d} \boldsymbol{l}}{\mathrm{d} t}=\boldsymbol{\tau} \dots(7.27)\)
Thus, the time rate of change of the angular momentum of a particle is equal to the torque acting on it. This is the rotational analogue of the equation \(\mathbf{F}=\mathrm{d} \mathbf{p} / \mathrm{d} t\), which expresses Newton’s second law for the translational motion of a single particle.

Torque and angular momentum for a system of particles

The angular momentum of a system of particles is the vector sum of the angular momenta of the particles of the system. particles. Thus, for a system of \(n\) particles,
\(
\mathbf{L}=\boldsymbol{l}_1+\boldsymbol{l}_2+\ldots+\boldsymbol{l}_n=\sum_{i=1}^n \boldsymbol{l}_i  (\vec{L}=\sum_i \overrightarrow{l_i}=\sum\left(\overrightarrow{r_i} \times \overrightarrow{p_i}\right))
\)
The angular momentum of the \(t^{\text {th }}\) particle is given by
\(
l_i=\mathbf{r}_i \times \mathbf{p}_i
\)
where \(\mathbf{r}_i\) is the position vector of the \(i^{\text {th }}\) particle with respect to a given origin and \(\mathbf{p}=(m_i, \mathbf{v_i})\) is the linear momentum of the particle. (The particle has mass \(m_i\) and velocity \(\mathbf{v}_i\) ) We may write the total angular momentum of a system of particles as
\(
\mathbf{L}=\sum \boldsymbol{l}_i=\sum_i \mathbf{r}_i \times \mathbf{p}_i
\)
This is a generalisation of the definition of angular momentum (Eq. 7.25a) for a single particle to a system of particles.
Using Eqs. (7.23) and (7.25b), we get
\(
\frac{\mathrm{d} \mathbf{L}}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{d} t}\left(\boldsymbol{\sum} \boldsymbol{l}_i\right)=\sum_t \frac{\mathrm{d} \boldsymbol{l}}{\mathrm{d} t}=\sum_t \tau_i \dots(7.28a)
\)
where \(\boldsymbol{\tau}_i\) is the torque acting on the \(i^{\text {th }}\) particle;
\(
\boldsymbol{\tau}_i=\mathbf{r}_i \times \mathbf{F}_i
\)
The force \(\mathbf{F}_i\) on the \(i^{\text {th }}\) particle is the vector sum of external forces \(\mathbf{F}_i^{e x t}\) acting on the particle and the internal forces \(\mathbf{F}_i^{\mathrm{int}}\) exerted on it by the other particles of the system. We may therefore separate the contribution of the external and the internal forces to the total torque
\(
\boldsymbol{\tau}=\sum_i \boldsymbol{\tau}_i=\sum_i \mathbf{r}_i \times \mathbf{F}_i \text { as }
\)
\(
\begin{array}{ll}
& \boldsymbol{\tau}=\boldsymbol{\tau}_{\text {ext }}+\boldsymbol{\tau}_{\text {int }}, \\
\text { where } & \boldsymbol{\tau}_{\text {ext }}=\sum_i \mathbf{r}_i \times \mathbf{F}_i^{\text {ext }} \\
\text { and } & \boldsymbol{\tau}_{\text {int }}=\sum_i \mathbf{r}_i \times \mathbf{F}_i^{\text {int }}
\end{array}
\)
We shall assume not only Newton’s third law of motion, 1.e. the forces between any two particles of the system are equal and opposite, but also that these forces are directed along the line joining the two particles. In this case, the contribution of the internal forces to the total torque on the system is zero since the torque resulting from each action-reaction pair of forces is zero. We thus have, \(\boldsymbol{\tau}_{\text {int }}=0\) and therefore \(\boldsymbol\tau=\boldsymbol\tau_{\text {ext }}\)

Since \(\boldsymbol\tau=\sum \boldsymbol\tau_i\), it follows from Eq. (7.28a) that
\(
\frac{\mathrm{d} \mathbf{L}}{\mathrm{d} t}=\boldsymbol{\tau}_{\text {ext }} \dots(7.28b)
\)

Thus, the time rate of the total angular momentum of a system of particles about a point (taken as the origin of our frame of reference) is equal to the sum of the external torques (i.e. the torques due to external forces) acting on the system taken about the same point. Eq. (7.28 b) is the generalisation of the single particle case of Eq. (7.23) to a system of particles. Note that when we have only one particle, there are no internal forces or torques. Eq. (7.28 b) is the rotational analogue of
\(
\frac{\mathrm{d} \mathbf{P}}{\mathrm{d} t}=\mathbf{F}_{e x t} \dots(7.17)
\)
Note that like Eq.(7.17), Eq.(7.28b) holds good for any system of particles, whether it is a rigid body or its individual particles have all kinds of internal motion.

Conservation of angular momentum

If \(\boldsymbol{\tau}_{e x t}=\mathbf{0}, \mathrm{Eq} .(7.28 \mathrm{~b})\) reduces to
\(
\frac{\mathrm{d} \mathbf{L}}{\mathrm{d} t}=0
\)
or \(\quad \mathbf{L}=\) constant……(7.29a)

Thus, if the total external torque on a system of particles is zero, then the total angular momentum of the system is conserved, i.e. remains constant. Eq. (7.29a) is equivalent to three scalar equations,
\(
L_x=K_1, L_y=K_2 \text { and } L_z=K_3 \quad(7.29 \mathrm{~b})
\)
Here \(K_1, K_2\) and \(K_3\) are constants; \(L_x, L_y\) and \(L_{\mathrm{z}}\) are the components of the total angular momentum vector \(\mathbf{L}\) along the \(x, y\) and \(z\) axes respectively. The statement that the total angular momentum is conserved means that each of these three components is conserved.

For a rigid body rotating about a fixed axis, we can write the equation
\(
\begin{array}{rlrl}
L & =I \omega \\
\text { or, } & \frac{d L}{d t} =I \frac{d \omega}{d t}=I \alpha \\
\text { or, } & \frac{d L}{d t} =\boldsymbol\tau_{e x t} 
\end{array}
\)

If the total external torque on a system is zero, its angular momentum remains constant. This is known as the principle of conservation of angular momentum.

Example 7.11: Find the torque of a force \(7 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\) about the origin. The force acts on a particle whose position vector is \(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\)

Solution:

Here \(\mathbf{r}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\mathbf{F}=7 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\).
We shall use the determinant rule to find the torque \(\boldsymbol{\tau}=\mathbf{r} \times \mathbf{F}\)
\(
\begin{aligned}
\tau= & \left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & -1 & 1 \\
7 & 3 & -5
\end{array}\right|=(5-3) \hat{\mathbf{i}}-(-5-7) \hat{\mathbf{j}}+(3-(-7)) \hat{\mathbf{k}} \\
& \text { or } \tau=2 \hat{\mathbf{i}}+12 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}
\end{aligned}
\)

Example 7.12: Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion.

Solution:

Let the particle with velocity \(\mathbf{v}\) be at point \(\mathrm{P}\) at some instant \(t\). We want to calculate the angular momentum of the particle about an arbitrary point \(O\).

The angular momentum is \(\mathbf{l}=\mathbf{r} \times m \boldsymbol{v}\). Its magnitude is \(mvr \sin \theta\), where \(\theta\) is the angle between \(\mathbf{r}\) and \(\mathbf{v}\) as shown in Fig. 7.19. Although the particle changes position with time, the line of direction of \(\mathbf{v}\) remains the same and hence \(\mathrm{OM}=r \sin \theta\) 1s a constant. Further, the direction of \(\mathbf{l}\) is perpendicular to the plane of \(\mathbf{r}\) and \(\mathbf{v}\). It is into the page of the figure. This direction does not change with time.
Thus, \(\mathbf{l}\) remains the same in magnitude and direction and is therefore conserved.

You cannot copy content of this page