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$$\begin{array}{l}\Rightarrow {\mathbf{F}}_{\mathrm{net}}=m\mathbf{a}\end{array}$$
Substituting the same in the equation above, we have :
$$\begin{array}{l}\Rightarrow \frac{d\mathbf{\ell}}{dt}=\mathbf{r}\mathbf{x}m\mathbf{a}=\mathbf{r}\mathbf{x}{\mathbf{F}}_{\mathrm{net}}=\mathbf{r}\mathbf{x}\sum {\mathbf{F}}_{i}\end{array}$$
We note here that all forces are acting at the same point (occupied by the particle). As such, position vectors for all forces are same. Thus, we can enclose the vector product inside the summation sign,
$$\begin{array}{l}\Rightarrow \frac{d\mathbf{\ell}}{dt}=\sum \mathbf{r}\mathbf{x}{\mathbf{F}}_{i}=\sum {\mathbf{t}}_{i}={\mathbf{t}}_{\mathrm{net}}\end{array}$$
In words, Newton's second law of motion for a particle in angular form is read as :
Problem : Consider projectile motion of a particle of mass "m" and show that time derivative of angular moment about point of projection is equal to torque on the particle about that point.
Solution : For this question, we express various vector quantities in terms of unit vectors, because two directions involved in projectile motion i.e. horizontal and vertical are mutually perpendicular. This makes it easy to represent vectors in the unit vector form.
For the problem in hand, we consider horizontal and vertical directions as "x" and "y" reference directions respectively. Now, the torque, τ, on the particle about the point of projection "O" is :
$$\begin{array}{l}\mathbf{\tau}=\mathbf{r}\mathbf{x}\mathbf{F}\end{array}$$
The position vector of the position of the projectile at a given time "t" is :
$$\begin{array}{l}\mathbf{r}=x\mathbf{i}+y\mathbf{j}\end{array}$$
We note here that the external force on the projectile is force due to gravity. Hence,
$$\begin{array}{l}\mathbf{F}=-\mathrm{mg}\mathbf{j}\end{array}$$
Putting in the equation of the torque, we have :
$$\begin{array}{l}\mathbf{t}=(x\mathbf{i}+y\mathbf{j})\mathbf{x}-\mathrm{mg}\mathbf{j}\end{array}$$
$$\begin{array}{l}\mathbf{t}=-x\mathrm{mg}\mathbf{k}\end{array}$$
Negative sign shows that torque is clockwise i.e. into the plane of the projection "xy" plane. Next, we evaluate torque about "O" in terms of angular momentum. Here, angular momentum is given as :
$$\begin{array}{l}\mathbf{\ell}=m(\mathbf{r}\mathbf{x}\mathbf{v})=m\left\{\right(x\mathbf{i}+y\mathbf{j}\left)\right)\mathbf{x}\mathbf{v}\end{array}$$
Differentiating with respect to time "t", we have :
$$\begin{array}{l}\mathbf{t}=\frac{d\mathbf{\ell}}{dt}=m\left\{\right(\frac{dx}{dt}\mathbf{i}+\frac{dx}{dt}\mathbf{j}\left)\right)\mathbf{x}\frac{d\mathbf{v}}{dt}\end{array}$$
But $\frac{d\mathbf{v}}{dt}$ is equal to acceleration. In the case of projectile, acceleration due to gravity is the acceleration of the projectile,
$$\begin{array}{l}\mathbf{a}=\frac{d\mathbf{v}}{dt}=-g\mathbf{j}\end{array}$$
Putting this value in the equation,
$$\begin{array}{l}\mathbf{t}=m\left\{\right(\frac{dx}{dt}\mathbf{i}+\frac{dx}{dt}\mathbf{j}\left)\right)\mathbf{x}-g\mathbf{j}\end{array}$$
$$\begin{array}{l}\mathbf{t}=-x\mathrm{mg}\mathbf{k}\end{array}$$
The quantities torque and angular momentum are both measured or calculated about a common point "O". If it is not so, then the relationship is not valid. This aspect should always be kept in mind while applying the law.
We have emphasized that the angular form of Newton's second law is defined about a point - not an axis and as such not limited to rotational motion. Hence, we can employ this law even in situation where motion is strictly along a straight line. We shall, as a matter of fact, work out an example to illustrate this.
Problem : A bunjee jumper of mass 60 kg jumps from the middle of a bridge 200 m in length. Find the torque as time derivative of angular momentum about a point "O" as shown in the figure, before the rope attached to the jumper becomes taught.
Solution : Initially, the jumper is at a horizontal distance of 100 m from the end of the bridge. We must observe here that the jumper is not tied to the bridge. The rope is loose during the fall. He or she falls through a vertical straight path. It is a linear motion for which we are required to calculate torque. It is also evident that we can treat the jumper as point mass as all body parts are moving with same motion.
Now, angular momentum of the jumper with respect to the point, O, is :
$$\begin{array}{l}\ell =mrv\mathrm{sin}\theta \end{array}$$
Here, r sinθ = L/2, where "L" is the length of the brifge.
$$\begin{array}{l}\ell =mv\frac{L}{2}\end{array}$$
However, we do not know the speed of the jumper. We note here that the force on the jumper is force due to gravity. He/she falls with a constant acceleration. Thus, applying equation of motion, speed of the jumper after time "t" is :
$$\begin{array}{l}v=u+at=0+gt=gt\end{array}$$
Substituting in the equation above :
$$\begin{array}{l}\Rightarrow \ell =mgt\frac{L}{2}\end{array}$$
Differentiating with respect to time,
$$\begin{array}{l}\Rightarrow \tau =\frac{d\mathbf{\ell}}{dt}=mg\frac{L}{2}=60x10x\frac{200}{2}=60000\phantom{\rule{2pt}{0ex}}N-m\end{array}$$
In a similar situation in the earlier module titled Angular momentum , we had found that a particle moving in a straight line parallel to one of the axes of coordinate system has constant angular momentum and zero torque. The difference has arisen as velocity was constant in that case, whereas velocity is not constant in this case. We need force (hence torque) to cause acceleration or deceleration. The result, therefore, is consistent with the basic percept of Newton's law.
Further, we can also calculate torque using relation $\mathbf{\tau}=\mathbf{r}\mathbf{x}\mathbf{F}$ , where F = mg. This relation should also give same result.
Newton's law for a particle in rotation is established by taking first derivative of the angular momentum of a particle about an axis of rotation. As derived in previous module titled Angular momentum . it is given by :
Differentiating with respect to time,
$$\begin{array}{l}{t}_{\mathrm{net}}=\frac{d\ell}{dt}=\frac{d\left(\mathrm{I\omega}\right)}{dt}\end{array}$$
Since particle moves around a circular path of a constant radius, $I=m{{r}_{\perp}}^{2}$ is a constant and the same can be taken out of differentiation,
This derivation underlines that the above relation is specific to rotation about an axis - not for general motion. On the other hand, the definition of torque as the time rate of angular momentum about a point is a general form of Newton's second law.
Since discussion for angular momentum and corresponding Newton's second law are considered for different situations, it would be of great help to summarize the main results and special points for each of the cases so far covered. For clarity, we summarize as follows :
1: The reference for measurement of various angular quantities is a fixed point in the coordinate system. All angular quantities are measured about the same point.
2: Angular momentum of a particle in motion is given as :
$$\begin{array}{l}\mathbf{\ell}=\mathbf{r}\mathbf{x}\mathbf{p}=m(\mathbf{r}\mathbf{x}\mathbf{v})\end{array}$$
3: Newton's second law of motion in angular form for a particle is expressed mathematically as :
$$\begin{array}{l}{\mathbf{t}}_{\mathrm{net}}=\frac{d\mathbf{\ell}}{dt}\end{array}$$
In particular we must note that the form of Newton's second expressed in terms of moment of inertia and angular acceleration is not valid for general motion. It is so because, moment of inertia is defined about an axis - not about a point. For this reason, we shall not use the following form for general motion,
$$\begin{array}{l}{t}_{\mathrm{net}}=I\alpha \end{array}$$
1: For rotational motion, the reference for measurement of various angular quantities is a fixed axis of rotation. All angular quantities are measured about the same axis of rotation.
2: Angular momentum of a particle in motion is given as :
$$\begin{array}{l}\ell ={r}_{\perp}p=m{r}_{\perp}v=m{r}_{\perp}x\omega {r}_{\perp}=m{{r}_{\perp}}^{2}\omega =I\omega \end{array}$$
where "I" and "ω" denote moment of inertia and angular velocity respectively. Here, " ${r}_{\perp}$ " is moment arm about the axis of rotation.
Since there are only two directions invoved (one in the direction of axis and other opposite to it), it is possible to represent the directional features of angular quantities in rotation with a preceding "plus" or "minus" sign for anti-clockwise and clockwise rotational motions respectively. This is also the reason why do we write equations of rotation in scalar form.
3: Newton's second law of motion is expressed mathematically as :
$$\begin{array}{l}{t}_{\mathrm{net}}=\frac{d\ell}{dt}=I\alpha \end{array}$$
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