Review of Laws of Motion
Particle Dynamics
It is the study of motion of bodies and the relationship of this motion with the forces producing it.
- Linear motion
- Projectile
- Gravitational
- Circular
- Rigid body rotation
Types of Forces
1. Conservative Force
A force is said to be conservative if the work done by it is independent of the path followed by the body.
Example: Central forces (forces which are directed towards or away from a fixed position), electrostatic force, gravitational force.
$$W(\neq) \;\text{Independent of path followed}$$2. Non-conservative Force
A force is said to be non-conservative if the work done by it depends on the path followed by the body.
Example: Frictional force, viscous force, etc. Velocity-dependent forces whose values depend on the magnitude and direction of the body's motion are non-conservative forces.
$$W(=) \;\text{depends on path followed}$$Work Done by a Conservative Force
Work done from position $A$ to $B$ can be written as:
$$W_{AB} = \int_A^B \vec{F}\cdot d\vec{r}$$And work done from position $B$ to $A$ can be written as:
$$W_{BA} = \int_B^A \vec{F}\cdot d\vec{r} = -\int_A^B \vec{F}\cdot d\vec{r}$$ $$\Rightarrow W_{BA} = -W_{AB}$$or,
$$W_{BA}+W_{AB} = 0$$$\therefore$ Total work done by a conservative force along a closed path is zero.
$$\oint \vec{F}\cdot d\vec{r} = 0$$Work–Energy Principle (Theorem)
Let us consider a body of mass $m$ travelling from position $A$ to $B$.
Let $u$ = initial velocity at $A$, $v$ = final velocity at $B$.
From the second law of motion,
$$F = ma$$or,
$$F = m\frac{dv}{dt}$$Multiplying both sides by $v$, we get,
$$F\cdot v = m\cdot v\cdot\frac{dv}{dt}$$Integrating both sides w.r.t. $t$,
$$\int_u^v F\cdot v\,dt = \int_u^v mv\frac{dv}{dt}\cdot dt$$or,
$$\int_u^v F\cdot\frac{dr}{dt}\cdot dt = \int_u^v mv\,dv$$or,
$$\int_u^v F\cdot dr = m\int_u^v v\,dv$$or,
$$\int_u^v F\cdot dr = m\left[\frac{v^2}{2}\right]_u^v$$or,
$$\int_u^v F\cdot dr = m\left(\frac{v^2}{2}-\frac{u^2}{2}\right)$$or,
$$\int_A^B F\cdot dr = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$$ $$\boxed{W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2}$$$\therefore$ “Work done from an initial position to a final position is equal to the change in kinetic energy between them.” This is called the work–energy principle.
Energy
The capacity of a body to do work is called energy.
There are two types of mechanical energy:
- Kinetic energy: Energy due to the motion.
- Potential energy: Energy due to the position and configuration (shape & size).
P.E.: Potential energy of a particle at a point $r$ is given by: “amount of work done in moving it from infinity to that point.”
$$(U)\,P.E. = \int_r^{\infty} F\cdot dr$$or,
$$U = -\int_{\infty}^{r} F\cdot dr$$Conservative Force as Negative Gradient of Potential Energy
We have,
$$P.E.(U) = -\int_{\infty}^{r} F\cdot dr \qquad \cdots (1)$$where $F$ = conservative force.
In a 3-dimensional co-ordinate system,
$$F = iF_x + jF_y + kF_z$$ $$r = ix+jy+kz$$ $$\Rightarrow dr = i\,dx+j\,dy+k\,dz$$Now, eq.(1) becomes,
$$U = -\int_{\infty}^{r} (iF_x+jF_y+kF_z)\cdot(i\,dx+j\,dy+k\,dz)$$or,
$$U = -\int_{\infty}^{r} F_x\,dx+F_y\,dy+F_z\,dz$$In partial differential form,
$$F_x = -\frac{\partial U}{\partial x}, \qquad F_y = -\frac{\partial U}{\partial y}, \qquad \text{and} \qquad F_z = -\frac{\partial U}{\partial z}$$ $$\left(\frac{dU}{dr}=F \;\Rightarrow\; dU=F\,dr \;\Rightarrow\; \int dU=\int F\,dr \;\Rightarrow\; U=\int F\,dr\right)$$$\therefore$
$$F = iF_x+jF_y+kF_z = -\left(i\frac{\partial U}{\partial x}+j\frac{\partial U}{\partial y}+k\frac{\partial U}{\partial z}\right)$$or,
$$F = -\left(i\frac{\partial}{\partial x}+j\frac{\partial}{\partial y}+k\frac{\partial}{\partial z}\right)U$$or,
$$F = -\text{grad}\,U$$or,
$$\boxed{F = -\nabla U}$$Necessary and Sufficient Condition for a Conservative Force (Curl of a Conservative Force)
We know, for a conservative force,
$$F = -\nabla U$$Taking curl on both sides of the equation,
$$\nabla\times F = -\nabla\times\nabla U$$ $$= -\begin{vmatrix} i & j & k \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ F_x & F_y & F_z \end{vmatrix}$$ $$= -\begin{vmatrix} i & j & k \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \dfrac{\partial U}{\partial x} & \dfrac{\partial U}{\partial y} & \dfrac{\partial U}{\partial z} \end{vmatrix}$$ $$= -\left[i\left(\frac{\partial^2 U}{\partial y\partial z}-\frac{\partial^2 U}{\partial z\partial y}\right) - j\left(\frac{\partial^2 U}{\partial x\partial z}-\frac{\partial^2 U}{\partial z\partial x}\right) + k\left(\frac{\partial^2 U}{\partial x\partial y}-\frac{\partial^2 U}{\partial y\partial x}\right)\right]$$Since $P.E.\ (U)$ is a perfect differential, so,
$$\frac{\partial^2 U}{\partial y\partial z}=\frac{\partial^2 U}{\partial z\partial y}, \qquad \frac{\partial^2 U}{\partial x\partial z}=\frac{\partial^2 U}{\partial z\partial x}, \qquad \text{and} \qquad \frac{\partial^2 U}{\partial x\partial y}=\frac{\partial^2 U}{\partial y\partial x}$$ $$\Rightarrow \nabla\times F = 0$$or,
$$\text{Curl}\,F = 0$$Hence, the curl of a conservative force is always zero.
Law of Conservation of Mechanical Energy for a Conservative Force
We know,
$$W = \int_A^B F\cdot dr$$Since $F = -\nabla U$, i.e. $F = \dfrac{\partial U}{\partial r}$ giving $F\,dx = \partial U$,
$$W = \int_A^B dU = -\big|U\big|_A^B = -(U_B-U_A) = U_A-U_B \qquad \cdots (1)$$From the work–energy principle,
$$W = \int_A^B F\cdot dr = \frac{1}{2}mv_B^2-\frac{1}{2}mv_A^2 \qquad \cdots (2)$$From eq.(1) and (2),
$$U_A-U_B = \frac{1}{2}mv_B^2-\frac{1}{2}mv_A^2$$or,
$$U_A+\frac{1}{2}mv_A^2 = U_B+\frac{1}{2}mv_B^2$$or, Total initial energy = Total final energy
$$\Rightarrow \text{Total energy}\ (E) = \text{constant}$$Linear Restoring Force
[Diagram: Mass–Spring System]
Let us consider the motion of a particle subject to a linear restoring force $F$ along the $x$-axis. If $i,j,k$ be the unit vectors along $x,y,z$-axis and the displacement of the particle at any instant $t$ be $r$, then,
From Hooke's law,
$$\text{restoring force}\ (F)\propto r$$ $$\Rightarrow F=-Cr$$where $C$ is the force constant and the negative sign indicates the restoring force opposes displacement.
We know,
$$F=-\text{grad}\,U$$ $$\Rightarrow U=-\oint(F_x\,dx+F_y\,dy+F_z\,dz)$$ $$= -\oint(-Cx\,dx-Cy\,dy-Cz\,dz)$$ $$= \oint(Cx\,dx+Cy\,dy+Cz\,dz)$$ $$= C\left[\int x\,dx+\int y\,dy+\int z\,dz\right]$$ $$= C\left[\frac{x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2}\right]+A \qquad (A=\text{integration constant})$$ $$= \frac{C}{2}(x^2+y^2+z^2)+A$$or,
$$U = \frac{1}{2}Cr^2+A \qquad (\because r^2=x^2+y^2+z^2)$$For $r=0$, $U=0$,
$$\Rightarrow A=0$$ $$\therefore \boxed{U = \frac{1}{2}Cr^2}$$The graph between potential energy and displacement is shown below.
[Diagram: Parabolic Curve]
The total energy of a particle under a conservative force is given by:
$$E=U+T \qquad (U\to P.E.,\ T\to K.E.)$$At maximum P.E., $T=0$.
$$\therefore E=U_{max}$$ $$E=U_{max}=\frac{1}{2}Ca^2$$For any displacement $x$,
$$E=\frac{1}{2}Cx^2+T$$ $$\frac{1}{2}Ca^2=\frac{1}{2}Cx^2+T$$ $$\therefore T=\frac{1}{2}C(a^2-x^2)$$At $x=0$,
$$T_{max}=\frac{1}{2}Ca^2$$At $x=a$,
$$T_{min}=0$$Due to the inertia of motion, the particle crosses the mean position. After that, K.E. decreases and P.E. increases until it becomes maximum. When P.E. is maximum, K.E. is zero. After that, the particle returns from the extreme position to the mean position. In this way, the particle vibrates between two extreme positions. The region under the two extreme positions of a vibrating particle is called the bounded region or potential well.
Potential Energy (P.E.) Curve
[Diagram: Potential Energy Curve]
The graph showing variation of P.E. with position is called the P.E. curve. It gives information about the motion of a particle without solving any equation. Let us take a motion of particle only along the $x$-direction.
- In the given figure, $Q$ and $N$ are maxima (maximum potential energy) and $C$ and $G$ are minima.
For a conservative force,
$$F=-\text{grad}\,U=-\frac{\partial U}{\partial x} \qquad \text{(for $x$-direction only)}$$ - Slope of the curve at any point gives the force acting on the particle at that point. For curves $CQ$ and $GN$, the slope is positive and force is negative. So potential energy $(U)$ increases on increasing displacement $x$.
- For curves $AC$ and $QG$, the slope is negative and force is positive. So potential energy $(U)$ decreases on increasing displacement $x$.
- In both curves, the force tends to pull the body into the region of lower potential energy.
- At points $C, G, Q$ and $N$, the slope of the graph is zero. So, force is zero. $\therefore$ These points are the positions of equilibrium.
But slight displacement of a particle at $Q$ and $N$ tends to move the particle to a lower potential energy. $\therefore$ These points are the positions of unstable equilibrium.
But slight displacement of a particle at $C$ and $G$ results in a force to bring it back to the original position.
Thus, these points are the positions of stable equilibrium. - In the horizontal portion of the curve beyond $N$, the slope of the curve as well as the force is zero, and the particle can remain in any displaced position, which is called the region of neutral equilibrium.
- For energy $E$, the particle has two turning points $B$ and $D$, and it oscillates inside the potential well $B, C, D$. And for energy $E_1$, the particle has four turning points $J, C, F$ and $H$. And it oscillates in either of the two potential wells $J,C,L$ and $F,G,H$.