The Foundation of Mechanics: Integral Calculus

Problem Statement

In kinematics, we often start with an expression for a particle's acceleration, derived from the forces acting upon it. However, the questions usually ask about the particle's speed, distance travelled, or position at a later time.

Integration is the foundational mathematical operation that allows us to reverse the process of differentiation, moving from acceleration to velocity, and from velocity to displacement. Without strong integration skills—including substitution, integration by parts, and trigonometric integration—solving complex mechanics differential equations is impossible.

Consider a particle of mass $m=1$ kg moving in a straight line along the $x$-axis. It is subjected to an attractive force towards the origin, proportional to the inverse square of its distance from the origin. The acceleration is given by: $a = -\frac{k}{x^2}$ where $k$ is a positive constant. The particle is initially held at rest at a position $x = D$ (where $D > 0$) and then released.

(a) By using the identity $a = \frac{d}{dx} \left(\frac{1}{2}v^2\right)$, find an expression for the velocity $v$ of the particle in terms of $x$, $k$, and $D$ as it moves towards the origin.

(b) Show that the time $T$ taken for the particle to reach a point $x = \frac{D}{2}$ is given by the integral: $T = \sqrt{\frac{D}{2k}} \int_{D/2}^{D} \sqrt{\frac{x}{D-x}} \, dx$

(Note: Evaluating this integral is a complex Extension 2 substitution usually involving $x = D \sin^2\theta$, but we will stop at setting it up for this exercise).

Hints

• Part (a): Replace $a$ with $\frac{d}{dx} \left(\frac{1}{2}v^2\right)$. Integrate both sides with respect to $x$. Don't forget the constant of integration. Use the initial condition (at $t=0$, $v=0$, $x=D$) to find this constant. When taking the square root for velocity, consider the direction of motion to choose the correct sign.
• Part (b): Once you have an expression for $v$ (which is $\frac{dx}{dt}$), rearrange the equation to make $t$ the subject. You will need to separate variables ($x$ and $t$) and integrate again. Pay close attention to your limits of integration and the sign of $v$.

Solutions

Part (a): Integrating for Velocity

1. We are given acceleration $a = -\frac{k}{x^2}$.
2. Use the specified identity for acceleration: $\frac{d}{dx} \left(\frac{1}{2}v^2\right) = -\frac{k}{x^2}$.
3. Integrate both sides with respect to $x$: $\int \frac{d}{dx} \left(\frac{1}{2}v^2\right) \, dx = \int -kx^{-2} \, dx$ $\frac{1}{2}v^2 = \frac{-kx^{-1}}{-1} + C$ $\frac{1}{2}v^2 = \frac{k}{x} + C$
4. Apply initial conditions: The particle is released from rest at $x=D$. Therefore, when $x = D$, $v = 0$. $0 = \frac{k}{D} + C \implies C = -\frac{k}{D}$
5. Substitute $C$ back into the equation: $\frac{1}{2}v^2 = \frac{k}{x} - \frac{k}{D}$ $\frac{1}{2}v^2 = k \left( \frac{D - x}{xD} \right)$ $v^2 = \frac{2k(D - x)}{xD}$
6. Take the square root to find $v$. Since the particle starts at $x=D$ ($D>0$) and is attracted to the origin, it moves in the negative $x$-direction. Therefore, velocity must be negative. $v = -\sqrt{ \frac{2k(D - x)}{xD} }$

Part (b): Setting up the Time Integral

1. We know that velocity $v = \frac{dx}{dt}$. Substitute this into our velocity equation: $\frac{dx}{dt} = -\sqrt{ \frac{2k(D - x)}{xD} }$
2. Separate the variables $x$ and $t$: $dt = -\sqrt{ \frac{xD}{2k(D - x)} } \, dx$ $dt = -\sqrt{\frac{D}{2k}} \sqrt{\frac{x}{D - x}} \, dx$
3. We want to find the time $T$ taken to move from $x = D$ to $x = \frac{D}{2}$. We integrate both sides. The time limits are from $0$ to $T$. The corresponding position limits are from $D$ to $\frac{D}{2}$. $\int_0^T 1 \, dt = \int_{D}^{D/2} -\sqrt{\frac{D}{2k}} \sqrt{\frac{x}{D - x}} \, dx$
4. Evaluate the left side: $T = -\sqrt{\frac{D}{2k}} \int_{D}^{D/2} \sqrt{\frac{x}{D - x}} \, dx$
5. To make the integral look neater (and match the required form), we can swap the limits of integration. Swapping the limits multiplies the integral by $-1$, which cancels out the negative sign in front. $T = \sqrt{\frac{D}{2k}} \int_{D/2}^{D} \sqrt{\frac{x}{D - x}} \, dx$ This matches the required expression.

Takeaways

• The Power of $v \frac{dv}{dx}$: The identity $a = \frac{d}{dx}(\frac{1}{2}v^2)$ (or $v \frac{dv}{dx}$) is incredibly powerful when acceleration is given as a function of position ($x$) rather than time ($t$). It provides a direct path via integration to find velocity as a function of position.
• Direction Matters: When taking the square root of $v^2$, you must actively choose the positive or negative root based on the physical context of the problem. A particle moving towards the origin from a positive position has negative velocity.
• Integration Limits: Be meticulous when assigning limits of integration. Ensure the lower limit of time corresponds strictly to the lower limit of position, and similarly for the upper limits. Swapping limits to absorb negative signs is a common and useful algebraic trick.

Further Readings

• HSC Integrals: https://vumaths.com/booklets/hsc-integrals/
• HSC Mechanics: https://vumaths.com/booklets/hsc-mechanics/
• HSC Differential Equations: https://vumaths.com/booklets/hsc-differential-equations/

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Integration is the key that unlocks mechanics. For exhaustive practice on tricky mechanics integrals, explore the booklets on Vu's Maths Hub. I regularly post solutions to difficult integrals and differential equations on YouTube. Follow me on LinkedIn for professional updates, or read my detailed articles on the HSC curriculum over at my Substack.