Unit M3: Mechanics 3

For motion along a straight line, displacement $x$, signed velocity $v$ and signed acceleration $a$ are linked by derivatives. Choose the form whose independent variable matches the information given; this avoids introducing an unnecessary unknown function.

The displacement form of acceleration follows from the chain rule:$$\frac{\mathrm dv}{\mathrm dt}=\frac{\mathrm dv}{\mathrm dx}\frac{\mathrm dx}{\mathrm dt}=v\frac{\mathrm dv}{\mathrm dx}.$$Use it when acceleration or velocity is expressed in terms of $x$. If the required quantity is time and $v=f(x)$, separate variables instead: $\mathrm dt=\mathrm dx/v(x)$.

Every indefinite integration needs a constant. Apply a stated condition such as $x=x_0$ and $v=v_0$ at $t=t_0$ only after integrating, or use definite integrals with those values as limits. Keep $v$ signed: a negative velocity represents motion in the negative $x$-direction, while speed is $|v|$.

Suppose $v=12/(x+2)$ metres per second and $x=1$ when $t=0$. First,$$a=v\frac{\mathrm dv}{\mathrm dx}=\frac{12}{x+2}\left(-\frac{12}{(x+2)^2}\right)=-\frac{144}{(x+2)^3}.$$At $x=4$, $a=-2/3\,\mathrm{m\,s^{-2}}$. For the elapsed time to reach $x=4$,$$t=\int_1^4\frac{\mathrm dx}{v}=\int_1^4\frac{x+2}{12}\,\mathrm dx=\left[\frac{x^2}{24}+\frac{x}{6}\right]_1^4=\frac98\text{ s}.$$The negative acceleration means velocity is decreasing here; the deceleration magnitude is $2/3\,\mathrm{m\,s^{-2}}$.

Check that the final variable matches the question, substitute the initial condition back into the integrated relation, and verify units: $v$ has units $\mathrm{m\,s^{-1}}$ and $a$ has units $\mathrm{m\,s^{-2}}$. If a derivation divides by $v$ or by another expression that can be zero, solve on intervals where that division is valid and inspect the zero case separately.

An elastic element has natural length $l$, the length at which it is undeformed. If its current length is $L$, its extension is $x=L-l$. Under Hooke's law, the elastic-force magnitude is proportional to this extension.

Here $T$ is tension in newtons and $\lambda$ is the modulus of elasticity, also measured in newtons. The ratio $x/l$ is dimensionless. When a spring is described using stiffness $k$ instead, the equivalent form is $T=kx$, with $k=\lambda/l$ and unit $\mathrm{N\,m^{-1}}$.

First calculate extension from the actual geometry, not from the total length alone. Apply Hooke's law to find tension, then place that force on the free-body diagram and resolve it in the chosen direction. If two equal sections pull symmetrically, include both resolved tensions; do not replace them by one tension.

A particle of mass $1.5$ kg hangs from a vertical light elastic string with natural length $0.8$ m and modulus $40$ N. At an instant the string has length $1.1$ m, so $x=0.3$ m and$$T=\frac{40(0.3)}{0.8}=15\text{ N}.$$Taking upward as positive, the particle's acceleration at that instant is$$a=\frac{T-mg}{m}=\frac{15-1.5(9.8)}{1.5}=0.20\,\mathrm{m\,s^{-2}}.$$Hooke's law supplies the force; Newton's second law supplies the acceleration.

Stretching an elastic string or deforming a spring stores elastic potential energy (EPE). For a Hooke-law element, the force rises linearly from zero, so the stored energy is the area under the force-extension relation.

Use the extension $x=L-l$ at the particular state being considered. For an elastic string, $E_{\mathrm e}=0$ whenever the string is slack; for a spring, $x^2$ gives positive stored energy for either extension or compression. Elastic energy is measured in joules.

Choose initial and final states, calculate the elastic extension separately at each state, and use one consistent height datum. If only gravity and ideal elastic forces do work,$$K_i+U_{g,i}+E_{\mathrm e,i}=K_f+U_{g,f}+E_{\mathrm e,f}.$$Otherwise add the work of any other force with its correct sign. Energy equations compare states; they do not require the acceleration to be constant.

A $1$ kg particle is attached below a vertical elastic string of natural length $1$ m and modulus $20$ N. It is released from rest with extension $0.10$ m and falls $0.20$ m, so the final extension is $0.30$ m. The initial and final EPE values are $0.10$ J and $0.90$ J. Conservation of mechanical energy gives$$0+0.10+1(9.8)(0.20)=\frac12v^2+0.90,$$hence $\tfrac12v^2=1.16$ and $v=1.52\,\mathrm{m\,s^{-1}}$ to three significant figures.

For motion along one signed axis, first write the resultant force with its direction, then choose the acceleration form that matches the variable in the force law. A varying force produces a differential equation rather than a constant-acceleration problem.

Define the coordinate origin and positive direction before inserting a force. Include every force component in the resultant. Separate variables or integrate with limits, use the supplied motion condition to determine the constant, and accept only positions for which $v^2\ge0$ and the force model is defined.

A $2$ kg particle moves in the positive $x$-direction under resultant force $F=12-2x$ N. At $x=0$, its speed is $1\,\mathrm{m\,s^{-1}}$. Since$$2v\frac{\mathrm dv}{\mathrm dx}=12-2x,$$integration between the initial position and $x$ gives$$v^2=1+12x-x^2.$$At $x=3$ m, $v=\sqrt{28}=2\sqrt7\,\mathrm{m\,s^{-1}}$. A turning point requires $v=0$; the positive solution is $x=6+\sqrt{37}$ m.

For an attractive inverse-square force with outward coordinate $x>0$, $F=-k/x^2$. Then$$mv\frac{\mathrm dv}{\mathrm dx}=-\frac{k}{x^2}$$integrates to $\tfrac12mv^2=k/x+C$. The plus sign on $k/x$ after integration is consistent with differentiating $k/x$ to recover $-k/x^2$.

Motion is simple harmonic when acceleration is proportional to displacement from a fixed equilibrium point and always directed back towards that point. With signed displacement $x$ from equilibrium, the defining equation is

To prove SHM in a mechanical system, write a signed equation of motion and simplify it in terms of displacement from equilibrium. Constant force terms must cancel at equilibrium. Reaching $\ddot x=-c x$ with $c>0$ proves SHM and identifies $\omega=\sqrt c$; merely obtaining a periodic-looking expression is not the required force-law proof.

Choose the phase from the initial state. Release from rest at $x=A$ allows $x=A\cos\omega t$; starting at equilibrium in the positive direction allows $x=A\sin\omega t$. For time spent in a region, solve the boundary displacement within one cycle and use the symmetry of the cosine or sine curve without double-counting endpoints.

If $x=0.12\cos(5t)$ metres, then $A=0.12$ m, $\omega=5\,\mathrm{rad\,s^{-1}}$ and $T=2\pi/5$ s. The maximum speed is $0.60\,\mathrm{m\,s^{-1}}$. Starting at the positive endpoint, the first time the particle reaches $x=0.06$ m satisfies $\cos(5t)=1/2$, so $t=\pi/15$ s.

A particle attached to an ideal spring or taut elastic string can perform SHM along the element's direction. Measure displacement from the equilibrium position: doing so makes the constant weight term cancel and exposes the restoring force.

For a vertical element, let its equilibrium extension be $e$. Equilibrium gives $ke=mg$. If downward displacement from equilibrium is $x$, the extension is $e+x$ and$$m\ddot x=mg-k(e+x)=-kx.$$Gravity shifts the equilibrium position but does not appear in $\omega$ after this cancellation.

A $0.5$ kg particle oscillates vertically on an elastic element with stiffness $18\,\mathrm{N\,m^{-1}}$. Then $\omega=\sqrt{18/0.5}=6\,\mathrm{rad\,s^{-1}}$ and $T=\pi/3$ s. If its amplitude is $0.08$ m, its speed at displacement $0.04$ m is$$v=6\sqrt{0.08^2-0.04^2}=0.416\,\mathrm{m\,s^{-1}}.$$If the element is a string, this calculation applies for the whole oscillation only after checking that the minimum extension remains non-negative.

Angular speed $\omega$ measures the rate at which angular position changes. Angles in circular-motion formulae are measured in radians, so one complete revolution is $2\pi$ radians.

For a point at radius $r$, arc length is $s=r\theta$. Differentiating gives tangential speed $v=r\omega$. All points on one rigid rotating body have the same angular speed, but a point farther from the axis has a larger linear speed.

A wheel completes $15$ revolutions in $6$ minutes. Its period is $360/15=24$ s and$$\omega=\frac{2\pi}{24}=\frac{\pi}{12}\,\mathrm{rad\,s^{-1}}.$$A point $8$ m from the axis therefore moves at $v=8(\pi/12)=2\pi/3\,\mathrm{m\,s^{-1}}$.

A particle moving in a circle accelerates even when its speed is constant, because its velocity direction changes. The acceleration points radially inward, towards the centre of the circle.

The inward acceleration is not an extra force. Choose inward as the positive radial direction and resolve the actual forces:$$\sum F_{\mathrm{inward}}=m\frac{v^2}{r}=mr\omega^2.$$Forces pointing away from the centre enter with a negative sign.

A $0.40$ kg particle moves at $3.0\,\mathrm{m\,s^{-1}}$ in a circle of radius $1.5$ m. Then$$a_r=\frac{3.0^2}{1.5}=6.0\,\mathrm{m\,s^{-2}},$$so the resultant inward force must be $0.40(6.0)=2.4$ N. Individual forces need not each equal $2.4$ N; their inward components must have that resultant.

For uniform motion in a horizontal circle, vertical acceleration is zero while the horizontal resultant towards the centre is $mr\omega^2$. Draw every real force, resolve vertically for equilibrium, and resolve horizontally inwards for circular motion.

Mark the circle's horizontal radius before resolving. Geometry determines $r$ and the angles of strings or reactions. Use limiting friction only when the speed is at a limiting value; otherwise the friction magnitude is an unknown no greater than $\mu R$. Check that every tension and reaction is non-negative.

A $0.50$ kg particle is a conical pendulum on a string of length $1.2$ m at $30^\circ$ to the vertical. Its circle has radius $r=1.2\sin30^\circ=0.60$ m. From$$T\cos30^\circ=mg,\qquad T\sin30^\circ=mr\omega^2,$$division gives $\tan30^\circ=r\omega^2/g$. Hence $\omega=3.07\,\mathrm{rad\,s^{-1}}$ and $T=mg/\cos30^\circ=5.66$ N using $g=9.8\,\mathrm{m\,s^{-2}}$.

In a vertical circle, speed changes with height while the required inward resultant remains $mv^2/r$. Use conservation of energy to find speed at a position, then use a separate radial equation to find tension or reaction.

Let $\theta$ be measured from the lowest point and let the bottom speed be $u$. The height above the bottom is $r(1-\cos\theta)$. With no non-conservative work,$$v^2=u^2-2gr(1-\cos\theta).$$For a particle on a taut string, resolving inward gives

A string must satisfy $T\ge0$; a smooth contact must satisfy reaction $R\ge0$. At the first loss of constraint, set the relevant force to zero. For a string to complete a full vertical circle, the limiting top condition is $v_{\mathrm{top}}^2=gr$, which with energy gives the minimum bottom speed $u=\sqrt{5gr}$. A rigid rod can push as well as pull, so it does not use the string-tension condition.

A particle on a string enters the bottom with $u^2=6gr$. At the top, energy gives $v^2=6gr-4gr=2gr$, so$$T+mg=2mg\quad\Rightarrow\quad T=mg>0.$$The string remains taut there. At the bottom, $T-mg=6mg$, so $T=7mg$; the larger bottom tension supplies both the inward acceleration and opposition to the downward weight.

The centre of mass is the point at which the body's total mass may be treated as concentrated when calculating moments. For a uniform body, symmetry locates every coordinate that lies on an axis or plane of symmetry; integration is needed for a remaining coordinate when the shape varies.

Choose axes that exploit symmetry. Write both the total mass and its first moment using the same density model and limits, then divide first moment by total mass. For a simple composite body, treat each part as a point mass at its own centre of mass:$$\bar x=\frac{\sum m_i x_i}{\sum m_i},\qquad \bar y=\frac{\sum m_i y_i}{\sum m_i}.$$A removed piece is entered with negative mass (or negative area or volume when density is common).

A uniform $6\times4$ rectangular lamina has a $2\times2$ square removed from its top-right corner. Taking the lower-left corner as the origin, use the full rectangle minus the hole:$$\bar x=\frac{24(3)-4(5)}{24-4}=2.6,\qquad \bar y=\frac{24(2)-4(3)}{24-4}=1.8.$$The centre shifts left and down, away from the removed mass, which checks the result.

A rigid body is in equilibrium only when the resultant force is zero and the resultant moment about any point is zero. Its weight acts vertically through its centre of mass, so the position of that vertical line controls suspension and toppling.

First find the centre of mass in body-fixed coordinates. For suspension, rotate the body until the line joining the suspension point to the centre of mass is vertical. For a supported body, identify the possible pivot edge and take moments about it. At the instant of toppling the reaction at the opposite side has fallen to zero and the resultant contact force acts through the pivot edge.

A uniform rectangular block has base length $b$ measured up the line of greatest slope and its centre of mass is a perpendicular distance $h$ above the plane. On a rough plane inclined at $\theta$, the vertical through the centre of mass meets the plane $h\tan\theta$ downhill from the middle of the base. It is stable against toppling while$$h\tan\theta<\frac b2,$$and is on the point of toppling when $h\tan\theta=b/2$. This test assumes the available friction prevents sliding first.