Unit M2: Mechanics 2

Motion in a vertical plane is two-dimensional, but a constant acceleration lets each perpendicular component be modelled with the same signed SUVAT relationships. Choose fixed unit vectors $\mathbf i$ horizontally and $\mathbf j$ vertically, then keep one common time $t$ for both components.

For free motion under gravity with upward $\mathbf j$, acceleration is constant and vertical: $\mathbf a=-g\mathbf j$. Hence horizontal velocity remains constant, while vertical velocity changes by $-gt$. Gravity changes neither the horizontal component directly nor the chosen coordinate axes.

A particle starts at the origin with velocity $(6\mathbf i+8\mathbf j)\,\mathrm{m\,s^{-1}}$ and acceleration $-9.8\mathbf j\,\mathrm{m\,s^{-2}}$. After $1$ s,$$\mathbf r=6\mathbf i+(8-4.9)\mathbf j=6\mathbf i+3.1\mathbf j\text{ m},$$and $\mathbf v=6\mathbf i-1.8\mathbf j\,\mathrm{m\,s^{-1}}$. The negative vertical velocity means downward motion; the particle may still be above its starting level.

A projectile is modelled as a particle that, after projection, moves freely under gravity. With air resistance neglected, its horizontal velocity is constant and its vertical acceleration is $-g$ when upward is positive. The two component motions are independent but describe the same particle at the same time.

Use the component whose displacement condition is known to find the physically relevant $t\ge0$, then substitute that same time into the other component. At greatest height $v_y=0$, not the whole velocity. At impact, use $|\mathbf v|=\sqrt{v_x^2+v_y^2}$ and obtain its direction from the signed components.

A particle is projected at $20\,\mathrm{m\,s^{-1}}$ at $30^\circ$ above horizontal ground and lands at its launch level. From $0=10t-4.9t^2$, the non-zero flight time is $t=2.04$ s. Its horizontal range is $(20\cos30^\circ)(2.04)=35.3$ m, and its greatest height is found from $0^2=10^2-2(9.8)h$, giving $h=5.10$ m.

When signed displacement $x$ is a differentiable function of time, its first derivative is velocity and its second derivative is acceleration. Differentiation reveals instantaneous motion; integration reconstructs velocity or displacement only after the relevant initial condition fixes the constant.

For $t\ge0$, let $x=t^3-6t^2+9t$ metres. Then$$v=3(t-1)(t-3),\qquad a=6t-12.$$The particle is instantaneously at rest at $t=1$ and $t=3$, and the velocity changes sign at both times. Since $x(0)=0$, $x(1)=4$, $x(3)=0$ and $x(4)=4$, the distance travelled from $t=0$ to $t=4$ is $4+4+4=12$ m, not $|x(4)-x(0)|=4$ m.

A position vector $\mathbf r(t)=x(t)\mathbf i+y(t)\mathbf j$ records a particle's coordinates relative to a fixed origin. Differentiate or integrate each component with respect to the same time variable; the constant of integration is itself a vector fixed by a position or velocity condition.

Suppose$$\mathbf v=(3t^2-3)\mathbf i+2t\mathbf j\quad\text{and}\quad\mathbf r(0)=2\mathbf i-\mathbf j.$$Integrating and applying the initial position gives$$\mathbf r=(t^3-3t+2)\mathbf i+(t^2-1)\mathbf j.$$Also $\mathbf a=6t\mathbf i+2\mathbf j$. At $t=1$, $\mathbf v=2\mathbf j\,\mathrm{m\,s^{-1}}$, so the motion is parallel to $\mathbf j$ and the speed is $2\,\mathrm{m\,s^{-1}}$.

The centre of mass of discrete particles is the mass-weighted mean of their positions. Choose an origin and signed coordinate axes first: particles farther from an axis contribute a larger moment, but only in proportion to their masses.

These equations state that the total first moment about an axis is unchanged if the whole distribution is replaced by its total mass at $(\bar x,\bar y)$. In one dimension use only the relevant coordinate. In two dimensions calculate the two coordinates separately with the same total mass.

Particles of masses $2$, $3$ and $5$ kg are at $(0,0)$, $(4,0)$ and $(2,6)$ metres. Their total mass is $10$ kg, so$$\bar x=\frac{2(0)+3(4)+5(2)}{10}=2.2,qquad \bar y=\frac{2(0)+3(0)+5(6)}{10}=3.$$The centre of mass is therefore $(2.2,3)$ m relative to the chosen origin.

For a uniform plane lamina, mass is proportional to area. Split a composite figure into standard shapes, place each shape's area at its known centre, and take area-weighted moments. An axis of symmetry contains the centre of mass; two independent symmetry axes locate it immediately.

Choose axes that make all component centres easy to measure. Record signed areas and centre coordinates in one table, then take moments about a parallel reference axis. The numerator has dimensions of length cubed and the denominator length squared, so the result must have dimensions of length. No integration is required in this syllabus objective.

A uniform $6\times4$ rectangle has the top-right $2\times2$ square removed. From the lower-left corner, the full rectangle has area $24$ and centre $(3,2)$; the removed square has signed area $-4$ and centre $(5,3)$. Hence$$\bar x=\frac{24(3)-4(5)}{20}=2.6,qquad \bar y=\frac{24(2)-4(3)}{20}=1.8.$$The missing top-right mass shifts the centre down and left.

A lamina's weight acts vertically downward through its centre of mass. Equilibrium requires zero resultant force and zero total moment. The geometry of the centre of mass therefore determines how a suspended, pivoted or supported lamina can rest.

For free suspension, first locate the centre of mass $G$ in body-fixed coordinates. The lamina rotates until $PG$ is vertical, with $G$ below $P$. If in a reference orientation $G$ is $3$ cm to the right and $4$ cm below $P$, the angle through which $PG$ differs from the downward vertical satisfies$$\tan\theta=\frac{3}{4},$$so $\theta=36.9^\circ$. Use the required orientation in the diagram to choose the correct sense.

If a pivoted lamina of weight $W$ has a horizontal moment arm $0.30$ m from its axis, and a horizontal holding force $F$ has perpendicular moment arm $0.50$ m, equilibrium gives$$F(0.50)=W(0.30),$$so $F=0.60W$. The pivot reaction is absent from this moment equation because its line of action passes through the pivot.

Energy methods compare two states without requiring the time taken between them. Kinetic energy depends on speed, gravitational potential energy depends on vertical height, and work transfers energy when a force acts through a displacement.

Here $m$ is mass, $v$ is speed, $h$ is height above any consistent datum, and $\phi$ is the angle between a constant force and the displacement. Energy and work are measured in joules. Only changes in potential energy matter, so choose a convenient zero height and preserve the sign of $\Delta h$.

The work-energy principle is equivalently $\Delta K=$ total work done by all forces. Using gravitational potential energy in the ledger avoids also counting gravity as work: choose one representation, not both. On an incline of length $s$ and angle $\theta$, the vertical change is $s\sin\theta$, while a constant resistance of magnitude $R$ dissipates energy $Rs$ in either direction of travel.

A $4$ kg particle moves $5$ m up an incline whose sine is $0.3$. Its initial speed is $8\,\mathrm{m\,s^{-1}}$ and a constant resistance of $6$ N opposes motion. With no driving work,$$\frac12(4)v^2+4g(5\times0.3)=\frac12(4)(8^2)-6(5).$$Thus $2v^2+58.8=98$, so $v=4.43\,\mathrm{m\,s^{-1}}$. The calculation accounts separately for increased potential energy and energy dissipated by resistance.

Power is the rate of doing work, measured in watts, with $1\,\mathrm{kW}=1000\,\mathrm W$. At a particular speed, a parallel driving force is $F=P/v$; therefore constant power produces a smaller driving force at a larger speed. For example, $12$ kW at $20\,\mathrm{m\,s^{-1}}$ corresponds to $600$ N before resistance and weight components are included in $F=ma$.

Linear momentum has the same direction as velocity: for a particle of mass $m$, $\mathbf p=m\mathbf v$. An impulse is the vector change in momentum, so its components change the corresponding velocity components independently.

Impulse is measured in $\mathrm{N\,s}=\mathrm{kg\,m\,s^{-1}}$. During an interaction, the impulses two particles exert on each other are equal and opposite. If the total external impulse on a chosen system is negligible, those internal impulses cancel and total vector momentum is conserved.

A $2$ kg particle has velocity $(3\mathbf i-\mathbf j)\,\mathrm{m\,s^{-1}}$ and receives impulse $(-4\mathbf i+6\mathbf j)\,\mathrm{N\,s}$. Then$$\mathbf v=\mathbf u+\frac{\mathbf J}{m}=(3\mathbf i-\mathbf j)+(-2\mathbf i+3\mathbf j)=\mathbf i+2\mathbf j.$$Its new speed is $\sqrt5\,\mathrm{m\,s^{-1}}$ and its direction is $\tan^{-1}2$ above the positive $\mathbf i$-axis.

For a direct impact, choose one positive direction along the line of motion and give every velocity a sign. During the short collision, conserve total signed momentum for the two-particle system and pair it with Newton's law of restitution.

The restitution equation shown assumes particle 1 approaches particle 2 from behind, so $u_1-u_2$ is the relative speed of approach and $v_2-v_1$ the relative speed of separation. Rebuild the equation from 'separation equals $e$ times approach' if directions differ; do not memorize signs without the geometry.

A $2$ kg particle moving at $5\,\mathrm{m\,s^{-1}}$ catches a $3$ kg particle moving in the same direction at $1\,\mathrm{m\,s^{-1}}$. If $e=0.5$, then$$13=2v_1+3v_2,\qquad v_2-v_1=2.$$Thus $v_1=1.4$ and $v_2=3.4\,\mathrm{m\,s^{-1}}$. Kinetic energy falls from $26.5$ J to$$\tfrac12(2)(1.4)^2+\tfrac12(3)(3.4)^2=19.3\text{ J},$$so the loss is $7.2$ J.

A successive-impact problem is a sequence of separate direct collisions. After each event, replace the affected velocities with their new values, preserve every unaffected velocity, then decide which objects can meet next from their positions and signed relative velocities.

A fixed wall supplies an external impulse, so particle momentum is not conserved across the wall impact. For a normal impact with wall coefficient $e_w$, a velocity $+u$ towards the wall becomes $-e_wu$ after rebound when positive points towards the wall. This syllabus does not include oblique impact with a plane surface.

Immediately after a wall rebound, particle $Q$ is $6$ m to the right of particle $P$. Let right be positive. If $P$ moves right at $1\,\mathrm{m\,s^{-1}}$ and $Q$ moves left at $3.2\,\mathrm{m\,s^{-1}}$, their closing speed is $1-(-3.2)=4.2\,\mathrm{m\,s^{-1}}$. They therefore collide after $6/4.2=1.43$ s. If $Q$ had instead moved right faster than $P$, the gap would grow and no second collision would occur.

When a coefficient is unknown, conditions such as 'a second collision occurs' become strict inequalities on signed speeds after the preceding event. Intersect the resulting condition with $0\le e\le1$. At a boundary where closing speed is zero, the particles never close a positive gap, so equality usually does not produce another collision.

The moment of a force about a point measures the force's turning effect about that point. Its magnitude is the force multiplied by the perpendicular distance from the point to the force's line of action—not necessarily the distance to the point where the force is applied.

Choose and state a sign convention, such as anticlockwise positive. A force whose line of action passes through the chosen point has $d_{\perp}=0$ and hence zero moment about that point. Moment has unit $\mathrm{N\,m}$; it is not a force and should not be labelled in newtons.

A horizontal rod is pivoted at $A$. A $30$ N downward force acts $2$ m from $A$, and an upward force $P$ acts $5$ m from $A$. Taking anticlockwise as positive, equilibrium of moments about $A$ gives$$5P-30(2)=0,$$so $P=12$ N. Any reaction at $A$ contributes no moment about $A$, which is why this pivot makes the equation efficient.

Before calculating, extend each force mentally into its line of action and find the shortest distance from the pivot. Check that every term has dimensions force $\times$ distance and that forces on opposite sides or with opposite turning effects receive opposite signs.

A rigid body in coplanar equilibrium has no translational acceleration and no angular acceleration. Draw an isolated free-body diagram, replacing every contact by the forces it can exert, then use two independent force balances and one moment balance.

Friction opposes the impending or actual relative motion at a contact. In ordinary equilibrium, $|F|\le\mu R$; write $F=\mu R$ only when the body is stated to be in limiting equilibrium or on the point of slipping. A smooth contact has no friction, not no reaction.

A reliable order is: (1) mark every weight, reaction, friction and tension; (2) choose axes that simplify components; (3) take moments about a point through which the most unknown forces act; (4) resolve horizontally and vertically; (5) solve and check directions, non-negative reactions and any friction inequality. Other pivots give equivalent equations but may keep more unknowns.

A uniform $5$ m ladder of weight $W$ rests at angle $\theta$ on rough horizontal ground and against a smooth vertical wall, where $\sin\theta=4/5$ and $\cos\theta=3/5$. Let the wall reaction be $H$, and let the ground supply vertical reaction $R$ and horizontal friction $F$. Taking moments about the foot,$$H(5\sin\theta)=W\left(\frac52\cos\theta\right),$$so $4H=\tfrac32W$ and $H=3W/8$. Force balance gives $F=H=3W/8$ and $R=W$. Therefore equilibrium requires $\mu\ge F/R=3/8$; if the ladder is on the point of slipping, $\mu=3/8$.

When a supported body is on the point of tilting about one contact, the reaction at the other contact has fallen to zero. Take moments about the remaining contact and keep only forces still acting. This is a different limiting condition from impending sliding, although a problem may require both ideas.