3. Further Mechanics

With initial speed $u$ at angle $\theta$ above the horizontal,u_x=u\cos\theta,\qquad u_y=u\sin\theta.The components share the same elapsed time, but horizontal velocity stays constant while vertical velocity changes under $-g$.

From launch point $O$,x=(u\cos\theta)t,\qquad y=(u\sin\theta)t-\tfrac12gt^2,$$$$v_x=u\cos\theta,\qquad v_y=u\sin\theta-gt.At any time, speed is $\sqrt{v_x^2+v_y^2}$ and the direction satisfies $\tan\alpha=|v_y|/v_x$, with ascent/descent stated.

Let $u=20\text{ m s}^{-1}$, $\theta=30^\circ$ and $g=10\text{ m s}^{-2}$ on level ground. Then $v_x=10\sqrt3$ and $v_y=10-10t$. Greatest height occurs at $t=1$:H=10(1)-5(1)^2=5\text{ m}.Returning to $y=0$ gives $t=2$, soR=(10\sqrt3)(2)=20\sqrt3\text{ m}. $$$$

Immediately before landing, $(v_x,v_y)=(10\sqrt3,-10)$, so the speed is $20\text{ m s}^{-1}$ and the direction is $30^\circ$ below the horizontal. Equal launch and landing heights create this symmetric speed result.

$$From$$x=u\cos\theta,t,\qquad y=u\sin\theta,t-\tfrac12gt^2,use $t=x/(u\cos\theta)$ to obtainy=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}=xT-\frac{gx^2}{2u^2}(1+T^2),\qquad T=\tan\theta.$$Add the initial height if the launch point is not the origin.$$

If a known-speed projectile passes through $(X,Y)$, substitute the point to get a quadratic in $T$:\frac{gX^2}{2u^2}T^2-XT+\left(Y+\frac{gX^2}{2u^2}\right)=0.Each admissible real root gives a possible launch angle $\theta=\tan^{-1}T$; often these are low and high paths.

$$If the angle is known instead, rearrange the same point condition:$$u^2=\frac{gX^2(1+T^2)}{2(XT-Y)},requiring $XT>Y$. Use intersections with ground, walls or targets only on the forward part of the flight, with $t=x/(u\cos\theta)\ge0$.

For a coplanar force $F$, the moment about $O$ isM_O=F d_{\perp}=Fr\sin\phi,where $d_{\perp}$ is the perpendicular distance from $O$ to the force's line of action and $\phi$ is the angle between the position line and force. Units are N m.

A $40\text{ N}$ force acts at a point $0.50\text{ m}$ from O and makes $30^\circ$ with the position line. Its moment magnitude is40(0.50)\sin30^\circ=10\text{ N m}.$$State the sign from the actual turning sense.$$

For a uniform triangular lamina, G is the centroid: it lies on every median, $\tfrac13$ of the altitude from the base and $\tfrac23$ from the opposite vertex. If the vertices are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$,G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right).$$$$

For vertices $(0,0)$, $(6,0)$ and $(2,3)$,G=\left(\frac{0+6+2}{3},\frac{0+0+3}{3}\right)=\left(\frac83,1\right).The vertical coordinate is one third of the height from the base $y=0$.

For component masses $m_i$ at $(x_i,y_i)$,\bar x=\frac{\sum m_ix_i}{\sum m_i},\qquad \bar y=\frac{\sum m_iy_i}{\sum m_i}.$$With common density and thickness, lamina masses are proportional to areas; for solids with common density, masses are proportional to volumes. A removed piece contributes negative mass and negative mass moment.$$

An L-lamina is a $4\times3$ rectangle with the top-right $2\times1$ rectangle removed. From the lower-left origin, the large rectangle has area 12 and centre $(2,1.5)$; the cut-out has area 2 and centre $(3,2.5)$. Thus\bar x=\frac{12(2)-2(3)}{10}=1.8,\qquad \bar y=\frac{12(1.5)-2(2.5)}{10}=1.3.$$$$

$$For coplanar forces on one rigid body, equilibrium requires$$\sum F_x=0,\qquad \sum F_y=0,\qquad \sum M_O=0$$about any point O. Conversely, if the resultant force and resultant moment are both zero, the rigid body is in equilibrium.$$

A horizontal $4\text{ m}$ beam has weight $60\text{ N}$ at its centre, an extra $20\text{ N}$ load $3\text{ m}$ from support A, and vertical reactions $R_A,R_B$. Moments about A give4R_B=60(2)+20(3),so $R_B=45\text{ N}$. Vertical balance gives $R_A+R_B=80$, hence $R_A=35\text{ N}$.

A rectangular crate of weight $W$, base width $b$ and height $h$ is pushed horizontally at its top on a rough horizontal floor. Before motion, $R=W$. Sliding would begin atP_s=\mu W.$$Toppling about the lower far edge would begin when$$P_t h=W\frac b2,\qquad P_t=\frac{Wb}{2h}. $$$$

$$Angular speed is the rate of angular displacement:$$\omega=\frac{d\theta}{dt},\qquad \omega=\frac{\theta}{t}\text{ for constant }\omega.$$With angles in radians,$$v=r\omega,\qquad \omega=\frac{2\pi}{T}=2\pi f.Its unit is rad s$^{-1}$.

A disc turns at $120$ revolutions per minute. Then $f=2\text{ Hz}$ and\omega=2\pi(2)=4\pi\text{ rad s}^{-1}.A point $0.30\text{ m}$ from the axis hasv=r\omega=0.30(4\pi)=1.2\pi\text{ m s}^{-1}. $$$$

$$Its magnitude is$$a=r\omega^2=\frac{v^2}{r}.$$The resultant of the real force components in the inward radial direction must therefore be$$F_{\text{in,resultant}}=ma.$$Proof of the acceleration formulas is not required.$$

A $0.50\text{ kg}$ particle moves at $6.0\text{ m s}^{-1}$ in a circle of radius $2.0\text{ m}$. Thena=\frac{6.0^2}{2.0}=18\text{ m s}^{-2},\qquad F_{\text{in,resultant}}=0.50(18)=9.0\text{ N}. $$$$

For a conical pendulum of string length $l$, let the string make angle $\theta$ with the downward vertical. Its circular radius is $r=l\sin\theta$. If the tension is $T$,T\cos\theta=mg,\qquad T\sin\theta=m\omega^2r.$$$$

$$Dividing the equations gives$$\tan\theta=\frac{\omega^2r}{g}=\frac{\omega^2l\sin\theta}{g},$$so$$\omega^2=\frac{g}{l\cos\theta},\qquad T=\frac{mg}{\cos\theta}.A real taut-string model requires $T>0$ and $0<\cos\theta\le 1$.

For a mass on a light string of radius $r$, let $\theta$ be measured from the downward vertical and let the bottom speed be $u$. At angle $\theta$,v^2=u^2-2gr(1-\cos\theta),$$and inward radial balance gives$$T-mg\cos\theta=\frac{mv^2}{r},\qquad T=m\left(\frac{v^2}{r}+g\cos\theta\right).$$$$

At the top, $\theta=\pi$ andT+mg=\frac{mv_{\text{top}}^2}{r}.A string just remains taut when $T=0$, so $v_{\text{top}}^2=gr$. Energy from bottom to top gives $u^2=v_{\text{top}}^2+4gr$, hence the minimum bottom speed for a complete circle isu_{\min}=\sqrt{5gr}. $$$$

For natural length $l$, current length $L$ and extension $x=L-l$, a Hookean elastic element has force magnitudeF=\frac{\lambda x}{l}=kx,where $\lambda$ is the modulus of elasticity in newtons and $k=\lambda/l$ is stiffness in N m$^{-1}$. The model applies within the stated elastic range.

An elastic string has $l=0.80\text{ m}$ and $\lambda=100\text{ N}$. At length $0.92\text{ m}$,x=0.92-0.80=0.12\text{ m},\qquad T=\frac{100(0.12)}{0.80}=15\text{ N}. $$$$

For a Hookean string or spring of natural length $l$, modulus $\lambda$ and extension or compression magnitude $x$, the stored elastic potential energy isE_{\text{elastic}}=\frac{\lambda x^2}{2l}=\frac12Fx.$$Proof of this formula is not required.$$

Between extensions $x_1$ and $x_2$, use the endpoint difference\Delta E_{\text{elastic}}=\frac{\lambda}{2l}(x_2^2-x_1^2).$$For several active elastic elements, calculate and add one energy term for each element in each state.$$

For $l=0.80\text{ m}$, $\lambda=100\text{ N}$ and $x=0.12\text{ m}$,E_{\text{elastic}}=\frac{100(0.12)^2}{2(0.80)}=0.90\text{ J}.At natural length, $x=0$ and the stored elastic energy is zero.

$$At the first turning point the speed is again zero, but the particle need not be in equilibrium. Loss of gravitational potential energy equals gain in elastic energy:$$mgx\sin\alpha=\frac{\lambda x^2}{2l}.Besides the initial root $x=0$, the turning extension isx=\frac{2mgl\sin\alpha}{\lambda}.The speed is greatest earlier, where tangential acceleration is zero and $\lambda x/l=mg\sin\alpha$.

A particle of mass $m$ moves positively with $v(0)=1$ and experiences resistance $mkv^3$. Thenm\frac{dv}{dt}=-mkv^3,\qquad v^{-3}dv=-k,dt.$$Hence$$-\frac{1}{2v^2}=-kt+C.Using $v=1$ at $t=0$ gives $C=-\tfrac12$, sov(t)=\frac{1}{\sqrt{1+2kt}}.Displacement follows from integrating $dx/dt=v(t)$ with its own position condition.

$$Newton's experimental law acts along the line of impact:$$e=\frac{\text{relative speed of separation}}{\text{relative speed of approach}},\qquad 0\le e\le1.If A approaches B with signed velocities $u_A>u_B$ and they separate with $v_B>v_A$, thenv_B-v_A=e(u_A-u_B).$$$$

Two particles have relative approach speed $5\text{ m s}^{-1}$ and relative separation speed $2\text{ m s}^{-1}$. Thereforee=\frac25=0.4.$$This determines only one relation between the two final velocities; their masses and momentum provide the other relation for an isolated pair.$$

For a direct impact of masses $m$ and $2m$, suppose A approaches stationary B at speed $u$. Let their signed velocities after impact be $w$ and $v$. Momentum and restitution givemu=mw+2mv,\qquad v-w=eu.$$Hence$$v=\frac{(1+e)u}{3},\qquad w=\frac{(1-2e)u}{3}.The sign of $w$ decides whether A reverses.