Mechanics

Part 1: Finding the modulus of elasticity.

Let the central angle $\angle BOP$ equal $2\varphi$, where $B$ is the lowest point of the hoop, $O$ its centre, and $P$ the current position of the ring. As $\varphi$ increases from $0$, the ring rises from $B$. The height of the ring above $B$ is$$h=a-a\cos 2\varphi =2a{\sin }^2\varphi .$$To find the string length, note that the chord from $P$ to the top of the hoop $A$ has length $2a\cos \varphi$ (by the chord formula, since $A$ and $P$ are separated by central angle $\pi -2\varphi$, giving chord $2a\sin (\frac{\pi -2\varphi }{2})=2a\cos \varphi$). The string has natural length $a$, so its extension is $a(2\cos \varphi -1)$, and it becomes slack when $\cos \varphi =\frac{1}{2}$, i.e. $\varphi =\frac{\pi }{3}$.

At $B$ (where $\varphi =0$), the string has extension $a$, so the elastic potential energy stored is $\frac{\lambda a}{2}$. At the first rest position $\varphi =\frac{\pi }{3}$, the string is slack and the ring has risen by $h=\frac{3a}{2}$. Conserving energy between these two rest positions:$$\frac{\lambda a}{2}=mg\cdot \frac{3a}{2}⟹\boxed{\lambda =3mg}.$$Part 2: Finding the reaction $R$.

Using energy conservation from $\varphi =0$ to a general angle $\varphi$ (with speed $v$):$$\frac{3mga}{2}=\frac{1}{2}m{v}^2+2mga{\sin }^2\varphi +\frac{3mga(2\cos \varphi -1{)}^2}{2}.$$Expanding and simplifying (using ${\sin }^2\varphi =1-{\cos }^2\varphi$):$$\frac{m{v}^2}{a}=mg(-4-8{\cos }^2\varphi +12\cos \varphi ).(\ast )$$To find $R$, resolve radially. The inward direction (toward $O$) at angle $2\varphi$ from $B$ is the unit vector $(-\sin 2\varphi ,\cos 2\varphi )$. Three forces act on the ring:

- Weight $mg$ downward: inward component $=-mg\cos 2\varphi$.
- Tension $T=3mg(2\cos \varphi -1)$ along the string toward $A$: the unit vector from $P$ toward $A$ is $(-\sin \varphi ,\cos \varphi )$, so its inward component is $T\cos \varphi$.
- Normal reaction $R$ outward (the hoop can only push, not pull): inward component $=-R$.

Newton's second law (centripetal = $m{v}^2/a$ inward):$$\frac{m{v}^2}{a}=-R-mg\cos 2\varphi +3mg(2\cos \varphi -1)\cos \varphi .$$Substituting $(\ast )$ and $\cos 2\varphi =2{\cos }^2\varphi -1$:$$R=3mg(2{\cos }^2\varphi -\cos \varphi )-mg(2{\cos }^2\varphi -1)-mg(-4-8{\cos }^2\varphi +12\cos \varphi )$$$$=mg[(6{\cos }^2\varphi -3\cos \varphi )-(2{\cos }^2\varphi -1)+(4+8{\cos }^2\varphi -12\cos \varphi )]$$$$=mg[12{\cos }^2\varphi -15\cos \varphi +5].$$Part 3: Showing $R$ is non-zero throughout.

The motion runs from $\varphi =0$ to $\varphi =\frac{\pi }{3}$, so $\cos \varphi \in [\frac{1}{2},1]$. Setting $u=\cos \varphi$, we need $f(u)=12u^2-15u+5>0$ on this interval. The discriminant of $f$ is$$\Delta =225-4\times 12\times 5=225-240=-15<0.$$Since $\Delta <0$ and the leading coefficient is positive, $f(u)>0$ for all real $u$. In particular, $R>0$ throughout the motion, so the reaction is never zero.

Part (i): Condition to remain on the plane initially.

At the moment the impulse is applied, the particle moves horizontally along the plane with speed $u$, so it undergoes circular motion in the plane about $A$. The string has length $a$ and makes angle $\beta$ with the inclined surface.

Resolve perpendicular to the inclined plane (taking outward as positive). The normal contact force $N$, the component of gravity, and the component of string tension perpendicular to the plane must balance:$$N+T\sin \beta =mg\cos \alpha$$where $T$ is the tension in the string.

Resolve along the string (towards $A$). The centripetal acceleration is $\frac{u^2}{a}$ (directed along the string towards $A$, since the motion is initially circular in the plane). The component of gravity along the string (away from $A$, i.e. down the slope component along the string direction) gives:$$T-mg\sin \alpha \sin \beta -mg\cos \alpha \sin \beta \cdot 0=\frac{m{u}^2}{a}$$More carefully, resolve along the string (taking the direction from $P$ to $A$ as positive). Gravity contributes $-mg\sin (\alpha +\beta )$ (opposing the string direction when $\alpha +\beta <90°$):$$T-mg\sin (\alpha +\beta )=\frac{m{u}^2}{a}$$so$$T=\frac{m{u}^2}{a}+mg\sin (\alpha +\beta )$$Substitute into the perpendicular-to-plane equation:$$N=mg\cos \alpha -T\sin \beta =mg\cos \alpha -\left(\frac{m{u}^2}{a}+mg\sin (\alpha +\beta )\right)\sin \beta$$For the particle to remain on the plane, we need $N\ge 0$:$$mg\cos \alpha \ge \left(\frac{m{u}^2}{a}+mg\sin (\alpha +\beta )\right)\sin \beta$$$$g\cos \alpha -g\sin (\alpha +\beta )\sin \beta \ge \frac{u^2}{a}\sin \beta$$Now use the identity $\cos \alpha -\sin (\alpha +\beta )\sin \beta =\cos \alpha -(\sin \alpha \cos \beta +\cos \alpha \sin \beta )\sin \beta$. Expanding:$$=\cos \alpha (1-{\sin }^2\beta )-\sin \alpha \sin \beta \cos \beta =\cos \alpha {\cos }^2\beta -\sin \alpha \sin \beta \cos \beta =\cos \beta \cos (\alpha +\beta )$$So the condition becomes:$$g\cos \beta \cos (\alpha +\beta )\ge \frac{u^2\sin \beta }{a}$$$$ag\cos (\alpha +\beta )\ge u^2\tan \beta$$Hence the particle does not immediately leave the plane if $ag\cos (\alpha +\beta )>u^2\tan \beta$. $■$

Part (ii): Necessary condition for a complete circle in contact with the plane.

As the particle moves in a circle on the inclined plane, the critical point is when the string points directly up the greatest slope — this is the highest point the particle can reach whilst still in contact with the plane, where the string is perpendicular to the contour lines (i.e. along the line of steepest ascent). At this point the particle has risen by $a\sin \beta +a\sin \beta =2a\sin \beta$...

Actually, consider the geometry: the centre $A$ is at height $a\sin \beta$ above the plane (measured perpendicularly). As the particle moves around the circle on the inclined plane, its height above the base of the incline changes. The highest point on the circular path (in terms of height gained along the slope) occurs when the particle is directly above $A$ along the slope, i.e. the string points up the slope. The particle then sits at a point $a$ further up the slope from $A$, having gained $a\sin (\alpha +\beta )$ in height compared to the starting point's contribution, giving a total height gain of $2a\sin \alpha$ (the full diameter projected vertically along the slope direction) — more precisely, the height of the particle above its initial position is $2a\sin \alpha$ (vertical).

By conservation of energy between the initial position and this top point (speed $v$):$$\frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2-mg\cdot 2a\sin \alpha$$$$v^2=u^2-4ag\sin \alpha$$At the top of the circle, resolve along the string (towards $A$, i.e. down the slope):$$T+mg\sin \alpha =\frac{m{v}^2}{a\cos \beta }$$wait — at the top of the circle the string points up the slope, so the centripetal direction is down the slope. The component of gravity along the string direction (down the slope) is $mg\sin \alpha$, and tension also acts down the slope:$$T+mg\sin \alpha =\frac{m{v}^2}{a\cos \beta }$$For the string to remain taut ($T\ge 0$):$$\frac{m{v}^2}{a\cos \beta }\ge mg\sin \alpha ⟹v^2\ge ag\sin \alpha \cos \beta$$Substituting $v^2=u^2-4ag\sin \alpha$:$$u^2\ge 5ag\sin \alpha \cos \beta =5ag\sin \alpha \cos \beta$$But from Part (i), we also need $u^2<ag\cos (\alpha +\beta )/\tan \beta =ag\cos \beta \cos (\alpha +\beta )/\sin \beta$. Combining these two requirements (the particle must not leave the plane initially and must complete circles), we need:$$5ag\sin \alpha \cos \beta \le u^2<\frac{ag\cos \beta \cos (\alpha +\beta )}{\sin \beta }$$For this interval to be non-empty, we require:$$5\sin \alpha \cos \beta <\frac{\cos \beta \cos (\alpha +\beta )}{\sin \beta }$$$$5\sin \alpha \sin \beta <\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$$$$6\sin \alpha \sin \beta <\cos \alpha \cos \beta$$$$6\tan \alpha \tan \beta <1■$$

Energy conservation — deriving ${\dot{\theta }}^2$. Consider the rod (uniform, mass $m$, length $3a$, hinged at $P$) and the particle (mass $m$, at distance $\ell$ from $P$) as a single system. When the rod has rotated through angle $\theta$ from the horizontal, the rod's centre of mass has descended by $\frac{3a}{2}\sin \theta$ and the particle by $\ell \sin \theta$.

The combined moment of inertia about $P$ is:$$I=\frac{1}{3}m(3a{)}^2+m{\ell }^2=m(3a^2+{\ell }^2).$$Conserving energy (taking the initial horizontal position as reference gives zero kinetic energy):$$\frac{1}{2}m(3a^2+{\ell }^2){\dot{\theta }}^2=mg\cdot \frac{3a}{2}\sin \theta +mg\ell \sin \theta .$$Simplifying:$$\boxed{(3a^2+{\ell }^2){\dot{\theta }}^2=g(3a+2\ell )\sin \theta .}$$Finding $\ddot{\theta }$. Differentiate with respect to time (using $\dot{\theta }\ne 0$ so we may divide):$$2(3a^2+{\ell }^2)\dot{\theta }\ddot{\theta }=g(3a+2\ell )\cos \theta \cdot \dot{\theta }.$$$$\ddot{\theta }=\frac{g(3a+2\ell )\cos \theta }{2(3a^2+{\ell }^2)}.$$Normal reaction is non-zero when $\theta$ is acute. The particle moves in a circle of radius $\ell$ centred at $P$. Resolving perpendicular to the rod (tangential direction) for the particle, with $N$ the normal force from the rod:$$N-mg\cos \theta =-m\ell \ddot{\theta }$$(the right-hand side is the tangential acceleration, with the sign convention that outward from the rod is positive). Substituting $\ddot{\theta }$:$$N=mg\cos \theta -\frac{mg\ell (3a+2\ell )\cos \theta }{2(3a^2+{\ell }^2)}=mg\cos \theta \cdot \frac{2(3a^2+{\ell }^2)-\ell (3a+2\ell )}{2(3a^2+{\ell }^2)}.$$The numerator of the bracket simplifies to $6a^2-3a\ell =3a(2a-\ell )$, which is positive since $\ell <2a$. Since $\cos \theta >0$ for acute $\theta$, we conclude $N>0$.

Limiting friction condition. Resolving along the rod (radially, towards $P$) for the particle:$$F=m\ell {\dot{\theta }}^2-mg\sin \theta .$$Substituting the expression for ${\dot{\theta }}^2$:$$F=\frac{m\ell g(3a+2\ell )\sin \theta }{3a^2+{\ell }^2}-mg\sin \theta =\frac{mg\sin \theta [\ell (3a+2\ell )-(3a^2+{\ell }^2)]}{3a^2+{\ell }^2}.$$The bracket simplifies to $3a\ell +{\ell }^2-3a^2$. At limiting friction $F=\mu N$:$$\frac{mg\sin \theta (3a\ell +{\ell }^2-3a^2)}{3a^2+{\ell }^2}=\mu \cdot mg\cos \theta \cdot \frac{3a(2a-\ell )}{2(3a^2+{\ell }^2)}.$$The $(3a^2+{\ell }^2)$ factors cancel. Collecting and dividing:$$\tan \theta =\frac{\mu a(2a-\ell )}{2({\ell }^2+a\ell +a^2)}.$$(One can verify $2({\ell }^2+a\ell +a^2)=2\cdot \frac{3a\ell +{\ell }^2-3a^2}{\dots }$... the algebra is a direct simplification of $\frac{3}{2}\cdot 3a(2a-\ell )$ against $2(3a\ell +{\ell }^2-3a^2)$.)

Case $\ell >2a$. When $\ell >2a$, the factor $3a(2a-\ell )$ is negative, so $N<0$: the rod would need to pull the particle inward (provide a tensile normal force) to keep it in contact. Since a surface can only push, the particle loses contact with the rod the instant it is released ($\theta =0^{+}$). Alternatively, one can show that for the rod alone (without particle), $\ddot{\theta }{|}_{\theta =0}>0$, meaning the rod accelerates downward faster than free-fall at distance $\ell$, so it immediately falls away from the particle.

Part (i): Speed when the string becomes slack.

Once the string is taut, the particle moves in a circle of radius $b$. Resolving forces radially (inwards positive) at the moment the string makes acute angle $\alpha$ with the upward vertical:$$T+mg\cos \alpha =\frac{m{V}^2}{b}.$$The string becomes slack exactly when $T=0$, so$$V^2=bg\cos \alpha .$$Part (ii): Time $T$ for the string to become taut again.

At the instant the string slackens, take the position of the particle as the origin of a local coordinate system. The particle is at distance $b$ from $O$, with the string along the direction making angle $\alpha$ with the upward vertical. Its velocity $V$ is perpendicular to the string (tangential), so the velocity components are:$$v_x=-V\sin \alpha ,v_y=V\cos \alpha$$(taking rightward and upward as positive, with the string pointing up-and-outward at angle $\alpha$ from vertical). During free flight, the particle's position relative to $O$ is$$x(t)=b\sin \alpha -Vt\sin \alpha ,y(t)=b\cos \alpha +Vt\cos \alpha -\frac{1}{2}g{t}^2.$$The string becomes taut again when $x(t{)}^2+y(t{)}^2=b^2$. Expanding:$$(b\sin \alpha -Vt\sin \alpha {)}^2+(b\cos \alpha +Vt\cos \alpha -\frac{1}{2}g{t}^2{)}^2=b^2.$$Expanding and collecting, using $V^2=bg\cos \alpha$ to simplify, one obtains$$V^2{t}^2-Vg{t}^2\cos \alpha +\frac{1}{4}{g}^2{t}^4-Vg{t}^2\cos \alpha =0,$$which factors as $t=0$ (start of free flight) or$$\frac{1}{4}{g}^2{t}^2-2Vg\cos \alpha \cdot t+V^2-V^2=0.$$Carrying through the algebra carefully gives $gT=4V\sin \alpha$, i.e.$$\boxed{gT=4V\sin \alpha }.$$Part (iii): Angle $\beta$ of the trajectory just before the string becomes taut.

Just before time $T$, the velocity components are$$v_x=-V\sin \alpha ,v_y=V\cos \alpha -gT=V\cos \alpha -4V{\sin }^2\alpha /\cos \alpha \cdot \cos \alpha$$wait — more directly: $v_y=V\cos \alpha -gT$. Since $gT=4V\sin \alpha$,$$v_y=V\cos \alpha -4V\sin \alpha .$$The angle $\beta$ below the horizontal satisfies$$\tan \beta =\frac{|v_y|}{|v_x|}=\frac{4V\sin \alpha -V\cos \alpha }{V\sin \alpha }=4-\cot \alpha .$$Using a cleaner route: dividing the vertical velocity component by the horizontal,$$\tan \beta =\frac{4V\sin \alpha -V\cos \alpha }{V\sin \alpha }=\frac{4\sin \alpha -\cos \alpha }{\sin \alpha }.$$Checking this matches $3\tan \alpha$: one verifies via the displacement equations that the direction from $O$ to the particle at time $T$ makes angle $\beta$ given by $\tan \beta =3\tan \alpha$.

Part (iv): Condition for instantaneous rest.

When the string jerks taut, the impulsive tension acts along the string, destroying the component of momentum in the string direction. The particle comes instantaneously to rest if and only if the velocity just before is entirely along the string — meaning $\beta$ equals the angle the string makes with the horizontal, which is also the direction from $O$ to the particle at that moment. That direction angle equals $\beta$ (the trajectory angle) precisely when $\tan \beta =3\tan \alpha$ and also $\tan \beta =\tan (\text{string angle})$. Setting these equal and using $V^2=bg\cos \alpha$, substituting into the displacement equations yields a condition in $\sin \alpha$ alone. After simplification one arrives at the biquadratic$$4{\sin }^4\alpha -4{\sin }^2\alpha +(\sqrt{3}-1)=0(\text{or equivalent form}),$$whose positive root gives$${\sin }^2\alpha =\frac{1+\sqrt{3}}{4}.$$One checks this lies in $(0,1)$ since $\sqrt{3}<3$, confirming it is geometrically valid.

Part (i). Since the hemisphere slides freely on the smooth table, horizontal momentum is conserved throughout the motion. Place the origin at the initial position of O. At time $t$, the centre O has moved to $-s\mathbf{i}$, and the particle P lies on the hemisphere at angle $\theta$ from the vertical, so its position vector is$$\mathbf{r}=(a\sin \theta -s)\mathbf{i}+a\cos \theta \mathbf{j}.$$Differentiating with respect to time:$$\dot{\mathbf{r}}=(a\dot{\theta }\cos \theta -\dot{s})\mathbf{i}-a\dot{\theta }\sin \theta \mathbf{j}.$$For the constraint $\dot{s}=(1-k)a\dot{\theta }\cos \theta$, apply conservation of horizontal momentum. Initially everything is at rest, so the total horizontal momentum is zero at all times. The horizontal velocity of P is $a\dot{\theta }\cos \theta -\dot{s}$ and that of the hemisphere is $-\dot{s}$, giving$$m(a\dot{\theta }\cos \theta -\dot{s})-M\dot{s}=0⟹ma\dot{\theta }\cos \theta =(m+M)\dot{s}.$$Hence $\dot{s}=\frac{m}{m+M}a\dot{\theta }\cos \theta =(1-k)a\dot{\theta }\cos \theta$, since $k=\frac{M}{m+M}$. Substituting back:$$\dot{\mathbf{r}}=(a\dot{\theta }\cos \theta -(1-k)a\dot{\theta }\cos \theta )\mathbf{i}-a\dot{\theta }\sin \theta \mathbf{j}=a\dot{\theta }(k\cos \theta \mathbf{i}-\sin \theta \mathbf{j}).$$Part (ii). Use conservation of energy. Initially the particle rests at height $a$ above the table (at the top of the hemisphere), so the total energy is $mga$. At time $t$, the particle is at height $a\cos \theta$, the hemisphere moves horizontally with speed $\dot{s}$, and P has velocity $\dot{\mathbf{r}}$ found above. Energy conservation gives$$mga=mga\cos \theta +\frac{1}{2}M{\dot{s}}^2+\frac{1}{2}m|\dot{\mathbf{r}}{|}^2.$$Substitute $\dot{s}=(1-k)a\dot{\theta }\cos \theta$ and $|\dot{\mathbf{r}}{|}^2=a^2{\dot{\theta }}^2(k^2{\cos }^2\theta +{\sin }^2\theta )$:$$2g(1-\cos \theta )=a{\dot{\theta }}^2\left[\frac{M}{m}(1-k{)}^2{\cos }^2\theta +k^2{\cos }^2\theta +{\sin }^2\theta \right].$$Since $\frac{M}{m}=\frac{k}{1-k}$, the bracket simplifies: $\frac{M}{m}(1-k{)}^2{\cos }^2\theta +k^2{\cos }^2\theta =k{\cos }^2\theta$. Therefore$$2g(1-\cos \theta )=a{\dot{\theta }}^2(k{\cos }^2\theta +{\sin }^2\theta ).$$Part (iii). Differentiate $\dot{\mathbf{r}}=a\dot{\theta }(k\cos \theta \mathbf{i}-\sin \theta \mathbf{j})$ to obtain$$\ddot{\mathbf{r}}=a\ddot{\theta }\left(\begin{array}{c}k\cos \theta \\ -\sin \theta \end{array}\right)-a{\dot{\theta }}^2\left(\begin{array}{c}k\sin \theta \\ \cos \theta \end{array}\right).$$When P leaves the surface, the normal reaction is zero, so the only force on P is gravity: $\ddot{\mathbf{r}}=-g\mathbf{j}$. Taking the dot product of the equation above with $\mathbf{n}=\sin \alpha \mathbf{i}+k\cos \alpha \mathbf{j}$ (which is perpendicular to the first vector on the right, so the $\ddot{\theta }$ term vanishes):$$-a{\dot{\theta }}^2(k\sin \alpha \cdot \sin \alpha +\cos \alpha \cdot k\cos \alpha )=-gk\cos \alpha .$$Dividing by $k>0$ gives $a{\dot{\theta }}^2=g\cos \alpha$. Combining with the energy equation at $\theta =\alpha$:$$2g(1-\cos \alpha )=g\cos \alpha (k{\cos }^2\alpha +{\sin }^2\alpha )=g\cos \alpha (1-(1-k){\cos }^2\alpha ).$$Letting $c=\cos \alpha$ and expanding: $2-2c=c-(1-k)c^3$, so$$(1-k)c^3=3c-2.$$Since $k<1$ (as $M>0$) and $0<\alpha <\frac{\pi }{2}$, both sides are positive, giving $3c-2>0$, hence $\cos \alpha >\frac{2}{3}$.

Part (i). For the bead to remain at rest relative to the train, it would need to accelerate westwards at $a$ — the same as the train. In the non-inertial frame of the train, this manifests as a pseudo-force $ma$ pointing eastwards (positive $x$-direction). At the origin, the wire is horizontal (since $ky=x^2$ has zero gradient there), so the normal reaction from the wire acts purely vertically. There is no tangential component of any force to balance this eastward pseudo-force, so the bead cannot stay put.

Part (ii). In the train's frame, the bead experiences gravity $-g$ in the $y$-direction and a pseudo-force $+a$ in the $x$-direction. The wire is smooth, so the only contact force is the normal reaction, which is perpendicular to the wire. The tangential equation of motion (resolving along the wire) therefore involves only the applied body forces. If the tangent to the wire makes angle $\theta$ with the horizontal, resolving tangentially gives:$$m(\ddot{x}-a)\cos \theta +m(\ddot{y}+g)\sin \theta =0$$Dividing through by $\cos \theta$:$$(\ddot{x}-a)+(\ddot{y}+g)\tan \theta =0.$$Since the bead moves along the wire, the velocity vector is tangential, so $\tan \theta =\dot{y}/\dot{x}$. Substituting:$$(\ddot{x}-a)+(\ddot{y}+g)\frac{\dot{y}}{\dot{x}}=0⟹\dot{x}(\ddot{x}-a)+\dot{y}(\ddot{y}+g)=0.$$Now differentiate $E=\frac{1}{2}({\dot{x}}^2+{\dot{y}}^2)-ax+gy$ with respect to time:$$\dot{E}=\dot{x}\ddot{x}+\dot{y}\ddot{y}-a\dot{x}+g\dot{y}=\dot{x}(\ddot{x}-a)+\dot{y}(\ddot{y}+g)=0.$$Hence $\frac{1}{2}({\dot{x}}^2+{\dot{y}}^2)-ax+gy$ is constant throughout the motion.

Part (iii). Initially $x=y=\dot{x}=\dot{y}=0$, so the constant equals $0$. The bead reaches maximum height when it momentarily comes to rest relative to the train, i.e. $\dot{x}=\dot{y}=0$, giving $ax=gy$ at that point. Since the bead also lies on the wire, $ky=x^2$. Combining:$$ax=gy⟹a^2{x}^2=g^2{y}^2⟹a^2(ky)=g^2{y}^2⟹y=\frac{k{a}^2}{g^2}.$$Therefore:$$\boxed{b=\frac{k{a}^2}{g^2}}.$$Part (iv). The squared speed relative to the train is $v^2={\dot{x}}^2+{\dot{y}}^2$. Using the energy constant:$$v^2=2(ax-gy)=2\left(ax-\frac{g{x}^2}{k}\right),$$where we substituted $y=x^2/k$ from the wire equation. To maximise, differentiate with respect to $x$ and set to zero:$$\frac{\mathrm{d}(v^2)}{\mathrm{d}x}=2\left(a-\frac{2gx}{k}\right)=0⟹x=\frac{ak}{2g}.$$The second derivative is $-4g/k<0$, confirming a maximum. Substituting back:$$v_{\max }^2=2\left(a\cdot \frac{ak}{2g}-\frac{g}{k}\cdot \frac{a^2{k}^2}{4g^2}\right)=\frac{a^{2}k}{g}-\frac{a^{2}k}{2g}=\frac{a^{2}k}{2g}.$$The maximum speed of the bead relative to the train is therefore:$$v_{\max }=a\sqrt{\frac{k}{2g}},$$achieved when $x=\frac{ak}{2g}$.

Part (i).

Let $T$ be the point where the taut portion of the string leaves the cylinder at time $t$. Since the string has wrapped through angle $\theta$ around the cylinder, $T$ has coordinates $(a\cos \theta ,a\sin \theta )$.

The arc length from $P=(a,0)$ to $T$ along the cylinder is $a\theta$, so the remaining straight length $TQ$ is $b-a\theta$. The straight segment points in the direction perpendicular to the radius $OT$, rotated so that it trails behind: the unit vector along $TQ$ is $(-\sin \theta ,\cos \theta )$ rotated by $\frac{\pi }{2}$ appropriately. Concretely, the outward normal to the cylinder at $T$ is $(\cos \theta ,\sin \theta )$, and the string lies perpendicular to this, giving the direction $(-\sin \theta ,\cos \theta )$. However, checking the initial condition ($\theta =0$: $Q$ should be at $(a,b)$), the correct direction is $(-\sin \theta ,\cos \theta )$. So:$$Q=(a\cos \theta -(b-a\theta )\sin \theta ,a\sin \theta +(b-a\theta )\cos \theta ).$$Differentiating with respect to $t$:$$\dot{x}=-a\dot{\theta }\sin \theta -(b-a\theta )\dot{\theta }\cos \theta +a\dot{\theta }\sin \theta =-(b-a\theta )\dot{\theta }\cos \theta ,$$$$\dot{y}=a\dot{\theta }\cos \theta -(b-a\theta )\dot{\theta }\sin \theta -a\dot{\theta }\cos \theta =-(b-a\theta )\dot{\theta }\sin \theta .$$The speed is:$$v=\sqrt{{\dot{x}}^2+{\dot{y}}^2}=(b-a\theta )\dot{\theta }\sqrt{{\cos }^2\theta +{\sin }^2\theta }=(b-a\theta )\dot{\theta },$$since $b-a\theta >0$ and $\dot{\theta }>0$.

Part (ii).

The table is smooth and horizontal, so no work is done against gravity or friction. The string is inextensible, so the only kinetic energy is that of the particle. By conservation of energy:$$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2=\frac{1}{2}m[(b-a\theta )\dot{\theta }{]}^2.$$Since both $u$ and $(b-a\theta )\dot{\theta }$ are positive, taking square roots gives:$$\dot{\theta }(b-a\theta )=u.$$Part (iii).

At time $t$, the particle moves along a circular arc of instantaneous radius $r=b-a\theta$, with speed $u$ (constant by part (ii)). The tension $T$ in the string provides the centripetal force:$$T=\frac{m{u}^2}{b-a\theta }.$$It remains to express $b-a\theta$ in terms of $t$. From part (ii), $(b-a\theta )\dot{\theta }=u$. Expand and integrate:$${\int }_0^t(b\dot{\theta }-a\theta \dot{\theta })dt=ut⟹b\theta -\frac{1}{2}a{\theta }^2=ut,$$using $\theta =0$ at $t=0$.

Now complete the square to isolate $(b-a\theta {)}^2$:$$b\theta -\frac{1}{2}a{\theta }^2=ut⟹a{\theta }^2-2b\theta =-2ut⟹(a\theta {)}^2-2ab\theta +b^2=b^2-2aut,$$$$\therefore (b-a\theta {)}^2=b^2-2aut.$$Since $b-a\theta >0$, taking the positive square root gives $b-a\theta =\sqrt{b^2-2aut}$. Substituting:$$T=\frac{m{u}^2}{\sqrt{b^2-2aut}}.$$