Newton's Laws & Dynamics

Question 1

(a) [3] FBD: weight $mg$ downward; string tension $T$ directed along string toward centre (radially inward). At angle $\theta$ from lowest point the radial (centripetal) direction points to centre. Radial Newton's-2nd-law: $T - mg\cos\theta = \frac{mv^2}{R}\quad(\theta\text{ from bottom, so weight radial component }=mg\cos\theta\text{ toward centre at bottom}).$ (1 for FBD forces, 1 for correct directions, 1 for radial equation.) At the top the geometry gives $T + mg = mv_{\text{top}}^2/R$.

(b) [6] At top, string can only pull (T≥0). Critical case $T=0$: $mg = \frac{mv_{\text{top,min}}^2}{R}\Rightarrow v_{\text{top,min}}=\sqrt{gR}.$ (2) Physical reason: a string cannot push; the minimum speed is when gravity alone supplies all centripetal force, i.e. $T=0$; any slower and required centripetal force $<mg$, string goes slack, ball leaves circular path. (2) Energy conservation top→bottom (height drop $2R$): $\tfrac12 mv_b^2 = \tfrac12 mv_t^2 + mg(2R)\Rightarrow v_{\text{bottom,min}}=\sqrt{gR+4gR}=\sqrt{5gR}.$ (2)

(c) [4] A rigid hoop/rod exerts a normal (constraint) force that can be either inward or outward (push or pull). So there is no $T\ge0$ restriction; the bead can move arbitrarily slowly, in principle $v_{\text{top}}\to 0$. (2) Minimum condition is only that it reaches the top: $v_{\text{top,min}}=0$. (1) By energy: $v_{\text{bottom,min}}=\sqrt{4gR}=2\sqrt{gR}$. (1)

(d) [4] With bottom launch $v_b^2=5gR$, at angle $\theta$ from bottom, height $=R(1-\cos\theta)$: $v^2 = v_b^2 - 2gR(1-\cos\theta)=5gR-2gR+2gR\cos\theta=3gR+2gR\cos\theta.$ Radial eq: $T-mg\cos\theta = mv^2/R \Rightarrow T = mg\cos\theta + m(3g+2g\cos\theta)=mg(3+3\cos\theta)$ measured from bottom. Measured from top ($\theta' = \pi-\theta$, $\cos\theta=-\cos\theta'$): height from bottom $=R(1+\cos\theta')$, $v^2=5gR-2gR(1+\cos\theta')=3gR-2gR\cos\theta'$, and radial component of weight toward centre at top is $-mg\cos\theta'$... standard result: $T(\theta') = mg(3 + 2\cos\theta')\quad(\theta'\text{ from top}).$ Check: top $\theta'=0$: $T=...$ using bottom form $\theta=\pi$: $T=mg(3+3(-1))=0$ ✓; bottom $\theta=0$: $T=6mg$. $T_{\text{top}}=0$ at critical launch, so ratio $T_{\text{bottom}}/T_{\text{top}}$ is undefined (÷0). Award the physically meaningful statement $T_{\text{bottom}}=6mg$, $T_{\text{top}}=0$. (2 for T(θ), 1 for values, 1 for correct note that ratio diverges.)

(e) [3] Pseudocode:

min_launch_speed(R, g, N):
  lo, hi = sqrt(g*R), sqrt(6*g*R)   # bounds
  repeat N times:
     v = (lo+hi)/2
     if survives(v): hi = v else lo = v
  return hi
survives(v0):
  integrate theta from 0..pi (bottom to top) via RK4
  at each step v^2 = v0^2 - 2 g R (1-cos theta)
  T = m*v^2/R + m*g*cos(theta)   # from-bottom form
  if v^2 <= 0 or T < 0: return False   # slack string / can't reach
  return True

Failure detection: string goes slack when $T<0$ or the ball fails to reach top ($v^2\le0$). Bisection converges to $\sqrt{5gR}$. (1 integration, 1 taut/slack test, 1 bisection.)

Question 2

(a) [4] FBD: normal $N\perp$ surface, weight $mg$ down. No friction. Components: Vertical: $N\cos\alpha = mg$. Horizontal (centripetal): $N\sin\alpha = mv_0^2/r$. (2) Divide: $\tan\alpha = \frac{v_0^2}{gr}\Rightarrow v_0=\sqrt{gr\tan\alpha}.$ (2)

(b) [7] Friction $f\le\mu N$. At $v_{\max}$ friction acts down the incline (prevents sliding outward). Resolve along/perp: Vertical: $N\cos\alpha - f\sin\alpha = mg$. Horizontal: $N\sin\alpha + f\cos\alpha = mv^2/r$, with $f=\mu N$. (2) $v_{\max}=\sqrt{gr\,\frac{\sin\alpha+\mu\cos\alpha}{\cos\alpha-\mu\sin\alpha}}=\sqrt{gr\,\frac{\tan\alpha+\mu}{1-\mu\tan\alpha}}.$ (2) At $v_{\min}$ friction acts up ($f=\mu N$ up), sign flips: $v_{\min}=\sqrt{gr\,\frac{\tan\alpha-\mu}{1+\mu\tan\alpha}}.$ (3)

(c) [4] Set $\mu=0$: both give $\sqrt{gr\tan\alpha}=v_0$. (2) For $v_{\min}$ real & positive need numerator $\tan\alpha-\mu>0$, i.e. $\tan\alpha>\mu$. If $\tan\alpha\le\mu$, friction alone holds the car at rest on the bank—no minimum speed; the car will not slide inward even at $v=0$. (2)

(d) [4] $\tan15^\circ=0.267949$. $v_0=\sqrt{9.81\cdot120\cdot0.267949}=\sqrt{315.45}=17.76\ \text{m s}^{-1}$. (1) $v_{\max}=\sqrt{1177.2\cdot\frac{0.267949+0.35}{1-0.35\cdot0.267949}}=\sqrt{1177.2\cdot\frac{0.617949}{0.906183}}=\sqrt{802.8}=28.33\ \text{m s}^{-1}$. (2) $v_{\min}=\sqrt{1177.2\cdot\frac{0.267949-0.35}{1+0.093782}}=\sqrt{1177.2\cdot(-0.075189)}$ → negative ⇒ no real minimum (since $\tan15^\circ=0.268<\mu=0.35$); $v_{\min}=0$, car safe from rest. (1, credit for recognising negativity.)

(e) [3] Friction angle $\phi=\arctan\mu$. Then $\frac{\tan\alpha+\mu}{1-\mu\tan\alpha}=\tan(\alpha+\phi)$, so $v_{\max}=\sqrt{gr\tan(\alpha+\phi)}$. Denominator $1-\mu\tan\alpha\to0$ when $\alpha+\phi\to90^\circ$, giving $v_{\max}\to\infty$. Physically, at that bank angle no speed can push the car off (analogue of angle-of-repose limit). (3)

Question 3

(a) [4] Work done by gravity moving $m$ from $\infty$ to $r$: $U(r)=-\int_\infty^r \vec F\cdot d\vec r$ with $F=-GMm/r^2$ (attractive): $U(r)=-\int_\infty^r\left(-\frac{GMm}{r'^2}\right)dr' = -\Big[\frac{GMm}{r'}\Big]_\infty^r=-\frac{GMm}{r}.$ (3) Zero chosen at $r=\infty$; $U<0$ because gravity is attractive—work must be done on the system (positive) to separate to infinity, so bound states have negative energy. (1)

(b) [5] Escape: total energy zero at infinity, so $\tfrac12 mv_e^2 - \frac{GMm}{R}=0\Rightarrow v_e=\sqrt{2GM/R}$. (2) Orbit at surface: gravity = centripetal: $\frac{GMm}{R^2}=\frac{mv_o^2}{R}\Rightarrow v_o=\sqrt{GM/R}$. (2) Ratio $v_e/v_o=\sqrt2$. Energetically: KE for orbit $=\tfrac12 mv_o^2=\tfrac12\frac{GMm}{R}$; escape needs KE $=\frac{GMm}{R}$, twice as much, hence $\sqrt2$ in speed. (1)

(c) [5] Depth: uniform sphere, only mass within radius $(R-d)$ acts: $M(r)=M(r/R)^3$, $g(r)=GM(r)/r^2=g_0 r/R$, with $r=R-d$: $g(d)=g_0(1-d/R)$ — linear. (2) Altitude: $g(h)=g_0 R^2/(R+h)^2 \approx g_0(1-2h/R)$ for small $h$. (2) Near surface with $d=h$: depth fractional change $=d/R$, altitude fractional change $\approx 2h/R$. Altitude decreases twice as fast. (1)

(d) [4] $v_e=\sqrt{2\cdot6.674\times10^{-11}\cdot5.972\times10^{24}/6.371\times10^{6}}$. $GM=3.9857\times10^{14}$; $2GM/R=1.2513\times10^{8}$; $v_e=1.1187\times10^{4}=11.19\ \text{km s}^{-1}$. (2) $v_o=\sqrt{GM/R}=\sqrt{6.256\times10^{7}}=7.910\times10^{3}=7.91\ \text{km s}^{-1}$. (1) Ratio $11.19/7.91=1.414=\sqrt2$ ✓ (3 s.f.). (1)

[
 {"claim":"v_bottom_min = sqrt(5 g R) from energy with v_top=sqrt(gR)",
  "code":"g,R=symbols('g R',positive=True); vtop2=g*R; vbot2=vtop2+2*g*(2*R); result= simplify(vbot2-5*g*R)==0"},
 {"claim":"Banking v0 = sqrt(g r tan a); vmax/vmin closed forms reduce to v0 at mu=