Rotational Mechanics

Q1 — Parallel axis theorem (10)

Set CM at origin. Let a mass element $m_i$ have position $(x_i, y_i)$ in the plane perpendicular to the axes. Take CM axis along $z$; the parallel axis passes through point $(d,0)$.

• $I_{CM} = \sum m_i (x_i^2 + y_i^2)$ (1)
• Distance to new axis: $I = \sum m_i \big[(x_i-d)^2 + y_i^2\big]$ (2)
• Expand: $I = \sum m_i (x_i^2 + y_i^2) - 2d\sum m_i x_i + d^2 \sum m_i$ (2)
• First term $= I_{CM}$ (1)
• $\sum m_i x_i = M x_{CM} = 0$ since origin at CM — key step (2)
• $\sum m_i = M$, so last term $= Md^2$ (1)
• Result: $I = I_{CM} + Md^2$ (1)

Q2 — Rod MI + theorems (12)

(a) Linear density $\lambda = M/L$. Element at $x$: $dI = \lambda x^2 dx$. $I_{CM} = \int_{-L/2}^{L/2} \lambda x^2 dx = \lambda \frac{x^3}{3}\Big|_{-L/2}^{L/2} = \frac{\lambda L^3}{12} = \frac{ML^2}{12}$ (5) (setup 2, integral 2, result 1)

(b) Parallel axis, $d=L/2$: $I_{end} = \frac{ML^2}{12} + M\frac{L^2}{4} = \frac{ML^2}{3}$. (2) Direct: $\int_0^L \lambda x^2 dx = \frac{\lambda L^3}{3} = \frac{ML^2}{3}$ ✓ (2)

(c) Perpendicular axis theorem: for a planar (lamina) body, $I_z = I_x + I_y$; restriction — body must lie in the $xy$-plane. (1) By symmetry two diameters give $I_x = I_y = I_d$. So $I_z = 2I_d \Rightarrow I_d = \frac{1}{2}MR^2$. (2)

Q3 — Rolling incline (10)

(a) Newton along incline: $Mg\sin\theta - f = Ma$. (1) Torque about CM: $fR = I\alpha = \beta MR^2 \alpha$. (1) Rolling: $a = R\alpha \Rightarrow \alpha = a/R$, so $f = \beta M a$. (2) Substitute: $Mg\sin\theta - \beta M a = Ma$ $\boxed{a = \frac{g\sin\theta}{1+\beta}}$ (2)

(b) Larger $\beta$ → smaller $a$. Sphere $a=\frac{5}{7}g\sin\theta$ > cylinder $\frac{2}{3}g\sin\theta$ > ring $\frac{1}{2}g\sin\theta$. Sphere first. (2)

(c) $\beta$ measures how much mass is far from axis; larger $\beta$ demands more energy/torque diverted into rotation, leaving less for translation → smaller $a$. Note: independent of $M$ and $R$. (2)

Q4 — Angular momentum (10)

(a) $\vec\tau_{ext}=d\vec L/dt$; if net external torque $=0$ then $\vec L$ is constant. (2) Internal forces come in equal-opposite pairs along the line joining particles, so their torques cancel → they can't change total $L$; a person pulling arms in only changes $I$ internally, and $L=I\omega$ stays fixed (so $\omega$ rises). (2)

(b) Conservation: $I_1\omega_1 = (I_1+I_2)\omega_f$. $\omega_f = \frac{4\times6}{6} = 4\ \mathrm{rad/s}$. (3) $KE_i = \tfrac12(4)(6^2)=72\ \mathrm J$; $KE_f=\tfrac12(6)(4^2)=48\ \mathrm J$. Fraction lost $=\frac{72-48}{72}=\frac{1}{3}\approx 0.333$. (3)

Q5 — Code + gyroscope (10)

(a) (5 marks: correct I loop 2, KE 1, comments/structure 2)

def rotational(m, r, omega):
    I = 0.0
    for mi, ri in zip(m, r):     # sum over point masses
        I += mi * ri**2          # I = Σ m_i r_i²  (definition)
    KE = 0.5 * I * omega**2      # rotational KE = ½ I ω²
    return I, KE

(b) Gravity torque $\tau = mgr$ (r = distance to CM) is horizontal, perpendicular to spin $\vec L$. Since $d\vec L = \vec\tau\,dt$ is perpendicular to $\vec L$, $\vec L$ changes direction (not magnitude) → tip traces a circle. In time $dt$, $L$ sweeps angle $d\phi = dL/L = \tau\,dt/L$, so $\Omega = \frac{d\phi}{dt} = \frac{\tau}{L}.$ (4) Application: control-moment gyroscopes / reaction wheels change spacecraft orientation without expelling propellant. (1)

Q6 — Ladder equilibrium (8)

(a) Forces: floor normal $N$, floor friction $f$, wall normal $W$ (horizontal, wall frictionless), weight $mg$ at centre. Translational: $N = mg$ (vertical), $f = W$ (horizontal). (2) Rotational (torques about base): $W\,\ell\sin\phi = mg\,\frac{\ell}{2}\cos\phi$. (2)

(b) $W = \frac{mg}{2}\cot\phi = f$. Since $f\le\mu N = \mu mg$: $\mu_{min} = \frac{f}{N} = \frac{1}{2}\cot\phi.$ (4)

[
{"claim":"Rod end MI via parallel axis = ML²/3","code":"M,L=symbols('M L',positive=True); Icm=M*L**2/12; Iend=Icm+M*(L/2)**2; result=simplify(Iend-M*L**2/3)==0"},
{"claim":"Rolling acceleration a=g sinθ/(1+β) ranks sphere>cyl>ring","code":"g,th=symbols('g theta',positive=True); a=lambda b: g*sin(th)/(1+b); result=(a(Rational(2,5))>a(Rational(1,2))) and (a(Rational(1,2))>a(1))"},
{"claim":"Disc drop: omega_f=4, fraction KE lost=1/3","code":"I1,w1,I2=4,6,2; wf=I1*w1/(I1+I2); KEi=Rational(1,2)*I1*w1**2; KEf=Rational(1,2)*(I1+I2)*wf**2; result=(wf==4) and (Rational(KEi-KEf,KEi)==Rational(1,3))"},
{"claim":"Ladder min friction = (1/2)cot(phi)","code":"m,g,l,phi=symbols('m g l phi',positive=True); N=m*g; W=(m*g*(l/2)*cos(phi))/(l*sin(phi)); mu=W/N; result=simplify(mu-cot(phi)/2)==0"}
]