Rotational Mechanics

Question 1

(a) Solid sphere inertia — 5 marks

Divide the sphere into thin disks of thickness $dz$ at height $z$; disk radius $y=\sqrt{R^2-z^2}$, density $\rho=\dfrac{M}{\frac43\pi R^3}$. (1) Disk moment about the vertical (diameter) axis... use standard method: mass element, integrate.

$dm=\rho\,\pi y^2\,dz$, and a disk's inertia about a diameter through its centre is $\tfrac14\,dm\,y^2$; by symmetry integrate the axis-along-$z$ contribution: $I=\int \tfrac12 y^2\,dm \ \text{(disk about central axis)}$ — cleaner shell method: Use spherical shells: $dm=\rho\,4\pi r^2\,dr$, each thin shell has $I=\tfrac23 r^2\,dm$. (2) $I=\int_0^R \tfrac23 r^2\,\rho\,4\pi r^2\,dr=\tfrac{8}{3}\pi\rho\int_0^R r^4\,dr=\tfrac{8}{3}\pi\rho\frac{R^5}{5}=\tfrac{8\pi\rho R^5}{15}.$ (1) Substitute $\rho=\dfrac{3M}{4\pi R^3}$: $I=\dfrac{8\pi R^5}{15}\cdot\dfrac{3M}{4\pi R^3}=\dfrac{2}{5}MR^2.$ (1)

(b) Tangent axis + parallel axis proof — 5 marks

Proof: Take CM axis; a mass element at position $\vec r'$ from CM. New axis parallel, offset $\vec d$. Position from new axis $=\vec r'-\vec d$ (perpendicular components). (1) $I=\sum m_i|\vec r_i'-\vec d|^2=\sum m_i r_i'^2 -2\vec d\cdot\sum m_i\vec r_i' + d^2\sum m_i.$ (1) Middle term $=0$ since $\sum m_i\vec r_i'=0$ (CM origin); first term $=I_{CM}$, last $=Md^2$. (1) $\boxed{I=I_{CM}+Md^2}.$ Tangent line: $d=R$: $I=\tfrac25MR^2+MR^2=\tfrac75MR^2.$ (2)

(c) Speed from energy conservation — 5 marks

$Mgh=\tfrac12Mv^2+\tfrac12I\omega^2,\quad \omega=v/R,\ I=\beta MR^2.$ (1) $Mgh=\tfrac12Mv^2(1+\beta)\Rightarrow v=\sqrt{\frac{2gh}{1+\beta}}.$ (2) Solid $\beta=\tfrac25$; hollow $\beta=\tfrac23$. (1) $\frac{v_\text{solid}}{v_\text{hollow}}=\sqrt{\frac{1+\tfrac23}{1+\tfrac25}}=\sqrt{\frac{5/3}{7/5}}=\sqrt{\frac{25}{21}}\approx1.091.$ (1)

(d) Acceleration and friction — 5 marks

Torque about contact point (friction gives no torque there): $Mg\sin\theta\cdot R=I_\text{contact}\alpha$, $I_\text{contact}=(1+\beta)MR^2$, $a=R\alpha$. (2) $a=\frac{g\sin\theta}{1+\beta}.$ (1) Friction from Newton's law: $Mg\sin\theta - f=Ma\Rightarrow f=Mg\sin\theta\left(1-\frac1{1+\beta}\right)=\frac{\beta}{1+\beta}Mg\sin\theta.$ (1) For solid $\beta=\tfrac25$: $f=\tfrac{2/5}{7/5}Mg\sin\theta=\tfrac27Mg\sin\theta$. Require $f\le\mu_s Mg\cos\theta$: $\mu_s\ge\frac{2}{7}\tan\theta.$ (1)

(e) Code — 4 marks

Constant acceleration down incline length $s=h/\sin\theta$, from rest: $s=\tfrac12 a t^2\Rightarrow t=\sqrt{2s/a}$.

import numpy as np
def arrival_time(beta, theta_deg, h, g=9.8):
    th = np.radians(theta_deg)
    a = g*np.sin(th)/(1+beta)
    s = h/np.sin(th)
    return np.sqrt(2*s/a)

(2) for code. Symbolic form: (2) $t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2h/\sin\theta}{g\sin\theta/(1+\beta)}}=\frac{1}{\sin\theta}\sqrt{\frac{2h(1+\beta)}{g}}.$

Question 2

(a) Conditions — 3 marks

$L$ conserved when net external torque about the axis is zero. (1) Axle is frictionless (no torque), gravity and normal force act vertically along/through the axis producing no torque about the vertical spin axis, and the person's walking is an internal force. (2)

(b) Final ω — 5 marks

$I_\text{disk}=\tfrac12MR^2=\tfrac12(4)(0.25)=0.5\ \mathrm{kg\,m^2}.$ (1) Initial (person at rim): $I_i=0.5+mR^2=0.5+60(0.25)=15.5.$ (1) Final (person at centre): $I_f=0.5$. (1) $\omega_f=\frac{I_i\omega_0}{I_f}=\frac{15.5\times6}{0.5}=186\ \mathrm{rad/s}.$ (2)

(c) ΔK — 6 marks

$K_i=\tfrac12I_i\omega_0^2=\tfrac12(15.5)(36)=279\ \mathrm J.$ (1) $K_f=\tfrac12I_f\omega_f^2=\tfrac12(0.5)(186^2)=\tfrac12(0.5)(34596)=8649\ \mathrm J.$ (1) $\Delta K=8649-279=+8370\ \mathrm J.$ (2) Positive: the person does positive work (their muscles) pushing inward against the outward "centrifugal" tendency; this internal work raises $K$. (1) $L$ is conserved because it depends only on external torque (zero); $K$ is not conserved because internal muscular work can add energy — conservation of $L$ and of mechanical $K$ are independent statements. (1)

(d) Variable inertia + torque — 6 marks

Total inertia $I(r)=\tfrac12MR^2+mr^2$; $L$ constant $\Rightarrow$ $\omega(r)=\frac{L}{\tfrac12MR^2+mr^2}.$ (2) Disk's own angular momentum $L_\text{disk}=\tfrac12MR^2\,\omega(r)$. Person walks outward $r=ut$, so $\tau_\text{on disk}=\frac{dL_\text{disk}}{dt}=\tfrac12MR^2\frac{d\omega}{dt}.$ (2) $\dfrac{d\omega}{dt}=\dfrac{d\omega}{dr}u=-\dfrac{2mr\,u\,L}{(\tfrac12MR^2+mr^2)^2}.$ $\boxed{\tau=-\frac{MR^2\,mr\,u\,L}{(\tfrac12MR^2+mr^2)^2}.}$ (2) (Negative: disk decelerates as person moves out.)

Question 3

(a) Precession derivation — 6 marks

Gravity torque about support: $\tau=Mg\ell$, horizontal, perpendicular to $\vec L_s$. (1) $\vec\tau=d\vec L/dt$ so $\vec L$ tip moves horizontally: $|d\vec L|=L_s\,d\phi$ over $dt$. (2) $\tau=L_s\frac{d\phi}{dt}=L_s\Omega\Rightarrow \Omega=\frac{\tau}{L_s}=\frac{Mg\ell}{I_s\omega_s}.$ (2) It precesses rather than falls because the torque is perpendicular to $\vec L$: it changes the direction of $\vec L$ (rotating it horizontally) rather than adding downward angular momentum. (1)

(b) Reaction wheel — 6 marks

Initial total $L=0$; wheel spun up $\Rightarrow$ body counter-rotates. (1) $\omega_w=3000\ \mathrm{rpm}=3000\cdot\frac{2\pi}{60}=314.159\ \mathrm{rad/s}.$ (1) $I_w\omega_w+I_\text{sc}\omega_\text{sc}=0\Rightarrow\omega_\text{sc}=-\frac{I_w\omega_w}{I_\text{sc}}=-\frac{0.8\times314.159}{1200}=-0.20944\ \mathrm{rad