Rotational Mechanics

Q1. (3 marks)

• Rigid body: a body in which the distance between any two constituent particles remains constant regardless of external forces (no deformation). (2)
• Degrees of freedom of a free rigid body = 6 (3 translational + 3 rotational). (1)

Q2. (4 marks) (a) $\alpha = \dfrac{\omega - \omega_0}{t} = \dfrac{20 - 0}{4} = 5\ \text{rad/s}^2$. (1) (b) $\theta = \omega_0 t + \tfrac12\alpha t^2 = 0 + \tfrac12(5)(16) = 40\ \text{rad}$. (2) (c) $a_t = r\alpha = 0.25 \times 5 = 1.25\ \text{m/s}^2$. (1)

Q3. (4 marks) Statement: $I = I_{CM} + Md^2$, where $I_{CM}$ is the MI about an axis through the centre of mass and $I$ about a parallel axis a distance $d$ away. (1) Proof: Take CM as origin. For an axis through CM, $I_{CM} = \sum m_i(x_i^2 + y_i^2)$. Parallel axis shifted by $d$ along $x$: $I = \sum m_i\big[(x_i - d)^2 + y_i^2\big] = \sum m_i(x_i^2+y_i^2) - 2d\sum m_i x_i + d^2\sum m_i.$ (2) Since origin is CM, $\sum m_i x_i = 0$, and $\sum m_i = M$, giving $I = I_{CM} + Md^2$. (1)

Q4. (4 marks) Statement: For a planar (laminar) body lying in the $xy$-plane, $I_z = I_x + I_y$. (1) Restriction: applies only to a plane lamina (two-dimensional body); the $z$-axis must be perpendicular to the plane at the point where $x,y$ axes meet. (1) Application: For a ring, $I_z$ (about central axis) $= MR^2$. By symmetry $I_x = I_y = I_{diam}$. So $MR^2 = 2I_{diam}$ → $I_{diam} = \tfrac12 MR^2$. (2)

Q5. (4 marks) (a) $I = \tfrac25 MR^2$. (1) $I = \tfrac25 (2)(0.1)^2 = 0.008\ \text{kg·m}^2$. (1) (b) $KE = \tfrac12 I\omega^2 = \tfrac12 (0.008)(10)^2 = 0.4\ \text{J}$. (2)

Q6. (4 marks)

• Relation: $\vec{\tau}_{ext} = \dfrac{d\vec{L}}{dt}$. (2)
• If the net external torque is zero, $\dfrac{d\vec{L}}{dt}=0$, so $\vec{L}$ is conserved. (1)
• Example: a spinning skater pulling arms in speeds up (I decreases, ω increases at constant L). (1)

Q7. (4 marks) (a) $v = R\omega$. (1) (b) $KE = \tfrac12 Mv^2 + \tfrac12 I\omega^2$, with $I = \tfrac12 MR^2$ and $\omega = v/R$: $\tfrac12 I\omega^2 = \tfrac12\Big(\tfrac12 MR^2\Big)\frac{v^2}{R^2} = \tfrac14 Mv^2.$ (2) $KE = \tfrac12 Mv^2 + \tfrac14 Mv^2 = \tfrac34 Mv^2.$ (1)

Q8. (5 marks) (a) $a = \dfrac{g\sin\theta}{1+k}$ where $I = kMR^2$. (2) (b) $k$ values: sphere $2/5=0.4$, disc $1/2=0.5$, ring $1$. Smaller $k$ → larger $a$ → reaches first. (2) Order (first to last): solid sphere, then solid disc, then ring. (1)

Q9. (4 marks) Let the floor normal $N_f$, wall normal (horizontal) $N_w$, floor friction $f$.

• $\sum F_x = 0$: $f = N_w$. (1)
• $\sum F_y = 0$: $N_f = W$. (1)
• $\sum \tau = 0$ (about foot): $N_w\, L\sin\theta = W\,\tfrac{L}{2}\cos\theta$ (torque of wall reaction balances weight torque). (2)

Q10. (4 marks)

• Gyroscopic effect: a spinning body has large angular momentum $\vec L$ along its spin axis; gravity applies a torque $\vec\tau$ perpendicular to $\vec L$. Since $\vec\tau = d\vec L/dt$, the change $d\vec L$ is perpendicular to $\vec L$, so the axis rotates (precesses) rather than falling. (2)
• Precession rate $\Omega = \tau/(I\omega)$ (mention). (1)
• Application: reaction/control-moment gyroscopes maintain and adjust spacecraft attitude (orientation) without expelling propellant. (1)

[
  {"claim":"Q2: alpha=5, theta=40, a_t=1.25","code":"alpha=(20-0)/4; theta=0.5*alpha*16; at=0.25*alpha; result=(alpha==5 and theta==40 and at==1.25)"},
  {"claim":"Q5: I=0.008 and KE=0.4 J","code":"I=Rational(2,5)*2*(Rational(1,10))**2; KE=Rational(1,2)*I*10**2; result=(I==Rational(1,125) and KE==Rational(2,5))"},
  {"claim":"Q7: rolling cylinder KE = 3/4 M v^2","code":"M,v,R=symbols('M v R',positive=True); I=Rational(1,2)*M*R**2; w=v/R; KE=simplify(Rational(1,2)*M*v**2+Rational(1,2)*I*w**2); result=(KE==Rational(3,4)*M*v**2)"},
  {"claim":"Q4: ring diameter MI = MR^2/2","code":"M,R=symbols('M R',positive=True); Iz=M*R**2; Idiam=Iz/2; result=(Idiam==M*R**2/2)"}
]