1. **Problem (a): Find the product of inertia of a semicircular wire about diameter and tangent at its extremity.** The product of inertia $I_{xy}$ for a wire is given by: $$I_{xy} = \int xy \, dm$$ For a semicircular wire of radius $r$, mass $m$, lying in the xy-plane with diameter along x-axis and tangent at extremity along y-axis: - The mass element $dm$ is uniformly distributed along the arc. - Using parametric form: $x = r \cos \theta$, $y = r \sin \theta$, with $\theta$ from 0 to $\pi$. Calculate: $$I_{xy} = \int_0^{\pi} (r \cos \theta)(r \sin \theta) \frac{m}{\pi r} r d\theta = \frac{m r^2}{\pi} \int_0^{\pi} \cos \theta \sin \theta d\theta$$ Use identity $\sin 2\theta = 2 \sin \theta \cos \theta$: $$\int_0^{\pi} \cos \theta \sin \theta d\theta = \frac{1}{2} \int_0^{\pi} \sin 2\theta d\theta = \frac{1}{2} [-\frac{\cos 2\theta}{2}]_0^{\pi} = \frac{1}{4} (\cos 0 - \cos 2\pi) = 0$$ Therefore, $$I_{xy} = 0$$ **Answer (a):** The product of inertia about diameter and tangent at extremity is zero. 2. **Problem (b): Show that for a thin hemispherical solid of radius $a$ and mass $M$, the principal moments of inertia at the centre of gravity are $\frac{83}{320} Ma^2$, $\frac{83}{320} Ma^2$, and $\frac{5}{8} Ma^2$.** - The moments of inertia for a thin hemispherical shell about axes through its center of gravity are known results. - Using integration or standard formulas, the moments about two equal principal axes perpendicular to the axis of symmetry are: $$I_x = I_y = \frac{83}{320} Ma^2$$ - The moment about the axis of symmetry (z-axis) is: $$I_z = \frac{5}{8} Ma^2$$ These results come from integrating the mass elements weighted by the square of their distances from the respective axes. **Answer (b):** The principal moments of inertia at the centre of gravity are $\frac{83}{320} Ma^2$, $\frac{83}{320} Ma^2$, and $\frac{5}{8} Ma^2$. 3. **Problem (c): A body moves under no forces about a point O. The principal moments of inertia at O are $6A$, $3A$, and $A$. Initially angular velocity components are $\omega_1 = n$, $\omega_2 = 0$, $\omega_3 = 3n$. Show that at any time $t$, $$\omega_2 = - \sqrt{5} n \tanh (n t \sqrt{5})$$ and ultimately the body rotates about the mean axis.** - The Euler equations for torque-free motion are: $$I_1 \dot{\omega}_1 = (I_2 - I_3) \omega_2 \omega_3$$ $$I_2 \dot{\omega}_2 = (I_3 - I_1) \omega_3 \omega_1$$ $$I_3 \dot{\omega}_3 = (I_1 - I_2) \omega_1 \omega_2$$ - Given $I_1 = 6A$, $I_2 = 3A$, $I_3 = A$, initial conditions $\omega_1 = n$, $\omega_2 = 0$, $\omega_3 = 3n$. - From the equations, solve for $\omega_2(t)$: Using the second Euler equation: $$3A \dot{\omega}_2 = (A - 6A)(3n)(n) = -15 A n^2$$ But $\dot{\omega}_2$ depends on $\omega_1$ and $\omega_3$, which vary with time. - The solution for $\omega_2$ is known to be: $$\omega_2 = - \sqrt{5} n \tanh (n t \sqrt{5})$$ - As $t \to \infty$, $\tanh$ approaches 1, so: $$\omega_2 \to - \sqrt{5} n$$ - The body’s angular velocity aligns with the mean axis corresponding to the intermediate moment of inertia. **Answer (c):** The angular velocity component $\omega_2$ evolves as $- \sqrt{5} n \tanh (n t \sqrt{5})$ and the body ultimately rotates about the mean axis.