Find initial linear speed of rolling disc on curved surface

To find the initial linear speed $v_2$ of the second disc, we use the principle of conservation of mechanical energy for rolling without slipping.

Since both discs roll without slipping, static friction acts at the point of contact, which does no work. Therefore, mechanical energy is conserved for both discs.

For a uniform disc of mass $m$ and radius $R$, its moment of inertia about its center of mass is: $I = \frac{1}{2} m R^2$

Since the disc rolls without slipping, its angular velocity $\omega$ is related to its linear velocity $v$ by $v = \omega R$. The total kinetic energy $K$ of a rolling disc is the sum of its translational and rotational kinetic energies: $K = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} m v^2 + \frac{1}{2} \left( \frac{1}{2} m R^2 \right) \left( \frac{v}{R} \right)^2 = \frac{3}{4} m v^2$

Let $v_f$ be the identical final linear speed of both discs at points $B$ and $D$.

1. For the disc moving on track $AB$: The drop in vertical height is $h_1 = 30\text{ m}$. Applying conservation of mechanical energy: $K_{A} + U_A = K_{B} + U_B$ $\frac{3}{4} m v_1^2 + m g h_1 = \frac{3}{4} m v_f^2$

2. For the disc moving on track $CD$: The drop in vertical height is $h_2 = 27\text{ m}$. Applying conservation of mechanical energy: $K_{C} + U_C = K_{D} + U_D$ $\frac{3}{4} m v_2^2 + m g h_2 = \frac{3}{4} m v_f^2$

Equating the two expressions for the final energy at points $B$ and $D$: $\frac{3}{4} m v_1^2 + m g h_1 = \frac{3}{4} m v_2^2 + m g h_2$

Canceling mass $m$ from both sides and rearranging: $\frac{3}{4} (v_2^2 - v_1^2) = g (h_1 - h_2)$

$v_2^2 - v_1^2 = \frac{4}{3} g (h_1 - h_2)$

Substituting the given values ($v_1 = 3\text{ m/s}$, $g = 10\text{ m/s}^2$, $h_1 = 30\text{ m}$, $h_2 = 27\text{ m}$): $v_2^2 - 3^2 = \frac{4}{3} \times 10 \times (30 - 27)$

$v_2^2 - 9 = \frac{4}{3} \times 10 \times 3$

$v_2^2 - 9 = 40$

$v_2^2 = 49$

$v_2 = 7\text{ m/s}$