Conservation of Energy and Approximation

The body rolls without slipping. The total kinetic energy is $KE = \frac{1}{2}mv^2(1+\beta)$, where $\beta$ is the rotational factor ($\beta = I/mR^2$).
Using conservation of energy, the velocity $v$ at a height $h$ (relative to flat track) is related to initial velocity $u$ by: $$ \frac{1}{2}mv^2(1+\beta) + mgh = \frac{1}{2}mu^2(1+\beta) $$ $$ v = u \sqrt{1 – \frac{2gh}{u^2(1+\beta)}} $$ Since the problem states that the change in potential energy is much smaller than the initial kinetic energy ($mgh \ll \frac{1}{2}mu^2$), we can use the binomial approximation $(1-x)^n \approx 1-nx$:

$$ v \approx u \left( 1 – \frac{gh}{u^2(1+\beta)} \right) $$

The time taken to traverse a small element $dx$ is $dt = \frac{dx}{v}$. Using the approximation $(1-\epsilon)^{-1} \approx 1+\epsilon$:

$$ dt \approx \frac{dx}{u} \left( 1 + \frac{gh}{u^2(1+\beta)} \right) = \frac{dx}{u} + \frac{gh}{u^3(1+\beta)} dx $$

Case 1: Elevated Section AB

Here $h$ is positive (above the reference). $$ \Delta t_1 = \int dt = \int \frac{dx}{u} + \frac{g}{u^3(1+\beta)} \int h \, dx $$ $$ \Delta t_1 = \frac{l}{u} + K \int h \, dx \quad \dots \text{(Equation 1)} $$ Where $l=20$ m is the length of the section, and $K$ is a constant.

Case 2: Depressed Section CD

The section is a mirror image, so the depth is $h$ but in the negative direction (potential energy decreases, velocity increases). The magnitude of $h$ is the same, but the sign in the energy equation flips. $$ v_2 \approx u \left( 1 + \frac{gh}{u^2(1+\beta)} \right) $$ $$ dt_2 \approx \frac{dx}{u} \left( 1 – \frac{gh}{u^2(1+\beta)} \right) $$ Integrating gives: $$ \Delta t_2 = \frac{l}{u} – K \int h \, dx \quad \dots \text{(Equation 2)} $$

Combining the Equations

Adding Equation 1 and Equation 2 eliminates the integral term involving $h$: $$ \Delta t_1 + \Delta t_2 = \frac{2l}{u} $$ $$ \Delta t_2 = \frac{2l}{u} – \Delta t_1 $$

Calculation:
Given $l = 20$ m, $u = 5.0$ m/s, and $\Delta t_1 = 5.0$ s. $$ \frac{2l}{u} = \frac{2 \times 20}{5.0} = 8.0 \text{ s} $$ $$ \Delta t_2 = 8.0 – 5.0 = 3.0 \text{ s} $$