Problem 27: Heat Developed by Falling Beam
1. Understanding the Physics:
Since the density of the beam $\rho$ is equal to the density of water $\rho_{water}$, the beam is neutrally buoyant. Once submerged, it will be in equilibrium anywhere. The “heat developed” corresponds to the loss in the total mechanical energy of the system due to viscous forces as the beam settles from the surface into the water.
Since the density of the beam $\rho$ is equal to the density of water $\rho_{water}$, the beam is neutrally buoyant. Once submerged, it will be in equilibrium anywhere. The “heat developed” corresponds to the loss in the total mechanical energy of the system due to viscous forces as the beam settles from the surface into the water.
2. Potential Energy Changes:
We analyze the change in potential energy of the beam and the displaced water.
We analyze the change in potential energy of the beam and the displaced water.
- Beam: Initially, it sits on top of the water. Finally, it is just submerged (top surface at level 0). The Center of Mass (COM) moves from height $+a/2$ to $-a/2$.
Loss in PE of Beam = $mg \Delta h = (\rho l a^2) g (a) = \rho l a^3 g$. - Water: As the beam enters, an equivalent volume of water is displaced. Based on the hint, this water effectively moves from the space occupied by the beam (depth $-a/2$) to the surface layer (height 0).
Gain in PE of Water = $m_{water} g \Delta h_{water} = (\rho l a^2) g (a/2) = \frac{1}{2} \rho l a^3 g$.
3. Energy Balance:
The heat developed ($H$) is the net loss in potential energy of the system. $$ H = \text{Loss in PE}_{beam} – \text{Gain in PE}_{water} $$ $$ H = \rho l a^3 g – \frac{1}{2} \rho l a^3 g $$ $$ H = \frac{\rho g l a^3}{2} $$
The heat developed ($H$) is the net loss in potential energy of the system. $$ H = \text{Loss in PE}_{beam} – \text{Gain in PE}_{water} $$ $$ H = \rho l a^3 g – \frac{1}{2} \rho l a^3 g $$ $$ H = \frac{\rho g l a^3}{2} $$
Answer:
$$ H = \frac{\rho g l a^3}{2} $$