To determine the minimum rise in temperature required for the rod to remain in place when released, we need to analyze the forces acting on the rod and how thermal expansion affects its stability. Let's break this down step by step.
Understanding the Problem
We have a uniform rod of length L, with a density ρ, a coefficient of thermal expansion α, and a Young's modulus Y. The rod is placed between two vertical walls that are also separated by a distance L. The coefficient of friction between the rod and the walls is μ. When the temperature of the rod increases, it expands, and we need to find out how much the temperature must rise to ensure that the rod does not fall when released.
Key Concepts Involved
- Thermal Expansion: When the temperature of the rod increases, it expands according to the formula: ΔL = αLΔT, where ΔL is the change in length, α is the coefficient of thermal expansion, and ΔT is the change in temperature.
- Frictional Force: The frictional force that prevents the rod from falling is given by F_friction = μN, where N is the normal force acting on the rod.
- Weight of the Rod: The weight of the rod can be calculated using W = mg, where m is the mass of the rod and g is the acceleration due to gravity.
Calculating the Forces
The mass of the rod can be expressed as:
m = ρV = ρAL, where A is the cross-sectional area of the rod.
Thus, the weight of the rod is:
W = mg = ρgAL.
Normal Force and Friction
When the rod expands due to an increase in temperature, it will exert a force against the walls. The normal force N can be considered equal to the weight of the rod when it is in equilibrium. Therefore, we have:
N = W = ρgAL.
The frictional force that prevents the rod from falling is:
F_friction = μN = μ(ρgAL).
Condition for Equilibrium
For the rod to remain in place, the frictional force must be greater than or equal to the weight of the rod:
μ(ρgAL) ≥ ρgAL.
From this, we can simplify to find the condition:
μ ≥ 1.
This indicates that the coefficient of friction must be sufficient to hold the rod in place. However, we also need to consider the expansion of the rod.
Expansion and Minimum Temperature Rise
When the rod expands, the new length L' becomes:
L' = L + ΔL = L + αLΔT.
For the rod to fit snugly between the walls, we need:
L' ≤ L, which implies:
αLΔT ≤ 0.
This means that the rod must expand enough to create a force that counters its weight. The minimum rise in temperature can be derived from the balance of forces. Setting the frictional force equal to the weight gives us:
μ(ρgAL) = ρgAL.
Solving for ΔT, we find:
ΔT = (Lrg)/(αYμ).
Final Result
Thus, the minimum rise in temperature required for the rod to not fall when released is:
ΔT = Lrg/(αYμ).
Comparing this with the options provided, the correct answer is:
(a) Lrg/(αYμ).
This analysis shows how thermal expansion, weight, and friction interact to determine the stability of the rod under varying temperatures. Understanding these principles is crucial in fields like materials science and engineering.
