Question

A uniform rod of length \(L\) and density \(\rho\) is being pulled along A smooth floor with a horizontal acceleration \(\alpha\) what is the magnitude of the stress at the transverse cross-section through the midpoint of the rod? (A) \((1 / 3) \operatorname{L} \rho \alpha\) (B) \(\operatorname{L} \rho \alpha\) (C) \((2 / 3) \mathrm{L} \rho \rho\) (D) \(\\{(\operatorname{L} \rho \alpha) / 2\\}\)

Short answer

\(σ = \frac{Lρα}{2}\)

Step-by-step solution

Calculate the mass and moment of inertia of the rod

To find the mass, we use the density and the volume of the rod: Mass(m) = Density(ρ) × Volume(V) Since the rod is uniform, we can assume that it has a constant cross-sectional area (A). Then the volume can be expressed as: Volume(V) = Length(L) × Area(A) So the mass would be: m = ρ × (L × A) Next, we need to find the moment of inertia (I) of the rod about its midpoint. Since it is a uniform rod, we know the formula for the moment of inertia about its midpoint is: I = (1/12) × m × L^2

Calculate the net torque at the midpoint of the rod

The net torque (τ) can be calculated by multiplying the force and the perpendicular distance from the pivot point (in this case the midpoint). Since the force is uniformly distributed over the length of the rod, we can simplify the force as: Force(F) = mass(m) × acceleration(α) As the rod has a uniform linear mass distribution, the force acting on its midpoint will be half of the total force. Hence, Force at midpoint(F_midpoint) = (1/2) × F = (1/2) × m × α The perpendicular distance (d) from the pivot point to the point of force application is half the length of the rod: d = L/2 Now we can calculate the net torque acting τ: τ = F_midpoint × d = (1/2) × m × α × (L/2)

Calculate the net angular acceleration

For a rigid body in rotation, torque τ is related to the angular acceleration (ω) and moment of inertia (I) as follows: τ = I × ω We have already calculated I and τ in step 1 and 2. Now, we can find the angular acceleration: ω = τ / I = ((1/2) × m × α × (L/2)) / ((1/12) × m × L^2) ω = (3/2) × α / L

Calculate the stress at the midpoint of the rod

Stress (σ) can be calculated using the formula: σ = Force / Area From step 2, we have the force at the midpoint (F_midpoint): σ = (1/2) × m × α / A Recall that the mass can be expressed in terms of density, length, and area: m = ρ × (L × A) We can rewrite the stress equation as: σ = (1/2) × (ρ × L × A) × α / A Finally, we can simplify the equation: σ = (ρ × L × α) / 2 Thus, the stress at the transverse cross-section through the midpoint of the rod is: \(σ = \frac{Lρα}{2}\) The correct answer is (D).

Key concepts

Understanding the Uniform Rod

In physics, a uniform rod is an idealized object that assists in simplifying problems regarding mechanics. It's a model used to represent objects that have mass evenly distributed along their length. This uniformity means each segment of the rod has an identical mass distribution.

• Length (\( L \)): The physical extent of the rod from one end to the other.
• Density (\( \rho \)): Mass per unit volume, sides can vary but the distribution along the length remains consistent.
• Cross-sectional area: While the thickness or cross-section might be constant, it doesn't affect the uniformity in mass per length.

The concept simplifies calculations because once you know a few basic properties, like length and density, you can predict behaviors and solve problems. For example, calculating the entire rod's mass becomes straightforward. This simplification aids in understanding more complex systems in physics, like those involving stress and strain.

Linear Mass Distribution Explained

In a uniform rod, the mass per unit length is constant, leading to what's known as a linear mass distribution. Such distribution plays a crucial role in calculating forces, torques, and also stress.

• Mass Distribution: Each segment along the rod's length holds an equal portion of total mass.
• Consistent Force Application: With linear distribution, it means forces are systematically applied, making it easier to calculate stresses.
• Simplified Calculations: Using linear mass distribution, various physical quantities tied to the rod become predictable.

By understanding this concept, one can predict where stresses may be more pronounced on the rod when external forces act on it. It makes it feasible to focus analyses only on more relevant sections of the rod, like when determining moments of inertia.

Moment of Inertia and Its Importance

The moment of inertia is a fundamental concept that helps understand how mass is distributed relative to an axis and how it affects rotational motion.In a uniform rod, calculating the moment of inertia about its midpoint is key:- The formula is\( I = \frac{1}{12} m L^2 \), where \( m \) is the mass and \( L \) is the length of the rod.- This formula accounts for how the mass is spread along the length relative to the midpoint.

• Equilibrium Analysis: Helps in analyzing static conditions.
• Rotational Dynamics: Moment of inertia is integral in understanding how a rod will behave under rotational forces.
• Design Decisions: Engineers use this to decide material stress allowances.

Comprehending the moment of inertia not only simplifies solving mechanical problems but also provides insight for more extensive engineering projects involving beams, shafts, or any elongated structures.