Problem Statement
A uniform rod of mass m and length l is hinged at one end and held horizontally. Find the angular acceleration immediately after release.
Given Information
- All numerical data are stated in the problem above; symbols are defined as they appear.
Physical Concepts & Formulas
These problems use Hooke’s law and the elastic moduli that relate stress and strain in a deformed solid.
- $\sigma = E\,\varepsilon$ — Hooke’s law (Young’s modulus)
- $\tau = G\,\gamma$ — shear stress and strain
- $U = \tfrac{1}{2}\dfrac{\sigma^2}{E}$ — elastic energy density
Step-by-Step Solution
Step 1 — Identify the governing principle: We begin by recognising which physical law controls the situation and why it is the correct starting point for this problem.
$$\tau = \text{mg}(l/2) (torque of gravity about hinge)$$
Step 2 — Set up the relevant equations: Next we write down the equations that follow from that principle, introducing the symbols we will carry through the algebra.
$$I = ml^{2}/3 (about hinge)$$
Step 3 — Apply the given conditions: We now substitute the specific conditions and constraints given in the problem so the equations describe this particular situation.
$$\alpha = \tau /I = \text{mg}(l/2)/(ml^{2}/3) = 3g/(2l)$$
Step 4 — Solve for the required quantity: With the equations specialised, we isolate and solve for the unknown the problem asks us to find.
$$Acceleration of free end i\text{mm}ediately: a_end = \alpha l = 3g/2 > g$$
Worked Calculation
$$\alpha = 3g/(2l)$$
$$a_tip = 3g/2$$
Answer
$$\boxed{\alpha = 3g/(2l); acceleration of tip = 3g/2}$$
This is the quantity the problem asked for, expressed in terms of the given data: $\alpha = 3g/(2l); acceleration of tip = 3g/2$.
Physical Interpretation
The tip of the falling rod initially accelerates at $3g/2$ — faster than free fall! This is why a hinged falling rod can move faster than a freely falling point at the same height. The magnitude of the answer is consistent with everyday physical experience for this class of problem in Irodov’s Part 1 — the result shows how the answer scales with the given quantities. If we doubled the dominant input, the boxed formula tells us exactly how the output would respond, and that scaling is the key physical insight this problem trains. Comparing the answer with the appropriate limiting cases (very small or very large values of the dominant parameter) recovers the familiar Newtonian or intuitive expectation, which is a useful sanity check.
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