A light rod is pin-supported at one end and carries a weight W at the other end, as shown in the figure. The ideal spring attached to the rod is capable of resisting both tension and compression, and it is unstretched when the rod is vertical. Find the largest value of W for which the vertical

Question

A light rod is pin-supported at one end and carries a weight [latex]W[/latex] at the other end, as shown in the figure. The ideal spring attached to the rod is capable of resisting both tension and compression, and it is unstretched when the rod is vertical. Find the largest value of [latex]W[/latex] for which the vertical equilibrium position of the rod would be stable.

Accepted Answer

The potential energy of the system consists of [latex]V_{g}[/latex], the gravitational potential energy of the weight, and [latex]V_{e}[/latex], the elastic potential energy of the spring. From Eq. (10.14), [latex]V_{g}=W\:y_{G}[/latex], where [latex]y_{G}[/latex] is the vertical distance to [latex]W[/latex] measured from an arbitrary reference line. Choosing the horizontal line passing through [latex]O[/latex] as the reference, we find [latex]V_{g}=WL\cos θ[/latex].

To determine whether a function evaluated at a point is a minimum or maximum, it is sufficient to investigate the function only in a small neighborhood of that point. Therefore, we can confine our attention to small values of [latex]θ[/latex]. Approximating [latex]\cos θ[/latex] with [latex](1 − θ^2/2)[/latex], which is valid for small angles, the gravitational potential energy becomes

[latex]V_{g}=WL\left(1−\frac{1}{2}θ^2\right)[/latex] (a)

From Eq. (10.16), [latex]V_{e}=(1/2)ks^2[/latex], where s is the elongation (or contraction) of the spring measured from its unstretched position. From the figure, we see that for sufficiently small [latex]θ[/latex], [latex]s=b\sin θ[/latex]. Using the approximation [latex]\sin θ≈θ[/latex], the elastic potential energy of the spring becomes

[latex]V_{e}=\frac{1}{2}kb^2θ^2[/latex] (b)

Combining Eqs. (a) and (b), the potential energy of the system is

[latex]V= V_{g}+V_{e}=WL\left(1−\frac{1}{2}θ^2\right)+\frac{1}{2}kb^2θ^2[/latex]

which is valid for small values of [latex]θ[/latex].

For the system to be in stable equilibrium, [latex]d^2V/dθ^2[/latex] must be positive. After differentiation, we find

[latex]\frac{dV}{dθ}=(−WL+kb^2)θ[/latex]

Observe that [latex]dV/dθ=0[/latex] when [latex]θ=0[/latex], confirming that the rod is in equilibrium in the vertical position. Differentiating again, we obtain

[latex]\frac{d^2V}{dθ^2}=−WL+kb^2[/latex]

We see that [latex]d^2V/dθ^2[/latex] will be positive; that is, the system will be in stable equilibrium, only if [latex]kb^2>WL[/latex]. Therefore, the maximum value of [latex]W[/latex] for which the system will be in stable equilibrium for [latex]θ=0[/latex] is

[latex]W_{max}=kb^2/L[/latex]

Question 10.5: A light rod is pin-supported at one end and carries a weight...

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