Question 11.12: (a) Find the moment of inertia of a thin homogeneous bar of ...
(a) Find the moment of inertia of a thin homogeneous bar of length L with respect to an axis perpendicular to the bar.
(b) A homogeneous bar of length L and mass m is supported at the ends by identical springs (spring constant k). The bar is moved at one end by a small displacement a and then released.
Solve the equation of motion and determine the normal frequencies and normal vibrations. Sketch the normal vibrations.
(a) If the bar is divided into small segments of length dx with the cross section f , we have elementary volumes dV = f dx. Let \varrho be the constant density of the bar; then we have
Θ_{A} =\int\limits_{0}^{L}{\varrho x^{2}(f dx)} =\varrho f\int\limits_{0}^{L}{ x^{2} dx} = \frac{1}{3} \varrho f L^{3}.
Since m =\varrho f L is the total mass of the bar, it follows that
Θ_{A} = \frac{1}{3} mL^{2}.According to Steiner’s theorem, the moment of inertia about an axis through the center of gravity is
Θ_{A} =Θ_{s} + m \left(\frac{L}{2}\right)^{2} ⇒ Θ_{s} = \frac{1}{12} mL^{2}.(b) Let b be the length of the spring before the motion (b is not the natural length of the spring, because of the existence of the gravitation field), and x_{1}, x_{2}, x be the lengths of the first and second spring, and the height of the center of gravity of the bar at the time t . Since the bar is rigid, we have x_{1} +x_{2} = 2x. Newton’s second law leads to
m\ddot{x} =−k(x_{1} − b)−k(x_{2} −b)or
m\ddot{x} =−k(x_{1} + x_{2}) +2kb.
The constraint condition leads to
m\ddot{x}=−2kx +2kb ⇒ \ddot{x} =−\frac{2k}{m} (x −b). (11.50)
We assume that there are only small displacements, so that sin ϑ ≈ ϑ. Then
x_{2} = x + \frac{L}{2}ϑ, x_{1} = x − \frac{L}{2}ϑ.
For the torque, we get
Θ\ddot{ϑ} =−\frac{k}{2} L(x_{2} −x_{1})=−\frac{1}{2}kL^{2}ϑ, since x_{2} − x_{1} = L ϑ.
From (a) Θ= (1/12)mL^{2}, we conclude
\ddot{ϑ}=−\frac{6k}{m}ϑ. (11.51)
The solutions of (11.50) and (11.51) are
x = A cos(ω_{1}t + B)+band
ϑ = C cos(ω_{2}t +D)with
ω_{1} =\sqrt{\frac{2k}{m}} and ω_{2} =\sqrt{\frac{6k}{m}}.The initial conditions at the time t = 0 are
x = b − \frac{a}{2}, ϑ= \frac{a}{L}, \dot{x}= 0, \dot{ϑ} = 0.Thus follows
\begin{matrix} B = D = 0, \\ A=−\frac{a}{2}, \\ C = \frac{a}{L},\end{matrix}\begin{cases}b − \frac{a}{2}= A\cos(B) +b, \\0=−Aω_{1} \sin(B),\\ \frac{a}{L} = C \cos(D),\\ 0=−Cω_{2} \sin(D),\end{cases}and, hence,
x = b − \frac{a}{2} cos \sqrt{\frac{2}{m}}t, ϑ = \frac{a}{L} cos \sqrt{\frac{6k}{m}}t.The normal modes are
X_{1} = x_{1} + x_{2} = 2b −a cos \sqrt{\frac{2k}{m}}t,X_{2} = x_{1} − x_{2} =−a cos \sqrt{\frac{6k}{m}}t.
