(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.

Question

Solve the equation of motion and determine the normal frequencies and normal vibrations. Sketch the normal vibrations.

Accepted Answer

(a) If the bar is divided into small segments of length [latex]dx[/latex] with the cross section f , we have elementary volumes [latex]dV = f dx. Let \varrho[/latex] be the constant density of the bar; then we have

[latex]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}[/latex].

Since [latex]m =\varrho f L[/latex] is the total mass of the bar, it follows that

[latex]Θ_{A} = \frac{1}{3} mL^{2}.[/latex]

According to Steiner’s theorem, the moment of inertia about an axis through the center of gravity is

[latex]Θ_{A} =Θ_{s} + m \left(\frac{L}{2} ight)^{2} ⇒ Θ_{s} = \frac{1}{12} mL^{2}.[/latex]

(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 [latex]x_{1}, x_{2}, x[/latex] 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 [latex]x_{1} +x_{2} = 2x[/latex]. Newton’s second law leads to

[latex]m\ddot{x} =−k(x_{1} − b)−k(x_{2} −b)[/latex]

or

[latex]m\ddot{x} =−k(x_{1} + x_{2}) +2kb[/latex].

The constraint condition leads to

[latex]m\ddot{x}=−2kx +2kb ⇒ \ddot{x} =−\frac{2k}{m} (x −b).[/latex] (11.50)

We assume that there are only small displacements, so that [latex]sin ϑ ≈ ϑ[/latex]. Then

[latex]x_{2} = x + \frac{L}{2}ϑ, x_{1} = x − \frac{L}{2}ϑ[/latex].

For the torque, we get

[latex]Θ\ddot{ϑ} =−\frac{k}{2} L(x_{2} −x_{1})=−\frac{1}{2}kL^{2}ϑ, since x_{2} − x_{1} = L ϑ[/latex].

From (a) [latex]Θ= (1/12)mL^{2}[/latex], we conclude

[latex]\ddot{ϑ}=−\frac{6k}{m}ϑ[/latex]. (11.51)

The solutions of (11.50) and (11.51) are

[latex]x = A cos(ω_{1}t + B)+b[/latex]

and

[latex]ϑ = C cos(ω_{2}t +D)[/latex]

with

[latex]ω_{1} =\sqrt{\frac{2k}{m}} and ω_{2} =\sqrt{\frac{6k}{m}}.[/latex]

The initial conditions at the time t = 0 are

[latex]x = b − \frac{a}{2}, ϑ= \frac{a}{L}, \dot{x}= 0, \dot{ϑ} = 0.[/latex]

Thus follows

[latex]\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}[/latex]

and, hence,

[latex]x = b − \frac{a}{2} cos \sqrt{\frac{2}{m}}t, ϑ = \frac{a}{L} cos \sqrt{\frac{6k}{m}}t.[/latex]

The normal modes are

[latex]X_{1} = x_{1} + x_{2} = 2b −a cos \sqrt{\frac{2k}{m}}t,[/latex]

[latex]X_{2} = x_{1} − x_{2} =−a cos \sqrt{\frac{6k}{m}}t.[/latex]

Question 11.12: (a) Find the moment of inertia of a thin homogeneous bar of ...

Related Answered Questions

Verified Answer:

Calculate the moment of inertia of a homogeneous massive cube about one of its edges. ...

Verified Answer:

A bar of length 2l and mass M is fixed at point A, so that it can rotate only in the vertical plane (see Fig. 11.20). The external force F acts on the center of gravity. Calculate the reaction force Fr at the point A! ...

Verified Answer:

A thin bar of length l and massM lies on a frictionless plate (the x,y-plane in Fig. 11.16). A hockey puck of mass m and velocity v knocks the bar elastically under 90◦ at the distance d from the center of gravity. After the collision the puck is at rest. (a) Determine the motion of the bar. ...

Verified Answer:

A cube with the edge length 2a and mass M glides with constant velocity v0 on a frictionless plate. At the end of the plate, it bumps against an obstacle and tilts over the edge (see Fig. 11.15). Find the minimum velocity v0 for which the cube still falls from the plate! ...

Verified Answer:

A cube of edge length s and mass M hangs vertically down from one of its edges. Find the period for small vibrations about the equilibrium position. How long is the equivalent thread pendulum? ...

Verified Answer:

Find the moment of inertia of a sphere about an axis through its center. The radius of the sphere is a, and the homogeneous density is e. ...

Verified Answer: