Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Your Ultimate AI Essay Writer & Assistant.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Your Ultimate AI Essay Writer & Assistant.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 19

Q. 19.12

Figure (a) shows a uniform sphere of radius R and mass m that is welded to the rod AB of length L (the mass of AB may be neglected). The clevis at B connects the rod to the vertical shaft BC. The assembly is initially rotating about the vertical at the angular velocity ω, with the sphere resting against the shaft. Assuming that ω is gradually increased, determine the critical angular velocity ω_{cr} at which contact between the sphere and rod is lost. Neglect friction.

19.12a

Step-by-Step

Verified Solution

The free-body diagram of the sphere and the rod, drawn at the instant when the assembly is rotating at ω = ω_{cr}, is shown in Fig. (b). The xyz-axes are assumed to be attached to rod AB with the origin at B (the x-axis is out of the paper). In addition to the weight mg of the sphere, the FBD also contains the reactions provided by the clevis at B: the pin force B and the two moment components M_y and M_z (M_x = 0 because the pin of the clevis is frictionless). If the angular velocity were less than the critical angular velocity, the FBD would also contain the normal force N that is exerted on the sphere by the vertical shaft. However, when ω = ω_{cr}, then N = 0.
It is convenient to choose the Z-axis to coincide with the vertical shaft, as shown in Fig. (b). The Euler angle θ, which was defined as the angle between the Z- and z-axes, is also shown in the figure. When the sphere is about to lose contact with the vertical shaft, its motion consists of a rotation about the Z-axis at the rate ω_{cr}. The spin rate is zero, because the sphere cannot rotate relative to the rod AB. Therefore, the motion of the sphere can be described as a steady precession with no spin; in other words, \dot \phi = ω_{cr}, \dot ψ = \dot θ = 0. As a result, the steady precession equation, Eq. (19.46), becomes

\begin{matrix}∑M_x = (I_z  −  I)\dot \phi^2 \sin  θ  \cos  θ  +  I_z\dot \phi\dot ψ  \sin  θ\\∑M_y = 0\qquad             ∑M_z = 0\end{matrix}               (19.46)

∑M_x = (I_z  −  I)ω_{cr}^2  \sin  θ  \cos  θ               (a)

The inertial properties of the sphere about point B are

I_z = \frac{2}{5}m  R^2      and      I = I_y = \frac{2}{5}m  R^2 + m  (L + R)^2

from which we obtain

I_z  −  I = −m  (L + R)^2               (b)

Referring to Fig. (b), we find that the moment of the external forces (the weight) about the x-axis is

∑M_x = mg  R               (c)

From the same figure we also deduce that

\sin  θ = \sin(π  −  θ ) = \frac{R}{L + R}               (d)

and

\cos  θ = −  \cos(π  −  θ ) = − \frac{\sqrt{(L + R)^2  −  R^2}}{L + R}               (e)

Substituting Eqs. (b)–(e) into Eq. (a) and solving for the critical angular velocity, we obtain

ω_{cr} =\frac{\sqrt{g}}{\sqrt[4]{(L + R)^2 − R^2}}

19.12b