Chapter 19
Q. 19.13
Figure (a) shows the same assembly that was described in Sample Problem 19.12. The assembly is initially stationary with the sphere resting against the shaft BC. A motor at C is then activated that drives the shaft with the constant angular acceleration α (the resulting angular velocity of the shaft is ω = αt). Letting t_0 be the time when the sphere loses contact with the shaft, (1) derive the equation of motion for the sphere in terms of the angle β for the period t ≥ t_0, and state the initial conditions; and (2) solve the equations numerically for the time interval t = t_0 to t = t_0 + 2 s and plot β versus t. Use m = 7 kg, L = R = 60 mm, and α = 5.5 rad/s².
Step-by-Step
Verified Solution
Part 1
The free-body diagram (FBD) of the rigid unit containing the sphere and the rod AB 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). This FBD displays the weight of the sphere and the reactions at B: the pin force B and the moment components M_y and M_z (the pin of the clevis does not provide a moment component about the x-axis). The fixed Z-axis is assumed to be directed along the vertical shaft as shown in Fig. (b), with θ being the Euler angle between the Z- and z-axes. Because the sphere and rod AB rotate as a rigid unit, the spin rate \dot ψ is zero. Furthermore, comparing Figs. (a) and (b) with Fig. 19.11, the precession rate is found to be \dot \phi = ω = αt.
The modified Euler equation that governs β is the first of Eqs. (19.44) (the remaining two equations could be used to find M_y and M_z).
\begin{matrix}∑M_x =& I\ddot θ + (I_z − I)\dot \phi^2 \sin θ \cos θ + I_z\dot \phi\dot ψ \sin θ\\∑M_y =& I\ddot \phi \sin θ + 2I\dot θ\dot \phi \cos θ − I_z \dot θ ( \dot ψ + \dot θ \cos θ ) &(19.44)\\∑M_z =& I_z(\ddot ψ + \ddot \phi \cos θ − \dot \phi\dot θ \sin θ )\\=& I_z\frac{d}{dt} (\dot ψ + \dot \phi \ \cos θ )\end{matrix}
∑M_x = I\ddot θ + (I_z − I)\dot \phi^2 \sin θ \cos θ + I_z\dot \phi\dot ψ \sin θ (a)
Using L = R and θ = π − β, the various terms in Eq. (a) become
∑M_x = 2mg R sin β (from the FBD)
I_z = \frac{2}{5}m R^2
I = I_y = \frac{2}{5}m R^2 + m(R + R)^2 = \frac{22}{5}m R^2
\ddot θ = −\ddot β \dot \phi = αt \dot ψ = 0
Substituting these expressions into Eq. (a), we get
2m g R \sin β = \frac{22}{5}m R^2(−\ddot β) + (−4m R^2)(αt)^2 \sin β(− \cos β) + 0
which, on canceling the mass m and rearranging terms, reduces to
\ddot β = \frac{10}{11}α^2t^2 \sin β \cos β − \frac{10g}{22R} \sin β (b)
When the numerical values α = 5.5 rad/s², g = 9.81 m/s², and R = 0.060 m are substituted into Eq. (b), the equation of motion becomes
\ddot β = 27.50t² sin β cos β − 74.32 sin β (for t ≥ t0) (c)
From the solution to Sample Problem 19.12, we know that ω_{cr}, the critical angular velocity at which contact between the sphere and vertical shaft is lost, is
ω_{cr} =\frac{\sqrt{g}}{\sqrt[4]{(L + R)^2 − R^2}} =\frac{\sqrt{9.81}}{\sqrt[4]{(0.120)^2 − (0.060)^2}}
= 9.716 rad/s
Consequently, the time at which contact is lost is t_0 = ω_{cr}/α = 9.716/5.5 = 1.7665 s. The initial value of β (the value when the sphere touches the vertical shaft) is β_0 = tan^{−1}[R/(L + R)] = tan^{−1}(R/2R) = tan^{−1}(1/2) = 30° = 0.5236 rad. Therefore, the initial conditions are
t_0 = 1.7665 s β_0 = 0.5236 rad \dot β_0 = 0 (d)
Part 2
The equivalent first-order equations and the initial conditions are (with x_1 = β and x_2 = \dot β)
\dot x_1 = x_2 x_1(1.7665) = 0.5236
\dot x_2 = 27.50t² sin β cos β − 74.32 sin β x_2(1.7665) = 0
MATLAB Verified Solution
Script Files
The MATLAB program that produced the plot in Fig. (c) is:
function example19_13
[t,x] = ode45(@f,[1.7665:0.01:3.7665],[0.5236 0]);
axes(’fontsize’,14)
plot(t,x(:,1)*180/pi,’linewidth’,1.5)
grid on
xlabel(’time (s)’); ylabel(’beta (deg)’)
function dxdt = f(t,x)
s = sin(x(1)); c = cos(x(1));
dxdt =[x(2); 27.5*tˆ2*s*c - 74.32*s];
end
end
Referring to Fig. (c), we see that β is made up of two parts—an average value that increases with time, and oscillations about the average value. As the angular speed ω of the shaft increases, we expect the average value of β to approach 90° and the amplitude of the oscillations to decrease (due to the increasing centrifugal force on the sphere). Both of these trends can be observed in Fig. (c).
