The 180 kg uniform plate shown in Fig. (a) rotates in the vertical plane about a pin at A. The plate is released from rest when θ = 0. (1) Show that the differential equation of motion for the plate is α = 0.588(4 cos θ −3 sin θ ) rad/s² . (2) Integrate the differential equation of motion

Question

The 180 kg uniform plate shown in Fig. (a) rotates in the vertical plane about a pin at A. The plate is released from rest when θ = 0. (1) Show that the differential equation of motion for the plate is α = 0.588(4 cos θ −3 sin θ ) rad/s² . (2) Integrate the differential equation of motion analytically to obtain the angular velocity of the plate as a function of θ. (3) Find the maximum value of θ.

Accepted Answer

Part 1

The mass of the plate is m = 180 kg, and the moment of inertia about its mass center G is (see Table 17.1)

Slender rod (A = cross-sectional area) [latex]\bar I_x = \bar I_y = \frac{1}{12}mL^2[/latex] [latex]\bar I_z ≈ 0 [/latex]	Cylinder [latex]\bar I_x = \bar I_y = \frac{1}{12}m(3R^2 + h^2)[/latex] [latex]\bar I_z = \frac{1}{2}mR^2 [/latex]
Thin ring (A = cross-sectional area) [latex]\bar I_x = \bar I_y = \frac{1}{2}mR^2[/latex] [latex]\bar I_z = mR^2 [/latex]	Sphere [latex]\bar I_x = \bar I_y = \bar I_z = \frac{2}{5}mR^2[/latex]
Rectangular prism [latex]\bar I_x = \frac{1}{12}m(b^2 + c^2)[/latex] [latex]\bar I_y = \frac{1}{12}m(c^2 + a^2)[/latex] [latex]\bar I_z = \frac{1}{12}m(a^2 + b^2)[/latex]	Cone [latex]\bar I_x = \bar I_y = \frac{3}{80}m(4R^2+ h^2)[/latex] [latex]\bar I_z = \frac{3}{10}mR^ 2[/latex]

Table 17.1 Mass Moments of Inertia of Homogeneous Bodies
(ρ = mass density)

[latex]\bar I =\frac {1}{12}m(b^2 + c^2) = \frac {1}{12}(180)(4^2 + 3^2) [/latex]= 375 kg · m²

Figure (b) shows the free-body diagram (FBD) and mass-acceleration diagram(MAD) for an arbitrary position of the plate. The FBD contains [latex]A_n[/latex] and [latex]A_t[/latex] , the unknown components of the pin reactions at A, and the 180(9.8) = 1764 N weight acting at G. The MAD consists of the inertia couple [latex]\bar Iα[/latex] and the components of the inertia vector [latex]m\bar a[/latex] acting at G. Because the path of G is a circle of radius r = 2.5 m centered at A, the normal component of [latex]\bar a[/latex] is [latex]\bar a_n[/latex] = rω² = 2.5ω² ft/s², and its tangential component is [latex]\bar a_t[/latex] = rα = 2.5α ft/s². Observe that the angular acceleration α is assumed to be clockwise in the MAD. The units for ω and α are rad/s and rad/s², respectively.

Because the angle θ completely determines the position of the plate, the plate has a single degree of freedom. Therefore, there is only one differential equation of motion. The most convenient method of deriving this equation is by equating moments about A in the FBD and MAD, thereby obtaining

[latex](∑M_A)_{FBD} = (∑MA)_{MAD}[/latex]
(1764 cos θ )(2) − (1764 sin θ )(3) = 375α + 180(2.5α)(2.5)

which reduces to

α = 0.588(4 cos θ − 3 sin θ ) rad/s² (a)

The identical result could be obtained by using the special case of the moment equation: [latex]∑M_A = I_Aα[/latex].

Part 2

To find the angular velocity as a function of θ, we substitute α from Eq. (a) into ω dω = α dθ, which yields

ω dω = 0.588(4 cos θ − 3 sin θ ) dθ

The result of integrating this equation is

[latex]\frac{ω²}{2}[/latex] = 0.588(4 sin θ + 3 cos θ ) + C

The constant of integration C is evaluated by applying the initial condition ω = 0 when θ = 0, which gives C = −3(0.588) rad²/s². Therefore, the above equation becomes

[latex]\frac{ω²}{2}[/latex] = 0.588(4 sin θ + 3 cos θ − 3)

from which the angular velocity is found to be

[latex]ω = ±1.08\sqrt {4 sin θ + 3 cos θ − 3} rad/s[/latex] (b)

Part 3

The maximum value of θ occurs when ω = 0. According to Eq. (b), this value is the nonzero root of the equation 4 sin θ +3 cos θ −3 = 0. The numerical solution of this equation yields

θ[latex]_{max}[/latex] = 106.3°

Only one equation of motion was used to determine the motion of the plate.
The remaining two independent equations of motion—for example, [latex]∑F_n = m\bar a_n[/latex] and [latex]∑F_t = m\bar a_t[/latex]—could now be utilized to find the components of the pin reaction, [latex]A_n[/latex] and [latex]A_t[/latex] , in terms of θ, ω, and α. By substituting for α from Eq. (a) and ω from Eq. (b), we would then obtain [latex]A_n[/latex] and [latex]A_t[/latex] as functions of θ only.

Question 17.10: The 180 kg uniform plate shown in Fig. (a) rotates in the ve...

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