The electric motor and its frame shown in Fig. (a) have a combined mass M. The unbalance of the rotor is equivalent to a mass m (included in M), located at the distance e from the center O of the shaft. The frame is supported by vertical guides, a spring of stiffness k, and a pin at A. (The pin at

Question

The electric motor and its frame shown in Fig. (a) have a combined mass M. The unbalance of the rotor is equivalent to a mass m (included in M), located at the distance e from the center O of the shaft. The frame is supported by vertical guides, a spring of stiffness k, and a pin at A. (The pin at A was inserted when the motor was at rest in the static equilibrium position.) The motor is running at the constant angular speed ω when the pin is withdrawn at the instant when the mass m is in the position shown. (1) Derive the differential equation of motion of the assembly. (2) Plot the transient, steady-state, and total displacement versus time, using the data M = 40 kg, m = 1.2 kg, e = 150 mm, k = 900 N/m, and ω = 8 rad/s.

Accepted Answer

The rotating unbalance causes the position of the mass center of the assembly to vary with time, thereby giving rise to vibrations in the vertical direction (the vertical guides prevent horizontal motion of the frame).

Part 1

The free-body diagram (FBD) of the assembly is shown in Fig. (b). Note that x was chosen to be the displacement of O from its static equilibrium position. The FBD contains the total weight Mg of the assembly, acting at its mass center G, the spring force k(x − Δ), where Δ is the static deflection of the spring, and the resultant horizontal force N exerted by the vertical guides. The mass-acceleration diagram in Fig. (b) displays the inertia vectors (M − m)[latex]\ddot x[/latex] and m[latex]\ddot x[/latex] associated with the vertical motion of the assembly, and the radial inertia vector meω² caused by the rotation of the mass m about O. Equating the resultant vertical forces on the FBD and the MAD, we get

[latex]+\big\uparrow − Mg − k(x − Δ) = (M − m)\ddot x + m\ddot x − meω^2 sin ωt[/latex] (a)

After eliminating Δ by utilizing the equilibrium equation Mg − k Δ = 0, the differential equation of motion becomes

M[latex]\ddot x[/latex] + kx = meω² sin ωt (b)

Comparing Eq. (b) with Eq. (20.23), we see that the equations are identical if we let c = 0 and replace the magnitude [latex]P_0[/latex] of the forcing function by meω² (the term meω² is sometimes referred to as the centrifugal force due to the unbalanced mass).

[latex]m\ddot x + c\dot x + kx = P_0 sin ωt[/latex] (20.23)

Part 2

With [latex]P_0[/latex] = meω² and ζ = 0 (no damping), Eqs. (20.26) and (20.27) yield

[latex]X = \frac{P_0/k}{\sqrt{[1 − (ω/p)^2]^2 + (2ζω/p)^2}}[/latex] (20.26)

[latex]φ = tan^{−1} [\frac{(2ζω/p)^2}{1 − (ω/p)^2}][/latex] (20.27)

[latex]X = \frac{meω^2/k}{1 − (ω/p)^2}[/latex] and Φ = 0

so that the particular solution of Eq. (b) is—see Eq. (20.25)—

[latex]x_p[/latex] = X sin (ωt − φ) (20.25)

[latex]x_p = \frac{meω^2/k}{1 − (ω/p)^2} sin ωt[/latex]

Since the system is undamped, the complementary solution is given by Eq. (20.4): [latex]x_c[/latex] = E sin(ωt + α). Therefore, the complete solution is

x = E sin (pt + α) (20.4)

[latex]x = x_c + x_p = E sin(ωt + α) + \frac{meω^2/k}{1 − (ω/p)^2} sin ωt[/latex] (c)

Using the given numerical values, we obtain

[latex]p = \sqrt{k/M} = \sqrt{900/40}[/latex] = 4.743 rad/s

[latex]\frac{meω^2/k}{1 − (ω/p)^2} = \frac{1.2(0.15)(8)^2/900}{1 − (8/4.743)^2}[/latex] = −0.006 938 m = −6.938 mm

Substituting these values into Eq. (c) yields

x(t) = E sin(4.743t + α) − 6.938 sin 8t mm (d)

Taking the time derivative, we obtain for the velocity

[latex]\dot x[/latex](t) = 4.743 E cos(4.743t + α) − 55.50 cos 8t mm/s (e)

Substituting the initial conditions (x = 0 and [latex]\dot x[/latex] = 0 when t = 0) into Eqs. (d) and (e) gives E sin α = 0 and 4.743 E cos α − 55.50 = 0, which yields α = 0 and E = 11.701 mm. Therefore the description of the motion is

x(t) = 11.70 sin 4.74t − 6.94 sin 8t mm (f)

Figure (c) shows plots of the transient vibration (with the period [latex]τ_t[/latex] = 2π/p = 2π/4.743 = 1.325 s), the steady-state vibration (period [latex]τ_s[/latex] = 2π/ω = 2π/8 = 0.785 s), and the superposition of the two. Because the coefficient of sin 8t in Eq. (f) is negative, the steady-state vibration is 180° out of phase with the rotation of the unbalanced mass.

Question 20.5: The electric motor and its frame shown in Fig. (a) have a co...

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