Figure (a) shows a 10-kg homogeneous disk of radius 0.2 m. The disk is at rest before the horizontal force P = 60 N is applied to its mass center G. The coefficients of static and kinetic friction for the surfaces in contact are 0.20 and 0.15, respectively. Determine the angular acceleration of the

Question

Figure (a) shows a 10-kg homogeneous disk of radius 0.2 m. The disk is at rest before the horizontal force P = 60 N is applied to its mass center G. The coefficients of static and kinetic friction for the surfaces in contact are 0.20 and 0.15, respectively. Determine the angular acceleration of the disk and the acceleration of G after the force is applied.

Accepted Answer

Two motions of the disk are possible: rolling without slipping, and rolling with slipping. We will solve the problem by assuming that the disk rolls without slipping. This assumption will then be checked by comparing the required friction force with its maximum static value.
The free-body diagram (FBD) and the mass-acceleration diagram (MAD) based on the no-slip assumption are shown in Fig. (b). The FBD contains the 10(9.8) = 98 N weight, the 60 N applied force, the normal force [latex]N_A[/latex], and the friction force F, assumed acting to the left. The MAD contains the inertia couple and inertia vector, where the angular acceleration α, measured in rad/s², has been assumed to be clockwise. The values of [latex]\bar Iα[/latex] and [latex]m\bar a[/latex] were computed as follows.

[latex]\bar Iα = \frac{mR^2}{2}α = \frac {10(0.2)^2}{2}α [/latex]= 0.2α N · m
[latex]m\bar a[/latex] = m Rα = 10(0.2)α = 2α N

Note that [latex]\bar a[/latex] = Rα is a valid kinematic equation because the disk is assumed to be rolling without slipping. There are a total of three unknowns in the FBD and MAD: F, N, and α, which can be computed using any three independent equations of motion.
A convenient solution is to first equate the resultant moment about C on the FBD to the resultant moment about C on the MAD and then utilize the force equations of motion.

[latex](∑M_C)_{FBD} = (∑M_C)_{MAD}[/latex] 60(0.2) = 0.2α + 2α(0.2)
α = 20 rad/s²

[latex]∑F_x = m\bar a_x [/latex] [latex]\underrightarrow{+}[/latex] 60 − F = 2α = 2(20)
F = 20 N

[latex]∑F_y = m\bar a_y +\big\uparrow N_A[/latex] − 98 = 0
[latex]N_A[/latex] = 98 N

Because α, F, and [latex]N_A[/latex] are all positive, their directions are as shown in Fig. (b).
Next we note that the maximum possible static friction force is [latex]F_{max} = μ_sN_A[/latex] = 0.20(98) = 19.6. The friction force required for rolling without slipping is, according to our solution, F = 20 N. Because F > [latex]F_{max}[/latex], we conclude that the disk does slip, and we must reformulate the problem.
The FBD and MAD for the case where the disk rolls and slips simultaneously are shown in Fig. (c). The friction force F in the FBD has been set equal to its kinetic value, [latex]μ_kN[/latex]. This force must be shown acting to the left in order to oppose slipping. The inertia couple [latex]\bar Iα[/latex] in the MAD is identical to that used in Fig. (b). However, the important difference here is that the magnitude of the inertia vector is now [latex]m\bar a = 10\bar a[/latex] N, where [latex]\bar a[/latex] is measured in m/s². Because the disk is slipping, the kinematic constraint [latex]\bar a[/latex] = Rα does not apply. Once again we see that there are three unknowns on the FBD and the MAD, except that now the unknowns are N, α, and [latex]\bar a[/latex].

The three unknowns can be calculated as follows (of course, any other three independent equations could also be used):

[latex]∑F_y = m\bar a_y      +\big\uparrow      N_A[/latex] − 98 = 0      N = 98 N           (a)
[latex]∑F_y = m\bar a_x      \underrightarrow{+}      60 − 0.15N_A = 10\bar a[/latex]           (b)
[latex]∑M_G = \bar Iα [/latex] [latex]0.2(0.15N_A)[/latex] = 0.2α           (c)

Substituting N = 50 lb from Eq. (a) into Eqs. (b) and (c) yields

[latex]\bar a[/latex] = 4.53 m/s² and α = 14.7 rad/s²

Question 17.8: Figure (a) shows a 10-kg homogeneous disk of radius 0.2 m. T...

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