The 40-kg unbalanced wheel in Fig. (a) is rolling without slipping under the action of a counterclockwise couple C0 = 20 N · m. When the wheel is in the position shown, its angular velocity is ω = 2 rad/s, clockwise. For this position, calculate the angular acceleration α and the forces exerted on

Question

The 40-kg unbalanced wheel in Fig. (a) is rolling without slipping under the action of a counterclockwise couple [latex]C_0[/latex] = 20 N · m. When the wheel is in the position shown, its angular velocity is ω = 2 rad/s, clockwise. For this position, calculate the angular acceleration α and the forces exerted on the wheel at C by the rough horizontal plane. The radius of gyration of the wheel about its mass center G is [latex]\bar k[/latex] = 200 mm.

Accepted Answer

The free-body and mass-acceleration diagrams for the wheel, shown in Fig. (b), were constructed as follows.

Free-body diagram The FBD consists of the applied couple [latex]C_0[/latex], the weight W = 40(9.8) = 392 N, and the normal and friction forces that act at the contact point C, denoted by [latex]N_C[/latex] and [latex]F_C[/latex], respectively. Observe that [latex]F_C[/latex] has been assumed to be directed to the right.

Mass-acceleration diagram In the MAD of Fig. (b) the angular acceleration α, measured in rad/s², has been assumed to be clockwise. The corresponding inertia couple shown on this diagram is

[latex]\bar Iα = m\bar k^2α[/latex] = 40(0.200)²α = 1.600α N · m

Because the wheel does not slip, the acceleration of its center O is [latex]a_O= R_α[/latex] = 0.250α m/s², directed to the right. Applying the relative acceleration equation between G and O, we obtain (the units of each term are m/s²)

[latex]\bar a = a_G = a_G + a_{G/O}[/latex]

from which we find [latex]\bar a_x[/latex] = 0.250α − 0.480 m/s² and [latex]\bar a_y[/latex] = 0.120α m/s². Multiplying these results by m = 40 kg, the components of the inertia vector become [latex]m\bar a_x[/latex] = (10.00α − 19.20) N, directed to the right, and [latex]m\bar a_y[/latex] = 4.80α N, directed downward.
The FBD and MAD in Fig. (b) now contain only three unknowns: [latex]N_C, F_C[/latex], and α, which can be found using any three independent equations of motion.
Because [latex]N_C[/latex] and [latex]F_C[/latex] act at C, it is convenient to use that point as a moment center, the corresponding moment equation being

[latex](∑M_C)_{FBD} = (∑M_C)_{MAD}[/latex]

− 20 + 392(0.120) = 1.600α + 0.250(10.00α − 19.20) + 0.120(4.80α)

The solution to this equation is

α = 6.820 rad/s²

Because α is positive, its direction is clockwise, as assumed. The forces at C can now be found from force equations of motion:

[latex]∑F_x = m\bar a_x \underrightarrow{+} F_C[/latex] = 10.00α − 19.20 = 10.00(6.820) − 19.20

and

[latex]∑F_y = m\bar a_y +\big\downarrow[/latex] 392 − [latex]N_C[/latex] = 4.80α = 4.80(6.820)

which yield

[latex]F_C[/latex] = 49.0 N and [latex]N_C[/latex] = 359.3 N

Because each force is positive, it is directed as shown in the FBD.

Question 17.7: The 40-kg unbalanced wheel in Fig. (a) is rolling without sl...

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