The cable connected to block B in Fig. (a) is wound tightly around disk A, which is free to rotate about the axle at its mass center G. The masses of A and B are 60 kg and 20 kg, respectively, and k = 400 mm for the disk. Determine the angular acceleration of A and the tension in the cable.

Question

The cable connected to block B in Fig. (a) is wound tightly around disk A, which is free to rotate about the axle at its mass center G. The masses of A and B are 60 kg and 20 kg, respectively, and [latex]\bar k[/latex] = 400 mm for the disk. Determine the angular acceleration of A and the tension in the cable.

Accepted Answer

The free-body and mass-acceleration diagrams of the system are shown in Fig. (b). The FBD contains the weights [latex]W_A[/latex] = 60(9.8) = 588 N and [latex]W_B[/latex] = 20(9.8) = 196 N together with the unknown pin reactions at G. The tension in the cable, being an internal force, does appear on this FBD.
The MAD displays the inertia couple of the disk and the inertia vector of the block. There is no inertia vector of the disk because its mass center G is stationary. The angular acceleration α of the disk is assumed to be directed clockwise. The corresponding inertia couple of the disk is

[latex]\bar I_α = m\bar k^2 α = 60(0.4)^2 α[/latex] = 9.600α N · m

which also is directed clockwise. Because the cable does not slip on the disk, the acceleration of the block is [latex]a_B = Rα[/latex], which results in the inertia vector

[latex]m_Ba_B = m_B(Rα) = 20(0.5α)[/latex] = 10α N [latex]\big\downarrow[/latex]

There are three unknowns on the FBD and the MAD: two components of the pin reaction at G and the angular acceleration α of the disk. Since the number of independent equations available from the FBD and MAD is also three, all the unknowns can be computed.
The angular acceleration α can be found by equating the resultant moments about G in the FBD and the MAD.

[latex](∑M_G)_{FBD} = (∑M_G)_{MAD}[/latex] 196(0.5) = 9.600α + 10α(0.5)
α = 6.712 rad/s²

To find the tension in the cable, we analyze the block separately (the disk could also be used). The FBD and MAD for the block are shown in Fig. (c), where T is the cable tension. Summing forces in the y-direction yields

[latex]∑F_y = ma_y +\big\downarrow [/latex] 196 − T = 10α = 10(6.712)
T = 129.0 N

Question 17.6: The cable connected to block B in Fig. (a) is wound tightly ...

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