The body shown in Fig. (a) consists of the homogeneous slender bar 1 that is rigidly connected to the homogeneous sphere 2. The body is rotating in the vertical plane about the pin at O. When the body is in the position where θ = 30°, its angular velocity is ω = 1.2 rad/s clockwise. At this instant

Question

The body shown in Fig. (a) consists of the homogeneous slender bar 1 that is rigidly connected to the homogeneous sphere 2. The body is rotating in the vertical plane about the pin at O. When the body is in the position where θ = 30°, its angular velocity is ω = 1.2 rad/s clockwise. At this instant, determine the angular acceleration α and the magnitude of the pin reaction at O.

Accepted Answer

The FBD and MAD of the body in the position θ = 30° are shown in Fig. (b). In these diagrams, the bar and the sphere are treated as separate entities, each with its own inertia couple and inertia vector. (An equivalent form of the MAD would be obtained by showing the inertia couple and inertia vector for the assembly; refer to Fig. 17.13.) Details of the diagrams are described in the following.

Free-body diagram The forces [latex]O_n[/latex] and [latex]O_t[/latex] are the components of the pin reaction relative to the n- and t-axes shown in the figure. The weights [latex]W_1[/latex] and [latex]W_2[/latex] of the bar and sphere, respectively, act at their mass centers [latex]G_1[/latex] and [latex]G_2[/latex]. The distances [latex]\bar r_1[/latex] = 0.4 m and [latex]\bar r_2[/latex] = 1.0 m, measured from O to the mass centers, are deduced from the dimensions in Fig. (a).

Mass-acceleration diagram In the MAD, we assume that the angular acceleration α, measured in rad/s², is clockwise. Using the fact that the body rotates about the fixed point O, kinematic analysis enables us to express the accelerations of [latex]G_1[/latex] and [latex]G_2[/latex] in terms of α and ω of the body. The inertia terms that appear in the MAD have been computed in the following manner.

For the slender bar:

[latex]\bar I_1α = \frac{m_1L^2}{12}α = \frac{30(0.8)^2}{12} \alpha[/latex] = 1.600α N · m
[latex]m_1\bar r_1ω^2[/latex] = 30(0.4)(1.2)² = 17.28 N
[latex]m_1\bar r_1α[/latex] = 30(0.4)α = 12.00α N

For the sphere:

[latex]\bar I_2α = \frac{2}{5}m_2R^2α = \frac{2}{5}(80)(0.2)^2α[/latex] = 1.280α N · m
[latex]m_2\bar r_2ω^2[/latex] = 80(1.0)(1.2)² = 115.2 N
[latex]m_2\bar r_2α[/latex] = 80(1.0)α = 80.00α N

In the MAD, the directions of the tangential components of the inertia vectors (those containing α) are consistent with the assumed clockwise direction of α. The normal components of the inertia vectors (those containing ω²) are directed toward the center of rotation O, regardless of the direction of ω.

From Fig. (b) we see that there are two unknowns in the FBD ([latex]O_n[/latex] and [latex]O_t[/latex]) and one unknown (α) in the MAD. Therefore, all that remains is to derive and solve the three independent equations of motion for the unknowns.

Equating moments about O on the FBD and the MAD in Fig. (b), we obtain

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

30(9.8)(0.4) cos 30° + 80(9.8)(1.0) cos 30°

= 1.600α + (12.00α)(0.4) + 1.280α + (80.00α)(1.0)
= 87.68α

from which we find that

α = 8.905 rad/s²

Because the acceleration of point O is zero, this result could also have been derived using the special case [latex]∑M_O = I_O α[/latex].
Using α = 8.914 rad/s² and referring to Fig. (b), the force equations in the t and n directions give

[latex]∑F_t = m\bar a_t[/latex]
[latex]+\swarrow O_t[/latex] + 30(9.8) cos 30° + 80(9.8) cos 30° = 12.00(8.905) +80.00(8.905)
[latex]O_t[/latex] = −114.3 N

and

[latex]∑F_n = m\bar a_n[/latex]
[latex] warrow^{+} O_n[/latex] − 30(9.8)sin 30° − 80(9.8)sin 30° = 17.28 + 115.2
[latex]O_n[/latex] = 671.5 N

Therefore, the magnitude of the pin reaction at O is

[latex]O =\sqrt {O_t^2 + O_n^2} =\sqrt { (−114.3)^2 + (671.5)^2}[/latex] = 681.2 N

Question 17.5: The body shown in Fig. (a) consists of the homogeneous slend...

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