The slider block has the motion shown. Determine the angular velocity and angular acceleration of the wheel at this instant.

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Accepted Answer

Rotation About A Fixed Axis. For wheel C, refer to Fig. a.

[latex]v_A = \omega_C r_C = \omega_C (0.15) \downarrow \ extbf{a}_A = \pmb{\alpha}_C imes extbf{r}_C - \omega^2_C extbf{r}_C \ \begin{aligned} extbf{a}_A &= (\alpha_C extbf{k}) imes (-0.15 extbf{i}) - \omega^2_C (-0.15 extbf{i}) \ &= 0.15 \omega^2_C extbf{i} - 0.15\alpha_C extbf{j} \end{aligned}[/latex]

General Plane Motion. The IC for crank AB can be located using [latex] extbf{v}_A[/latex] and [latex] extbf{v}_B[/latex] as shown in Fig. b. Here

[latex]r_{ A/IC } = 0.3 m \ r_{ B/IC } = 0.4 m[/latex]

Then the kinematics gives

[latex]v_B = \omega_{ AB } r_{ B/IC }; \quad\quad 4 = \omega_{ AB } (0.4) \quad\quad \omega_{ AB } = 10.0 rad/s \curvearrowleft \ v_A = \omega_{ AB } r_{ A/IC }; \quad\quad \omega_C (0.15) = 10.0(0.3) \quad\quad \omega_C = 20.0 rad/s \curvearrowleft[/latex]

Applying the relative acceleration equation by referring to Fig. c,

[latex] extbf{a}_B = extbf{a}_A + \pmb{\alpha}_{ AB } imes extbf{r}_{ B/A } - \omega_{ AB }^2 extbf{r}_{ B/A } \ \begin{aligned} 2 extbf{i} = 0.15 (20.0^2) extbf{i} - 0.15\alpha_C extbf{j} + (\alpha_{ AB } extbf{k} &) imes (0.3 extbf{i} - 0.4 extbf{j}) \ & -10.0^2 (0.3 extbf{i} - 0.4 extbf{j}) \end{aligned} \ 2 extbf{i} = (0.4\alpha_{ AB } + 30) extbf{i} + (0.3\alpha_{ AB } - 0.15\alpha_C + 40) extbf{j}[/latex]

Equating i and j components,

[latex]2 = 0.4\alpha_{ AB } + 30; \quad \alpha_{ AB } = -70.0 rad/s^2 = 70.0 rad/s^2 \curvearrowright \ 0 = 0.3(-70.0) + 0.15\alpha_C + 40; \quad \alpha_C = -126.67 rad/s^2 = 127 rad/s \curvearrowright[/latex]

The negative signs indicate that [latex]\pmb{\alpha}_C[/latex] and [latex]\pmb{\alpha}_{ AB }[/latex] are directed in the sense that those shown in Fig. a and c.

Question : The slider block has the motion shown. Determine the angular...