The mechanism shown is called a swinging block slider crank. First used in various steam locomotive engines in the 1800s, this mechanism is often found in door-closing systems. Let H = 1.25 m, R = 0.45 m, and r denote the distance between B and O. Assuming that the speed of B is constant and equal

Question

The mechanism shown is called a swinging block slider crank. First used in various steam locomotive engines in the 1800s, this mechanism is often found in door-closing systems. Let H = 1.25 m, R = 0.45 m, and r denote the distance between B and O. Assuming that the speed of B is constant and equal to 5 m/s, determine [latex]\dot{r}, \dot{\phi}, \ddot{r} ext {, and } \ddot{\phi} ext { when } heta=180^{\circ}[/latex].

Accepted Answer

We start by defining a polar coordinate system with origin at O and radial direction along the segment OB so that the coordinate r measures the distance from O to B. We take as transverse coordinate, the angle [latex]\phi[/latex]. Observe that for 0 < θ < 180°, r is growing. For θ = 180° r achieves its maximum value [latex]r_{\max }[/latex] = R + H, and for 180° < θ < 360°, r decreases. Therefore, the rate of change of r for θ = 180° must be equal to zero, i.e.,

[latex]\left.\dot{r} ight|_{ heta=180^{\circ}}=0 .[/latex] (1)

In polar coordinates, the velocity of B is expressed as

[latex]\vec{v}_B=\dot{r} \hat{u}_r+r \dot{\phi} \hat{u}_\phi.[/latex] (2)

Therefore, in view of Eq. (1), for θ = 180° , recalling that B is moving counterclockwise with the constant speed [latex]v_0[/latex] = 5 m/s, we have

[latex]\left.\vec{v}_B ight|_{ heta=180^{\circ}}=r_{\max }\left(\left.\dot{\phi} ight|_{ heta=180^{\circ}} ight) \hat{u}_\phi=-\left.v_0 \hat{u}_\phi \quad \Rightarrow \quad \dot{\phi} ight|_{ heta=180^{\circ}}=-\frac{v_0}{R+H}=-2.941 \ rad / s,[/latex] (3)

where we have used the fact that R = 0.45 m and H = 1.25 m. Expressing the above result to 3 significant figures, we have

[latex]\left.\dot{\phi} ight|_{ heta=180^{\circ}}=-2.94 \ rad / s.[/latex]

Next we recall that the general expression of the acceleration in polar coordinate is

[latex]\vec{a}_B=\left(\ddot{r}-r \dot{\phi}^2 ight) \hat{u}_r+(r \ddot{\phi}+2 \dot{r} \dot{\phi}) \hat{u}_\phi.[/latex] (4)

Recalling that for θ = 180° , r = R + H, [latex]\dot{r}=0 ext { and } \dot{\phi}[/latex] takes on the expression in the second of Eqs. (3), we have that, for θ = 180° , Eq. (4) reduces to

[latex]\left.\vec{a}_B ight|_{ heta=180^{\circ}}=\left(\left.\ddot{r} ight|_{ heta=180^{\circ}}-\frac{v_0^2}{R+H} ight) \hat{u}_r+\left[\left.(R+H) \ddot{\phi} ight|_{ heta=180^{\circ}} ight] \hat{u}_\phi .[/latex] (5)

We now observe that, due to the fact that B is in uniform circular motion along a circle with center at A, the acceleration of B is always directed toward A and, for θ = 180° , we must have

[latex]\left.\vec{a}_B ight|_{ heta=180^{\circ}}=-\frac{v_0^2}{R} \hat{u}_r.[/latex] (6)

Setting Eqs. (5) and (6) equal to each other component by component, we have

[latex]\left.\ddot{r} ight|_{ heta=180^{\circ}}-\frac{v_0^2}{R+H}=\left.\frac{-v_0^2}{R} \Rightarrow \ddot{r} ight|_{ heta=180^{\circ}}=v_0^2\left(\frac{1}{R+H}-\frac{1}{R} ight) ext {, }[/latex]

and

[latex]\left.(R+H) \ddot{\phi} ight|_{ heta=180^{\circ}}=0.[/latex] (7)

Recalling that [latex]v_0[/latex] = 5 m/s, R = 0.45 m, and H = 1.25 m, we can then express our results in numerical form as follows:

[latex]\left.\ddot{r} ight|_{ heta=180^{\circ}}=-40.8 \ m / s ^2 \quad ext { and }\left.\quad \ddot{\phi} ight|_{ heta=180^{\circ}}=0 ext {. }[/latex]

Question 2.248: The mechanism shown is called a swinging block slider crank....

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