The piston head at C is constrained to move along the y axis. Let the crank AB be rotating counterclockwise at a constant angular speed $\dot{\theta}$ = 2000 rpm, R = 3.5 in., and L = 5.3 in. Obtain the angular velocity of the connecting rod BC by differentiating the relative position vector of C with respect to B when θ = 35°.

Hint: You will also need to determine the velocity of B and enforce the constraint that demands that C move only along the y axis.

Question

The piston head at C is constrained to move along the y axis. Let the crank AB be rotating counterclockwise at a constant angular speed $\dot{\theta}$ = 2000 rpm, R = 3.5 in., and L = 5.3 in. Obtain the angular velocity of the connecting rod BC by differentiating the relative position vector of C with respect to B when θ = 35°.

Hint: You will also need to determine the velocity of B and enforce the constraint that demands that C move only along the y axis.

Accepted Answer

Referring to the figure on the right, the can describe the position of C relative to B as follows:

[latex]\vec{r}_{C / B}=L \hat{u}_{C / B}[/latex] (1)

Then, using the concept of time derivative of a vector, we have that

[latex]\vec{v}_{C / B}=L \dot{\hat{u}}_{C / B}=L \vec{\omega}_{B C} imes \hat{u}_{C / B}[/latex] (2)

where the angular velocity [latex]\vec{\omega}_{B C}[/latex] , which is the angular velocity of the vector [latex]\vec{r}_{C / B}[/latex] , is the angular velocity of the connecting rod. Next, again referring to the figure on the right, we must have

[latex]\vec{\omega}_{B C}=-\dot{\phi} \hat{k} \quad ext { and } \quad \hat{u}_{C / B}=\sin \phi \hat{\imath}+\cos \phi \hat{\jmath}.[/latex] (3)

Substituting Eqs. (3) into Eq. (2) and carrying out the cross-product, we have

[latex]\vec{v}_{C / B}=L \dot{\hat{u}}_{C / B}=L \dot{\phi}(\cos \phi \hat{\imath}-\sin \phi \hat{\jmath}).[/latex] (4)

To enforce the condition that C can only move only in the y direction, we must compute the velocity of B and then apply the relative kinematics equation [latex]\vec{v}_C=\vec{v}_B+\vec{v}_{C / B}[/latex] . Since B is in uniform circular motion along a circle of radius R and center A, we have

[latex]\vec{v}_B=-\dot{ heta} R(\cos heta \hat{\imath}+\sin heta \hat{\jmath}).[/latex] (5)

Therefore, combining Eqs. (5) and Eq. (3), we have

[latex]\vec{v}_C=\vec{v}_B+\vec{v}_{C / B}=(L \dot{\phi} \cos \phi-\dot{ heta} R \cos heta) \hat{\imath}-(L \dot{\phi} \sin \phi+\dot{ heta} R \sin heta) \hat{\jmath}.[/latex] (6)

Since [latex]v_{C x}=0[/latex], we have

[latex]L \dot{\phi} \cos \phi-\dot{ heta} R \cos heta=0 \quad \Rightarrow \quad \dot{\phi}=\frac{R \cos heta}{L \cos \phi} \dot{ heta} \quad \Rightarrow \quad \vec{\omega}_{B C}=-\frac{R \cos heta}{L \cos \phi} \dot{ heta} \hat{k},[/latex] (7)

where we have used the definition in the first of Eqs. (3). In order to complete our calculation, we need to determine the angle [latex]\phi[/latex]. Using trigonometry, we see that

[latex]L \sin \phi=R \sin heta \quad \Rightarrow \quad \phi=\left.\sin ^{-1}[(R / L) \sin heta] \quad \Rightarrow \quad \phi ight|_{ heta=35^{\circ}}=22.26^{\circ},[/latex] (8)

where we have used the fact that R = 3.5 in. and L = 5.3 in.. Hence, evaluating the last of Eqs. (7) using the value of [latex]\phi[/latex] in Eq. (8) and recalling that [latex]\dot{ heta}[/latex] = 2000 rpm, for θ = 35° we have

[latex]\vec{\omega}_{B C}=(-1170 \ rpm ) \hat{k}.[/latex]

Question 2.255: The piston head at C is constrained to move along the y axis...

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