The peg is constrained to move in the slots of the fixed bar CD and rotating bar AB. When θ=30°, the angular velocity and angular acceleration of arm AB are θ˙=2rad/s and θ¨=3rad/s^2, respectively. Determine the magnitudes of the velocity and acceleration of the peg P at this instant.

Question

The peg is constrained to move in the slots of the fixed bar CD and rotating bar AB. When heta = 30°, the angular velocity and angular acceleration of arm AB are \dot{ heta } = 2 rad/s and \ddot{ heta } = 3 rad/{ s }^{ 2 }, respectively. Determine the magnitudes of the velocity and acceleration of the peg P at this instant.

Accepted Answer

Time Derivatives:

[latex]r = 4\space \sec heta \ \dot{ r } = (4 \sec heta( an heta)\dot{ heta } ) ft/s \quad\quad\quad\quad \dot{ heta } = 2 rad/s \ \begin{aligned} \ddot{ r } &= 4[\sec heta(tan heta)\ddot{ heta } + \dot{ heta }(\sec heta \sec^2 heta \dot{ heta } + an heta \sec heta( an heta)\dot{ heta } )] \quad\quad\quad \ddot{ heta } = 3 rad/s^2 \ &= 4[\sec heta( an heta) heta + heta^2 (\sec^3 heta° + an^2 heta°\sec heta°)] ft/s^2 \end{aligned}[/latex]

When [latex] heta = 30°[/latex],

[latex]r|_{ heta = 30° } = 4 \sec 30° = 4.619 ft \ \dot{ r }|_{ heta = 30° } = (4 \sec 30° an 30°)(2) = 5.333 ft/s \ \ddot{ r }|_{ heta = 30° } = 4[(\sec 30° an 30°)(3) + 2^2 (\sec^3 30° + an^2 30° \sec 30°)] = 38.79 ft/s^2[/latex]

Velocity:

[latex]v_r = \dot{ r } = 5.333 ft/s \quad\quad\quad v_ heta = r \dot{ heta } = 4.619(2) = 9.238 ft/s[/latex]

Thus, the magnitude of the peg’s velocity is

[latex]v = \sqrt{ v_r^2 + v_ heta^2 } = \sqrt{ 5.333^2 + 9.238^2 } = 10.7 ft/s[/latex]

Acceleration:

[latex]a_r = \ddot{ r } - r \dot{ heta }^2 = 38.79 - 4.619(2^2) = 20.32 ft/s^2 \ a_ heta = r \ddot{ heta } + 2 \dot{ r } \dot{ heta } = 4.619(3) + 2(5.333)(2) = 35.19 ft/s^2[/latex]

Thus, the magnitude of the peg’s acceleration is

[latex]a = \sqrt{ a_r^2 + a_ heta^2 } = \sqrt{ 20.32^2 + 35.19^2 } = 40.6 ft/s^2[/latex]

Question : The peg is constrained to move in the slots of the fixed bar...

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