Question 17.SP.5: Each of the two slender rods shown is 0.75 m long and has a ......
Each of the two slender rods shown is 0.75 m long and has a mass of 6 kg. If the system is released from rest with β = 60°, determine (a) the angular velocity of rod AB when β = 20°, (b) the velocity of point D at the same instant.
Kinematics of Motion When β = 20°. Since \mathbf{v}_B is perpendicular to the rod AB and \mathbf{v}_D is horizontal, the instantaneous center of rotation of rod BD is located at C. Considering the geometry of the figure, we obtain
BC = 0.75 m CD = 2(0.75 m) sin 20° = 0.513 m
Applying the law of cosines to triangle CDE, where E is located at the mass center of rod BD, we find EC = 0.522 m. Denoting by ω the angular velocity of rod AB, we have
\begin{aligned}\bar{v}_{A B} & =(0.375 m ) \omega & \overline{ \mathbf{v} }_{A B} & =0.375 \omega \searrow \\v_B & =(0.75 m ) \omega & \mathbf{v} _B & =0.75 \omega \searrow\end{aligned}
Since rod BD seems to rotate about point C, we write
\begin{aligned}& v_B=(B C) \omega_{B D} \quad(0.75 m ) \omega=(0.75 m ) \omega_{B D} \quad \omega_{B D}=\omega \uparrow \\& \bar{v}_{B D}=(E C) \omega_{B D}=(0.522 m ) \omega \quad \overline{ \mathbf{v} }_{B D}=0.522 \omega \searrow\end{aligned}
Position 1. Potential Energy. Choosing the datum as shown, and observing that W = (6 kg)(9.81 m/s²) = 58.86 N, we have
V_1=2 W \bar{y}_1=2(58.86 N )(0.325 m )=38.26\text{ J}
Kinetic Energy. Since the system is at rest, T_1 =0.
Position 2. Potential Energy
V_2=2 W \bar{y}_2=2(58.86 N )(0.1283 m )=15.10\text{ J}
Kinetic Energy
\begin{aligned}I_{A B} & =\bar{I}_{B D}=\frac{1}{12} m l^2=\frac{1}{12}(6\text{ kg})(0.75 m )^2=0.281 \text{ kg} \cdot m ^2 \\T_2 & =\frac{1}{2} m \bar{v}_{A B}^2+\frac{1}{2} \bar{I}_{A B} \omega_{A B}^2+\frac{1}{2} m \bar{v}_{B D}^2+\frac{1}{2} \bar{I}_{B D} \omega_{B D}^2 \\& =\frac{1}{2}(6)(0.375 \omega)^2+\frac{1}{2}(0.281) \omega^2+\frac{1}{2}(6)(0.522 \omega)^2+\frac{1}{2}(0.281) \omega^2 \\& =1.520 \omega^2\end{aligned}
Conservation of Energy
\begin{aligned}T_1+V_1 & =T_2+V_2 \\0+38.26\text{ J}& =1.520 \omega^2+15.10\text{ J}\\\omega & =3.90\text{ rad}/ s \quad \omega_{A B}=3.90\text{ rad}/ s \downarrow\end{aligned}
Velocity of Point D
\begin{aligned}v_D=(C D) \omega=(0.513 m )(3.90\text{ rad}/ s )= & 2.00 m / s \\& \text{v}_D=2.00 m / s \rightarrow\end{aligned}
