Question 17.SP.6: Each of the two slender rods shown is 0.75 m long and has a ......

STRATEGY: You have two positions and are interested in velocities, so use the conservation of energy. You will also need to use kinematics to relate the velocity terms in the kinetic energy expression.

MODELING: Choose the system to be both bars and model them as rigid bodies.

ANALYSIS: To illustrate that the order in which you solve a problem doesn’t matter, let’s start with kinematics.

Kinematics of Motion When β = 20°. Because v_{B} is perpendicular to the rod AB and v_{D} is horizontal, the instantaneous center of rotation of rod BD is located at C (Fig. 1). From the geometry of the figure, you obtain

BC = 0.75 m                 CD = 2(0.75 m) sin 20° = 0.513 m

Apply the law of cosines to triangle CDE, where E is located at the mass center of rod BD. You find EC = 0.522 m. Denoting the angular velocity of rod AB by ω, you have (Fig. 2)

\overline{ v }_{A B}=(0.375~m ) \omega \quad \overline{ v }_{A B}=0.375 \omega \searrow

\begin{array}{ll}v_B=(0.75~m ) \omega & v _B=0.75 \omega \quad \searrow\end{array}

Because rod BD seems to rotate about point C, you have

v_B=(B C) \omega_{B D} \qquad(0.75~m ) \omega=(0.75~m ) \omega_{B D} \qquad \omega _{B D}=\omega \circlearrowleft

\bar{v}_{B D}=(E C) \omega_{B D}=(0.522~m ) \omega \qquad \overline{ v }_{B D}=0.522 \omega \searrow

Conservation of Energy. Because there are no springs in the system

T_1+V_{g_1}=T_2+V_{g_2}

You first need to determine the energy at the two positions.

Position 1.

Potential Energy. Choose the datum as shown in Fig. 3, and observe that W = (6 kg)(9.81 m/s²) = 58.86 N. Then you have

V_{g_1}=2 W \bar{y}_1=2(58.86~N )(0.325~m )=38.26~J

Kinetic Energy. Initially, the system is at rest, so T_1= 0.

Position 2.

Potential Energy. Compute the new height of the mass centers of the rods to be \bar{y}_2 = 0.75 sin(20) = 0.1283 m (Fig. 4).

V_{g_2}=2 W \bar{y}_2=2(58.86~N)(0.1283~m)=15.10~J

Kinetic Energy.

\bar{I}_{A B}=\bar{I}_{B D}=\frac{1}{12}m l^2=\frac{1}{12}(6~kg )(0.75~m )^2=0.281~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 ω²

Conservation of Energy. Now you can write

T_1+V_{g_1}=T_2+V_{g_2}

0 + 38.26 J = 1.520 ω² + 15.10 J

ω = 3.90 rad/s                \omega_{AB} = 3.90 rad/s ↻ ◂

Velocity of Point D.

v_D = (CD)ω = ( 0.513 m )( 3.90 rad/s ) = 2.00 m/s

v_D = 2.00 m/s → ◂

REFLECT and THINK: The only step in which you need to use forces is when calculating the gravitational potential energy in each position. However, it is good engineering practice to show the complete free-body diagram in each case to identify which, if any, forces do work. Rather than use the instantaneous center of rotation, you could have also used vector algebra to relate the velocities of the various objects.