Question

A counterclockwise couple $C\left(\theta \right)$ acts on a uniform 1.5 kg bar AB as shown in Fig. 1. Given two cases:
Case (a) $C\left(\theta \right)$ = $5.4\sin \theta$ N-m,
Case (b) $C\left(\theta \right)$ varies as shown in Fig. 2.
The total work done on the bar as it rotates in the vertical plane about A from $\theta$ = $0^{\circ }$ to $\theta$ $={180}^{\circ }$ in the above two cases are $W_a$ and $W_b$ respectively. Then,
(Take $g=10m/s^2$)

• A. $W_a$ = 3.22 N-m
• B. $W_a$ = 4.2 N-m
• C. $W_b$ = 4.2 N-m
• D. $W_b$ = 1.86 N-m

Answer 1

Answer: (B), (D)

The correct options are
B $W_a$ = 4.2 N-m
D $W_b$ = 1.86 N-m

Case (a)
The total work done on the bar is the sum of the work done by the weight W and the couple C $\left(\theta \right)$. The work done by W can be found from $(U_{1-2}{)}_W=-W(\Delta h)$ where $\Delta$h is the upward vertical distance moved by the center of gravity of the bar between $\theta$ = 0 (position 1) and $\theta$ = 180${}^o$ (position 2). As the bar is homogeneous, $\Delta h=450mm=0.45m$
$\Rightarrow (U_{1-2}{)}_W$ = - 1.5 x 0.45 x 10 $\simeq$ - 6.6 N-m
By couple $(U_{1-2}{)}_C={\int }_o^{\pi }C(\theta )d\theta ={\int }_o^{\pi }$ 5.4 sin $\theta$ d $\theta$
$[-5.4cos\theta {]}_o^{\pi }$
= 10.8 N-m
$\Rightarrow$ Total work done on the bar
=$U^{1-2}=(U_{1-2}{)}_c+(U_{1-2}{)}_W$
= 10.8 - 6.6 = 4.2 N-m
Case (b)
By couple $(U_{1-2}{)}_c$ = Area under C - $\theta$ triangle
=$\frac{1}{2}\left(5.4\right)\left(\pi \right)=8.48$ N - m
Total work done on the bar
$=\left(U_{1-2}\right)=(U_{1-2}{)}_C+(U_{1-2}{)}_W$
= 8.48 - 6.62 = 1.86 N-m