Question 605: Centre of Mass and Rotational Motion

Question

A rod $l \mathrm{~m}$ long is acted upon by a couple as shown in the figure. The moment of couple is $	au \mathrm{~Nm}$. If the force at each end of the rod, then magnitude of each force is
$$\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5ight)$$

Accepted Answer

Length of rod = $l$
Moment of couple,
$$\begin{aligned}
 & 	au=Fl \sin 	heta \
	herefore & F=\frac{	au}{l \sin 	heta}=\frac{	au}{l \sin 30^{\circ}}=\frac{	au}{l 	imes \frac{1}{2}}
\end{aligned}$$
Force, $\quad F=\frac{2 	au}{l}$

Answer

$\frac{	au}{l}$

Answer

$\frac{l}{2 	au}$

Answer

$\frac{2 	au}{l}$

Answer

$\frac{2}{	au}$

Centre of Mass and Rotational Motion

A rod l \mathrm{~m} long is acted upon by a couple as shown in the figure. The moment of couple is \tau \mathrm{~Nm}. If the force at each end of the rod, then magnitude of each force is $\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5\right)

Choose the correct answer: