Question 310: Centre of Mass and Rotational Motion

Question

An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If AB = BC, and the angle made by AB with downward vertical is $$	heta $$, thrown :

Accepted Answer

Assume mass of each part = m
Here, $${{{C_1}R} \over {{C_1}A}} = \sin 	heta $$
$$ 	herefore $$   C
1
R = C
1
A(sin$$	heta $$)
= $${1 \over 2}$$ sin$$	heta $$
$${{BY} \over {AB}} = \sin 	heta $$
$$ \Rightarrow $$   BY = Lsin$$	heta $$
As  By = MP
$$ 	herefore $$   MP = Lsin$$	heta $$
$${{{C_2}M} \over {B{C_2}}}$$ = cos$$	heta $$
$$ \Rightarrow $$   C
2
M = BC
2
cos$$	heta $$
$$ \Rightarrow $$   C
2
M = $${L \over 2}\cos 	heta $$
$$ 	herefore $$   C
2
P = C
2
M $$-$$ MP = $${L \over 2}\cos 	heta $$ $$-$$ Lsin$$	heta $$
Now balancing torque about hinge paint A,
mg(C
1
R) = mg(C
2
P)
$$ \Rightarrow $$   mg$$\left( {{L \over 2}\sin 	heta } ight)$$ = mg$$\left( {{L \over 2}\cos 	heta  - L\sin 	heta } ight)$$
$$ \Rightarrow $$   $${{\sin 	heta } \over 2}$$ = $${{\cos 	heta } \over 2} - $$ sin$$	heta $$
$$ \Rightarrow $$   $${{3\sin 	heta } \over 2}$$ = $${{\cos 	heta } \over 2}$$
$$ \Rightarrow $$   tan$$	heta $$ = $${1 \over 3}$$

Answer

tan$$	heta $$ = $${1 \over {2\sqrt 3 }}$$

Answer

tan$$	heta $$ =  $${1 \over 2}$$

Answer

tan$$	heta $$ =  $${2 \over {\sqrt 3 }}$$

Answer

tan$$	heta $$ =  $${1 \over 3}$$

Centre of Mass and Rotational Motion

An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If AB = BC, and the angle made by AB with downward vertical is $\theta $, thrown :

Choose the correct answer: