A wire of uniform cross-section is bent in the shape shown in the figure. The coordinates of the center of mass of each side are shown in the figure. The coordinates of the center of mass of the system are

Question

A wire of uniform cross-section is bent in the shape shown in the figure. The coordinates of the center of mass of each side are shown in the figure. The coordinates of the center of mass of the system are (A) ((15 l/14), (6 l/7)) (B) ((15 l/14), l) (C) (l, (l/2)) (D) (l, l)

Tardigrade Admin · Accepted Answer

X cm =(m × 0+m l+m × 2 l+(m/2) (3 l/2)/3.5 m ) X cm =(15 m l/4 × (7)2 m )=(15 l/14) Y cm =(m × 0+m l+m l+(m/2) 2 l/(7)2 m) ⇒ Y cm =(3 m l/(7 m)2)=(6 l/7)

Q. A wire of uniform cross-section is bent in the shape shown in the figure. The coordinates of the center of mass of each side are shown in the figure. The coordinates of the center of mass of the system are

$(\frac{15 l}{14}, \frac{6 l}{7})$

$(\frac{15 l}{14}, l)$

$(l, \frac{l}{2})$

$(l, l)$

$X_{c m} = \frac{m \times 0 + m l + m \times 2 l + \frac{m}{2} \frac{3 l}{2}}{3.5 m}$
$X_{c m} = \frac{15 m l}{4 \times \frac{7}{2} m} = \frac{15 l}{14}$
$Y_{c m} = \frac{m \times 0 + m l + m l + \frac{m}{2} 2 l}{\frac{7}{2} m}$
$\Rightarrow Y_{c m} = \frac{3 m l}{\frac{7 m}{2}} = \frac{6 l}{7}$

Q. A wire of uniform cross-section is bent in the shape shown in the figure. The coordinates of the center of mass of each side are shown in the figure. The coordinates of the center of mass of the system are

(1415l​,76l​)

(1415l​,l)

(l,2l​)

(l,l)

Xcm​=3.5mm×0+ml+m×2l+2m​23l​​ Xcm​=4×27​m15ml​=1415l​ Ycm​=27​mm×0+ml+ml+2m​2l​ ⇒Ycm​=27m​3ml​=76l​

$(\frac{15 l}{14}, \frac{6 l}{7})$

$(\frac{15 l}{14}, l)$

$(l, \frac{l}{2})$

$(l, l)$

$X_{c m} = \frac{m \times 0 + m l + m \times 2 l + \frac{m}{2} \frac{3 l}{2}}{3.5 m}$
$X_{c m} = \frac{15 m l}{4 \times \frac{7}{2} m} = \frac{15 l}{14}$
$Y_{c m} = \frac{m \times 0 + m l + m l + \frac{m}{2} 2 l}{\frac{7}{2} m}$
$\Rightarrow Y_{c m} = \frac{3 m l}{\frac{7 m}{2}} = \frac{6 l}{7}$