Question

The position vector of the centre of mass ⃗r
cm of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:
(A) ⃗r cm = (13/8)Lxˆ + (5/8)Lyˆ
(B) ⃗r cm = (5/8)Lxˆ + (13/8)Lyˆ
(C) ⃗r cm = (3/8)Lxˆ + (11/8)Lyˆ
(D) ⃗r cm = (11/8)Lxˆ + (3/8)Lyˆ

Answer 1

Answer:

$x_{cm} = \binom{2ml + 2ml + \frac{5ml}{2} }{4m} = \frac{13}{8}l$

$y_{cm} = \frac{2m \times l + m \times \frac{l}{2} + m \times 0}{4l} = \frac{5 l}{8}$

The position vector of the centre of mass ⃗r

cm of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:

(A) ⃗r cm = (13/8)Lxˆ + (5/8)Lyˆ ✓✓

(B) ⃗r cm = (5/8)Lxˆ + (13/8)Lyˆ

(C) ⃗r cm = (3/8)Lxˆ + (11/8)Lyˆ

(D) ⃗r cm = (11/8)Lxˆ + (3/8)Lyˆ

question

Answer 2

Answer:

The position vector of the centre of mass ⃗r

cm of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:

(A) ⃗r cm = (13/8)Lxˆ + (5/8)Lyˆ✅✅✅

(B) ⃗r cm = (5/8)Lxˆ + (13/8)Lyˆ

(C) ⃗r cm = (3/8)Lxˆ + (11/8)Lyˆ

(D) ⃗r cm = (11/8)Lxˆ + (3/8)Lyˆ