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Center of Mass question

2025 · 7 Apr · Shift 1 · Q54
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Center of Mass question

2025 · 7 Apr · Shift 1 · Q54

JEE MainPhysicsCenter of MassMCQ+4 / −1
A rod of length 5 L is bent right angle keeping one side length as 2 L . JEE Main 2025 (Online) 7th April Morning Shift Physics - Center of Mass and Collision Question 2 English The position of the centre of mass of the system : (Consider L=10 cm\mathrm{L}=10 \mathrm{~cm}L=10 cm)
  1. A
    4i^+9j^4 \hat{i}+9 \hat{j}4i^+9j^​
  2. B
    2i^+3j^2 \hat{i}+3 \hat{j}2i^+3j^​
  3. C
    5i^+8j^5 \hat{i}+8 \hat{j}5i^+8j^​
  4. D
    3i^+7j^3 \hat{i}+7 \hat{j}3i^+7j^​
View written solutionFree

Correct answer: A

  1. Interpret the figure/text

A uniform rod of total length 5L5L5L is bent at a right angle such that the two straight parts have lengths:

  • one side =2L=2L=2L
  • the other side =3L=3L=3L

Take the bend as the origin, with the 2L2L2L part along the xxx-axis and the 3L3L3L part along the yyy-axis.

So the system consists of two uniform rod segments:

  • Horizontal segment: length 2L2L2L
  • Vertical segment: length 3L3L3L

Since the rod is uniform, mass is proportional to length.

  1. Centre of mass of each segment

Segment 1: horizontal rod of length 2L2L2L

Its centre is at

(x1,y1)=(L,0)(x_1,y_1)=\left(L,0\right)(x1​,y1​)=(L,0)

Mass proportional to length:

m1∝2Lm_1 \propto 2Lm1​∝2L

Segment 2: vertical rod of length 3L3L3L

Its centre is at

(x2,y2)=(0,3L2)(x_2,y_2)=\left(0,\frac{3L}{2}\right)(x2​,y2​)=(0,23L​)

Mass proportional to length:

m2∝3Lm_2 \propto 3Lm2​∝3L
  1. Use centre of mass formula

Let linear mass density be λ\lambdaλ. Then

m1=2λL,m2=3λLm_1=2\lambda L, \qquad m_2=3\lambda Lm1​=2λL,m2​=3λL

Total mass:

M=5λLM=5\lambda LM=5λL

xxx-coordinate of COM

xcm=m1x1+m2x2M=(2λL)(L)+(3λL)(0)5λL=2L5x_{cm}=\frac{m_1x_1+m_2x_2}{M} =\frac{(2\lambda L)(L)+(3\lambda L)(0)}{5\lambda L} =\frac{2L}{5}xcm​=Mm1​x1​+m2​x2​​=5λL(2λL)(L)+(3λL)(0)​=52L​

Actually simplifying carefully,

xcm=2λL25λL=2L5x_{cm}=\frac{2\lambda L^2}{5\lambda L}=\frac{2L}{5}xcm​=5λL2λL2​=52L​

With L=10 cmL=10\text{ cm}L=10 cm,

xcm=2×105=4 cmx_{cm}=\frac{2\times 10}{5}=4\text{ cm}xcm​=52×10​=4 cm

yyy-coordinate of COM

ycm=m1y1+m2y2M=(2λL)(0)+(3λL)(3L2)5λLy_{cm}=\frac{m_1y_1+m_2y_2}{M} =\frac{(2\lambda L)(0)+(3\lambda L)\left(\frac{3L}{2}\right)}{5\lambda L}ycm​=Mm1​y1​+m2​y2​​=5λL(2λL)(0)+(3λL)(23L​)​ =92λL25λL=9L10=\frac{\frac{9}{2}\lambda L^2}{5\lambda L}=\frac{9L}{10}=5λL29​λL2​=109L​

With L=10 cmL=10\text{ cm}L=10 cm,

ycm=9×1010=9 cmy_{cm}=\frac{9\times 10}{10}=9\text{ cm}ycm​=109×10​=9 cm
  1. Final position vector

Therefore,

r⃗cm=4i^+9j^\vec r_{cm}=4\hat i+9\hat jrcm​=4i^+9j^​
  1. Check options
  • A: 4i^+9j^4\hat i+9\hat j4i^+9j^​ ✅
  • B: 2i^+3j^2\hat i+3\hat j2i^+3j^​ ❌
  • C: 5i^+8j^5\hat i+8\hat j5i^+8j^​ ❌
  • D: 3i^+7j^3\hat i+7\hat j3i^+7j^​ ❌

So the correct option is A.

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