SATHEE: Centre of Mass 1 Question 4

Centre of Mass 1 Question 4

4. The position vector of the centre of mass $r_{c m}$ of an asymmetric uniform bar of negligible area of cross-section as shown in figure is

(2019 Main, 12 Jan I)

(a) $r = \frac{13}{8} L \hat{x} + \frac{5}{8} L \hat{y}$

(b) $r = \frac{11}{8} L \hat{x} + \frac{3}{8} L \hat{y}$

(d) $r = \frac{5}{8} L \hat{x} + \frac{13}{8} L \hat{y}$

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Answer:

Correct Answer: 4. (a)

Solution:

Coordinates of centre of mass (COM) are given by

and

$\begin{aligned} X_{C O M} & = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} \\ Y_{C O M} & = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}} \end{aligned}$

For given system of rods, masses and coordinates of centre of rods are as shown.

So, $X_{C O M} = \frac{2 m L + m 2 L + m \frac{5 L}{2}}{4 m} = \frac{13}{8} L$

and $Y_{C O M} = \frac{2 m L + m \times \frac{L}{2} + m \times 0}{4 m} = \frac{5 L}{8}$

So, position vector of $C O M$ is

$\begin{aligned} r_{c o m} & = X_{C O M} \hat{x} + Y_{C O M} \hat{y} \\ = \frac{13}{8} L \hat{x} + \frac{5}{8} L \hat{y} \end{aligned}$

Centre of Mass 1 Question 4

4. The position vector of the centre of mass $r_{c m}$ of an asymmetric uniform bar of negligible area of cross-section as shown in figure is

Answer:

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Centre of Mass 1 Question 4

4. The position vector of the centre of mass rcm of an asymmetric uniform bar of negligible area of cross-section as shown in figure is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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4. The position vector of the centre of mass $r_{c m}$ of an asymmetric uniform bar of negligible area of cross-section as shown in figure is