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Calculation of Center of Mass

Practice Questions:

Question 1: easy

Two objects of mass $10\text{ kg}$ and $20\text{ kg}$ respectively are connected to the two ends of a rigid rod of length $10\text{ m}$ with negligible mass. The distance of the centre of mass of the system from the $10\text{ kg}$ mass is :

(2022)

1. $5\text{ m}$

2. $\frac{10}{3}\text{ m}$

3. $\frac{20}{3}\text{ m}$

4. $10\text{ m}$

View Answer

Centre of mass formula is $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$. Taking $10\text{ kg}$ at origin and $20\text{ kg}$ at $10\text{ m}$, we get $x_{cm} = \frac{10(0) + 20(10)}{10+20} = \frac{20}{3}\text{ m}$. Option (c) is correct.

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Question 2: easy

Two particles of mass $5\text{ kg}$ and $10\text{ kg}$ respectively are attached to the two ends of a rigid rod of length $1\text{ m}$ with negligible mass. The centre of mass of the system from the $5\text{ kg}$ particle is nearly at a distance of :

(2020)

1. $50\text{ cm}$

2. $67\text{ cm}$

3. $80\text{ cm}$

4. $33\text{ cm}$

View Answer

Use the centre of mass formula $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$. Substituting $m_1 = 5\text{ kg}$, $x_1 = 0$, $m_2 = 10\text{ kg}$, $x_2 = 100\text{ cm}$, we get $x_{cm} = \frac{10 \times 100}{15} = 66.67\text{ cm} \approx 67\text{ cm}$. Option (b) is correct.

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Question 3: easy

Which of the following statements are correct?

A. Centre of mass of a body always coincides with the centre of gravity of the body

B. Centre of gravity of a body is the point at which the total gravitational torque on the body is zero

C. A couple on a body produce both translational and rotational motion in a body

D. Mechanical advantage greater than one means that small effort can be used to lift a large load

(2017-Delhi)

1. A and B

2. B and C

3. C and D

4. B and D

View Answer

Centre of gravity is the point where total gravitational torque is zero (Statement B is correct). Mechanical advantage greater than one implies a small effort lifts a large load (Statement D is correct). Thus, statements B and D are correct, making option (d) the right choice.

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Question 4: moderate

Two spherical bodies of mass $M$ and $5M$ and radii $R$ and $2R$ are released in free space with initial separation between their centres equal to $12R$. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is :

(2015)

1. $4.5R$

2. $7.5R$

3. $1.5R$

4. $2.5R$

View Answer

The centre of mass remains stationary. Initial distance of COM from mass $M$ is $\frac{5M \times 12R}{M + 5M} = 10R$. At collision, distance between centers is $3R$, and the distance of $M$ from COM is $\frac{5}{6} \times 3R = 2.5R$. Thus, distance covered by $M$ is $10R - 2.5R = 7.5R$. Option (b) is correct.

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Question 5: moderate

Three masses are placed on the $x$-axis : $300\text{ g}$ at origin, $500\text{ g}$ at $x = 40\text{ cm}$ and $400\text{ g}$ at $x = 70\text{ cm}$. The distance of the center of mass from the origin is :

(2012 Mains)

1. $40\text{ cm}$

2. $45\text{ cm}$

3. $50\text{ cm}$

4. $30\text{ cm}$

View Answer

Use the centre of mass formula $x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$. Substituting the given values: $x_{cm} = \frac{300(0) + 500(40) + 400(70)}{300 + 500 + 400} = \frac{48000}{1200} = 40\text{ cm}$. Option (a) is correct.

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Question 6: easy

Two bodies of mass $1text{ kg}$ and $3text{ kg}$ have position vectors $\hat{i} + 2\hat{j} + \hat{k}$ and $-3\hat{i} – 2\hat{j} + \hat{k}$, respectively. The center of mass of this system has a position vector:

(2009)

1. $2\hat{i} - \hat{j} + \hat{k}$

2. $-2\hat{i} - \hat{j} + \hat{k}$

3. $-\hat{i} + \hat{j} + \hat{k}$

4. $-2\hat{i} + 2\hat{k}$

View Answer

Concept: Center of mass position vector formula. Formula: $\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$. Solution: Substituting the given masses and position vectors yields $-2\hat{i} - \hat{j} + \hat{k}$.

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