The centre of mass of a system of particles depends on
The position of the centre of mass of a system of particles is defined as \( \vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i} \). It clearly depends on individual masses, their coordinates (positions), and consequently the relative distances between them.
Two particles \(A\) and \(B\) initially at rest, move towards each other under mutual force of attraction. At an instance when the speed of \(A\) is \(v\) and speed of \(B\) is \(3v\), the speed of centre of mass is
Since no external force acts on the two-particle system, the acceleration of the centre of mass is zero. Since the system started from rest, the speed of the centre of mass remains zero.
Two particles of masses \(2\text{ kg}\) and \(6\text{ kg}\) located at the point \((1\text{ m}, 1\text{ m}, 1\text{ m})\) and \((2\text{ m}, 2\text{ m}, 1\text{ m})\) respectively. The distance of centre of mass from \(2\text{ kg}\) mass will be
Distance between masses is \(d = \sqrt{(2-1)^2+(2-1)^2+0^2} = \sqrt{2}\text{ m}\)
Distance of center of mass from \(m_1\) is \(r_1 = \frac{m_2 d}{m_1+m_2} = \frac{6\sqrt{2}}{8} = \frac{3\sqrt{2}}{4}\text{ m}\).
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)
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.
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)
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.
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)
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.
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)
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.
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)
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.
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)
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 $-2\hat{i} - \hat{j} + \hat{k}$.