The center of mass of a system of particles w | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the masses be $m_1 = 1 \, g, m_2 = 2 \, g, m_3 = 3 \, g$ and $m_4 = 4 \, g$.Since the center of mass of the first three particles is at the or

Vedclass · Accepted Answer

$5/2$

Vedclass · Answer

(B) Let the masses be $m_1 = 1 \, g, m_2 = 2 \, g, m_3 = 3 \, g$ and $m_4 = 4 \, g$.Since the center of mass of the first three particles is at the origin $(0, 0, 0)$,their combined position vector is $\vec{R}_{123} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3} = (0, 0, 0)$,which implies $m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3 = 0$.The position vector of the fourth particle is $\vec{r}_4 = \alpha(\hat{i} + 2\hat{j} + 3\hat{k})$.The center of mass of the new system is $\vec{R}_{cm} = (1, 2, 3)$.The formula for the center of mass is $\vec{R}_{cm} = \frac{(m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3) + m_4\vec{r}_4}{m_1 + m_2 + m_3 + m_4}$.Substituting the known values: $(1, 2, 3) = \frac{0 + 4\alpha(\hat{i} + 2\hat{j} + 3\hat{k})}{1 + 2 + 3 + 4} = \frac{4\alpha(\hat{i} + 2\hat{j} + 3\hat{k})}{10}$.Comparing the $x$-components: $1 = \frac{4\alpha}{10} \implies 10 = 4\alpha \implies \alpha = \frac{10}{4} = 2.5 = 5/2$.Similarly,for $y$-components: $2 = \frac{8\alpha}{10} \implies 20 = 8\alpha \implies \alpha = 2.5 = 5/2$.Thus,the value of $\alpha$ is $5/2$.

Similar Questions

$A$ slender uniform rod of length $L$ is balanced vertically at a point $P$ on a horizontal surface having some friction. If the top of the rod is displaced slightly to the right,the position of its centre of mass at the time when the rod becomes horizontal is:

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

$A$ uniform thin rod of length $L$ and mass $M$ is placed along the $x$-axis with one end at the origin. Find the position of the centre of mass of the rod.

If the linear mass density of a rod of length $L$ varies as $\lambda = kx^3$,determine the position of its centre of mass (where $x$ is the distance from one of its ends and $k$ is a constant):

$A$ thin bar of length $L$ has a mass per unit length $\lambda$ that increases linearly with distance $x$ from one end. If its total mass is $M$ and its mass per unit length at the lighter end $(x=0)$ is $\lambda_0$,then the distance of the centre of mass from the lighter end is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module