Question

Mass of $10\;kg$ is raised through a height of $7\;cm$ then up to what height $30\;kg$ should be shifted so that the center of mass rises by $1cm$.

Answer 1

The Centre of mass is defined as the point at which the entire mass of the system is equally distributed. In this question, two masses are given. We need to find out how much height we need to lift the second mass so that the center of mass rises by $1cm$. We have the formula for the center of mass. We need to substitute the values given in the question in the respective formula and we will arrive at the desired answer.

Complete answer:
The formula for the centre of mass is given as,
$COM = \dfrac{{{m_1}{y_1} + {m_2}{y_2}}}{{{m_1} + {m_2}}}$
Here, ${m_1}$and ${m_2}$are the two masses
${y_1}$and${y_2}$are the two distances.
Given that, the first mass ${m_1} = 10\;kg$and the second mass ${m_2} = 30kg$. Also, the mass $10\;kg$ is lifted $7\;cm$. Therefore, ${y_1} = 7cm$. Also, the center of mass should be raised $1cm$. Therefore, $COM = 1cm$.
$\therefore 1 = \dfrac{{10 \times 7 + 30{y_2}}}{{10 + 30}}$
$\Rightarrow 1 = \dfrac{{70 + 30{y_2}}}{{40}}$
$\Rightarrow 40 = 70 + 30{y_2}$
$\Rightarrow - 30 = 30{y_2}$
$\Rightarrow {y_2} = - 1cm$.
Therefore the second weight should be decreased by $1cm$.

Note:
When there is a single object then its center of mass is actually the point where all the mass of that object is concentrated. Also if there is a system of particles even then the point at which the entire mass of the system is concentrated is called the center of mass. Note that the center of mass is only applicable for rigid bodies and is not applicable for non-rigid bodies. For the symmetrical and regular-shaped objects, the center of mass will be present within the body. One of the most interesting things about the center of mass is that only at this point Newton's laws of motion are applicable perfectly.