Concepts of Physics

IIT JEE Physics (1978 to 2018: 41 Years) Topic-wise Complete Solutions

Centre of Mass and its Motion

Worked out Problems

Problem: A composite body is formed by joining a solid cylinder and a solid cone of the same radius. The length of the cylinder is $b$ and height of the cone is $h$. If the centre of mass of the composite body is located in the plane between the solid cone and the solid cylinder then the ratio $b/h$ is

(A) $1/\sqrt{6}$ (B) $1/2$ (C) $1/\sqrt{5}$ (D) $1/3$

Solution: Let the origin be located at the plane between the solid cone and the solid cylinder. Let $x$ axis is along the symmetric axis towards the right.The centre of mass of the solid cylinder is at a distance $x_1=-b/2$. The centre of mass of the solid cone is at a distance $x_2=h/4 $ (at distance h/4 from the base).

Centre of Mass

Let $\rho$ be the mass density and $r$ be the radius of the cylinder and the cone. The mass of the cylinder is $m_1=\rho (\pi r^2 b)$ and mass of the cone is $m_2=\rho (\frac{1}{3} \pi r^2 h)$. For the centre of mass of the composite body to be at origin, \begin{align} x_{cm}=\frac{m_1x1+m_2 x_2}{m_1+m_2}=\frac{\rho (\pi r^2 b) (-b/2) + \rho (\frac{1}{3} \pi r^2 h) (h/4) }{\rho (\pi r^2 b)+ \rho (\frac{1}{3} \pi r^2 h)}=0. \nonumber \end{align} Simplify to get $b/h=1/\sqrt{6}$.

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Problems from IIT JEE

Problem (IIT JEE 2009): Centre of Mass Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is $m$. The mass of the ink used to draw the outer circle is $6m$. The coordinates of the centres of the different parts are: outer circle $(0,0)$, left inner circle $(-a,a)$, right inner circle $(a,a)$, vertical line $(0,0)$ and horizontal line $(0,-a)$. The $y$-coordinate of the centre of mass of the ink in this drawing is,

  1. $a/10$
  2. $a/8$
  3. $a/12$
  4. $a/3$

Solution: Due to uniform line thickness, the centre of mass of circle lies on its centre and that of a line segment lies in its middle. The $y$ coordinate of the centre of mass of the ink given in the drawing is given by, \begin{align} y_\text{cm}=\frac{\sum m_iy_i}{\sum m_i}=\frac{6m(0)+m(a)+m(a)+m(0)+m(-a)}{6m+m+m+m+m}=\frac{a}{10}.\nonumber \end{align}

Problem (IIT JEE 2003): Two point masses $m_1$ and $m_2$ are connected by a spring of spring constant $k$ and natural length $l_0$. The spring is compressed such that the two point masses touch each other and then are fastened by a string. Then the system is moved with a velocity $v_0$ along positive $x$-axis. When the system reaches the origin the string breaks $(t=0)$. The position of the point mass $m_1$ is given by $x_1=v_0 t-A(1-\cos\omega t)$ where $A$ and $\omega$ are constants. Find the position of the second block as a function of time. Also find the relation between $A$ and $l_0$.

Solution: Consider $m_1$, $m_2$, and the spring together as a system. At $t=0$, the position and velocity of the centre of mass of the system are, \begin{align} \label{pqb:eqn:1} &x_\text{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}=0,\\ \label{pqb:eqn:2} &v_\text{cm}=\frac{\mathrm{d}x_\text{cm}}{\mathrm{d}t}=\frac{m_1v_1+m_2v_2}{m_1+m_2}=v_0. \end{align} There is no external force on the system. Hence, $v_\text{cm}$ remains constant at $v_0$. Integrate above equation with the initial condition $x_\text{cm}=0$ at $t=0$ to get, \begin{align} \label{pqb:eqn:3} x_\text{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}=v_0 t. \end{align} Given, the displacement of $m_1$ at time $t$ is, \begin{align} \label{pqb:eqn:4} x_1=v_0t-A(1-\cos\omega t). \end{align} Substitute $x_1$ from fourth equation into third equation to get the displacement of $m_2$ as, \begin{align} \label{pqb:eqn:5} x_2=v_0t+\frac{m_1}{m_2}A(1-\cos\omega t). \end{align} The forces on $m_1$ and $m_2$ become zero whenever the distance between the two particles is equal to the natural length of the spring i.e., \begin{align} \label{pqb:eqn:6} x_2-x_1=A(m_1/m_2+1)(1-\cos\omega t)=l_0. \end{align} Newton's second law gives force on $m_1$ as, \begin{align} m_1\frac{\mathrm{d}^2x_1}{\mathrm{d}t^2}=-A\omega^2\cos\omega t, \end{align} which becomes zero whenever $\cos\omega t=0$. Substitute $\cos\omega t=0$ in above equation to get $l_0=\left(m_1/m_2+1\right)A$.