Cavity Problems

By Jitender Singh on Feb 02, 2023

Problems on Centre of Mass

Problem (JEE Mains 2007): A circular disc of radius $r$ is removed from a bigger circular disc of radius $2r$ such that the circumferences of the discs coincide. The centre of mass of the new disc is $\alpha r$ from the centre of the bigger disc. The value of $\alpha$ is

Solution: Let $\sigma$ be the surface mass density. The mass of the complete disc is $m_0=4\pi r^2\sigma$, the mass of the removed disc is $m_1=\pi r^2\sigma$, and the mass of the remaining disc (shaded portion) is $m_2=m_0-m_1=3\pi r^2\sigma$.

The centre of the complete disc is $\mathrm{C_0}$ and the centre of the removed disc is $\mathrm{C_1}$. Let $\mathrm{C_0}$ be the origin and $x$-axis be towards $\mathrm{C_1}$. The centre of mass of the complete disc, removed disc, and remaining disc are at $\mathrm{C_0}$, $\mathrm{C_1}$, and $\mathrm{C_2}$. The coordinates of these points are $x_0=0$, $x_1=r$, and $x_2=-\alpha r$.

The complete disc comprises of "removed disc and remaining disc". Its centre of mass is the same as the centre of mass of the "removed disc and remaining disc" i.e., \begin{align} x_0&=\frac{m_1x_1+m_2x_2}{m_1+m_2} \\ &=\frac{(\pi r^2\sigma) (r)+(3\pi r^2\sigma) (-\alpha r)}{4\pi r^2\sigma} \nonumber\\ &=\frac{(1-3\alpha)r}{4}\\ &=0. \nonumber \end{align} Note that $x_0=0$. Thus, $\alpha=1/3$.

Problem (IIT JEE 1980): A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter 42 cm is removed from one edge of the plate as shown in the figure. Find the position of the centre of mass of the remaining portion.

Solution: The radius of the circular plate is $r_2=56/2=28$ cm and that of the removed circular portion is $r_1=42/2=21$ cm. Let $\mathrm{C_0}$ be the centre of mass of the original circular plate, $\mathrm{C_1}$ be the centre of mass of the removed circular portion, and $\mathrm{C_2}$ be the centre of mass of the remaining portion.

Let $\sigma$ be the mass per unit area of the plate. The mass of the circular plate is $m_0=\sigma (\pi r_2^2)$, mass of the removed portion is $m_1=\sigma (\pi r_1^2)$, and mass of the remaining portion is \begin{align} m_2&=m_0-m_1 \\ &=\sigma\pi(r_2^2-r_1^2). \end{align} By symmetry, $\mathrm{C_0}$ is at the geometrical centre of the circular plate, $\mathrm{C_1}$ is at the geometrical centre of the removed portion, and $\mathrm{C_2}$ lies on the line joining $\mathrm{C_0}$ and $\mathrm{C_1}$. Let origin of the coordinate system be at $\mathrm{C_0}$ and $x$ axis is towards $\mathrm{C_1}$. The $x$ coordinate of $\mathrm{C_0}$ is $x_0=0$, that of $\mathrm{C_1}$ is $x_1=r_2-r_1=28-21=7$ cm, and that of $\mathrm{C_2}$ is $x_2=-a$. The centre of mass of the circular plate is given by \begin{align} x_0&=\frac{m_1x_1+m_2x_2}{m_1+m_2} \nonumber\\ &=\frac{\sigma\pi r_1^2(r_2-r_1)+\sigma\pi(r_2^2-r_1^2)(-a)}{\sigma\pi r_2^2}\nonumber\\ &=\frac{r_1^2(r_2-r_1)-a(r_2^2-r_1^2)}{r_2^2}=0.\tag*{($\because\, x_0=0$).} \nonumber \end{align} Solve to get, \begin{align} a&=r_1^2/(r_2+r_1) \\ &=(21)^2/(28+21) \\ &=9,\mathrm{cm}. \end{align}

Problem (JEE Mains 2020): As shown in the figure, when a spherical cavity (centred at O) of radius 1 is cut out of a uniform sphere of radius $r$ (centred at C), the centre of mass of the remaining (shaded) part of the sphere is at G, i.e., on the surface of the cavity. The radius $r$ can be determined by the equation,

$(r^2+r+1)(2-r)=1$
$(r^2+r-1)(2-r)=1$
$(r^2-r-1)(2-r)=1$
$(r^2-r+1)(2-r)=1$

Solution: Let $\rho$ be the mass density. The mass of the complete sphere is $m_0=\frac{4}{3}\pi r^3\rho$, the mass of the removed sphere is $m_1=\frac{4}{3}\pi\rho$, and the mass of the remaining sphere is \begin{align} m_2&=m_0-m_1\\ &=\frac{4}{3}\pi\rho(r^3-1). \end{align}

Let the origin be at C, and the $x$-axis is towards O. The centre of mass of the complete sphere, removed sphere, and remaining sphere are at C, O, and G. The coordinates of these points are $x_0=0$, $x_1=r-1$, and $x_2=-(2-r)$.

The complete sphere comprises of "removed sphere and remaining sphere". Its centre of mass is the same as the centre of mass of the "removed sphere and remaining sphere" i.e., \begin{align} x_0&=\frac{m_1x_1+m_2x_2}{m_1+m_2}\nonumber\\ &=\frac{\tfrac{4}{3}\pi \rho (r-1)+\tfrac{4}{3}\pi \rho (r^3-1) (r-2)}{\tfrac{4}{3}\pi r^3\rho} \nonumber\\ &=\frac{(r-1)\left(1-(r^2+r+1)(2-r)\right)}{r^3}\nonumber\\ &=0. \nonumber \end{align} (Because $x_0=0$). Thus, $(r^2+r+1)(2-r)=1$.

Aliter: A trick for "cavity problems" is to take the mass density of the cavity as $-\rho$ in the centre of mass formula. The remaining portion comprises of (i) a sphere of radius $r$ and density $\rho$ centred at C and (ii) another sphere of radius $1$ and density $-\rho$ centred at O. The centre of mass of the remaining portion is obtained by, \begin{align} x_2&=\frac{\frac{4}{3}\pi r^3\rho (0)+\frac{4}{3}\pi (-\rho)(r-1)}{\frac{4}{3}\pi r^3\rho+\frac{4}{3}\pi (-\rho)} \\ &=-(2-r).\nonumber \end{align}

Problems on Moment of Inertia

Problem (IIT JEE 2005): From a circular disc of radius $R$ and mass $9M$, a small disc of radius $R/3$ is removed. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through $O$ is,

Moment of Inertia Problem from IIT JEE 2005

$4MR^2$
$\frac{40}{9}MR^2$
$10MR^2$
$\frac{37}{9}MR^2$

Solution: The moment of inertia of disc of mass $9M$ and radius $R$ about an axis perpendicular to its plane and passing through its centre O is, \begin{align} I_\text{total}&=\frac{1}{2}(9M)R^2 \\ &=\frac{9}{2}MR^2.\nonumber \end{align} The mass of removed disc is $\frac{9M}{\pi R^2}\frac{\pi R^2}{9}=M$. The parallel axis theorem gives moment of inertia of the removed disc about axis passing through $O$ as, \begin{align} I_\text{removed}&=\frac{1}{2}M\left(\tfrac{R}{3}\right)^{\!2}+Md^2 \\ &=\tfrac{1}{18}MR^2+M\left(\tfrac{2R}{3}\right)^{\!2} \\ &=\tfrac{1}{2}MR^2.\nonumber \end{align} Using, $I_\text{total}=I_\text{remaining}+I_\text{removed}$, we get $I_\text{remaining}=4MR^2$.

Problem (IIT JEE 2012) A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is $I_O$ and $I_P$, respectively. Both these axes are perpendicular to the plane of lamina. The ratio $\frac{I_P}{I_O}$ to the nearest integer is__________

Solution: Let $\sigma$ be the mass per unit area of lamina and the point Q be the centre of the cavity.

Let subscript t denotes the total disc (without cavity) and subscript c denotes only cavity. The moment of inertia $I_\text{O}$ is related to $I_\text{O,t}$ and $I_\text{O,c}$ by \begin{align} I_\text{O,t}=I_\text{O}+I_\text{O,c}. \end{align} The moment of inertia of a disc of mass $m$ and radius $r$ about its symmetry axis is given by $\frac{1}{2}mr^2$. Apply the theorem of parallel axes to the cavity to get \begin{align} \label{fca:eqn:1} I_\text{O,c}&=I_\text{Q,c}+m_c R^2 \nonumber\\ &=\frac{1}{2}m_c R^2+m_c R^2 \\ &=\frac{3}{2}\pi\sigma R^4. \end{align} The moment of inertia of the total disc is \begin{align} \label{fca:eqn:2} I_\text{O,t}=\frac{1}{2}m_t\,(2R)^2=8\pi\sigma R^4. \end{align} Using above equations, we get \begin{align} I_\text{O}&=I_\text{O,t}-I_\text{O,c} \\ &=\tfrac{21}{2}\pi\sigma R^4. \nonumber \end{align} Similarly, moments of inertia about an axis passing through the point P are \begin{align} I_\text{P,c}&=I_\text{Q,c}+m_c(\sqrt{5}R)^2\\ &=\frac{11}{2}\pi\sigma R^4, \nonumber\\ I_\text{P,t}&=I_\text{O,t}+m_t (2R)^2\\ &=24\pi\sigma R^4. \nonumber \end{align} Thus, \begin{align} I_\text{P}=I_\text{P,t}-I_\text{P,c}=\frac{37}{2}\pi\sigma R^4. \end{align} The ratio is \begin{align} \frac{I_\text{P}}{I_\text{O}}=\frac{37}{13}=2.84\approx 3. \end{align}

Cavity Problems

Problems on Centre of Mass

Problems on Moment of Inertia

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