Rigid Body Rotation
Rigid Body
The body whose constituent particles remain at their respective position during rotational and translational motion is called a rigid body.
In a rigid body, particles are compactly arranged and the inter-particle distance is fixed.
Moment of Inertia
The moment of inertia of a rigid body rotating about an axis is the sum of the product of mass and square of distance of each mass from the axis of rotation.
$$I = m_1r_1^2+m_2r_2^2+m_3r_3^2+\cdots+m_nr_n^2$$ $$I = \sum_{i=1}^{n} m_ir_i^2$$Radius of Gyration
The perpendicular distance between the centre of mass and the axis of rotation is called the radius of gyration.
The product of mass of the rigid body and the square of the radius of gyration gives the moment of inertia.
$$\therefore I = MK^2$$or,
$$\sum_{i=1}^n m_ir_i^2 = MK^2$$or,
$$K^2 = \frac{\sum_{i=1}^{n} m_ir_i^2}{M}$$or,
$$K^2 = \frac{m_1r_1^2+m_2r_2^2+m_3r_3^2+\cdots+m_nr_n^2}{m_1+m_2+m_3+\cdots+m_n}$$If $m_1=m_2=m_3=\cdots=m$, then,
$$K^2 = \frac{m(r_1^2+r_2^2+r_3^2+\cdots+r_n^2)}{n\cdot m}$$or,
$$K^2 = \frac{r_1^2+r_2^2+r_3^2+\cdots+r_n^2}{n}$$ $$\therefore K = \sqrt{\frac{r_1^2+r_2^2+r_3^2+\cdots+r_n^2}{n}}$$Dimensionally, $K$ = average distance.
Theorem of Perpendicular Axis
The moment of inertia of a plane sheet (lamina) about an axis perpendicular to the plane of lamina is equal to the sum of the moments of inertia of the lamina about two mutually perpendicular axes intersecting each other at a point where the perpendicular axis passes through it.
Let us consider a plane lamina. We have to calculate the moment of inertia about an axis passing through $O$ and perpendicular to the plane of the lamina.
Let the reference particle lie at point $P$ which is at distance $r$ from the axis of rotation. Using geometry,
$$r^2 = x^2+y^2 \qquad \cdots (1)$$From the definition of moment of inertia,
$$I = \sum mr^2$$or,
$$I = \sum m(x^2+y^2)$$or,
$$I = \sum mx^2+\sum my^2$$ $$\boxed{I = I_x+I_y}$$Theorem of Parallel Axis
The moment of inertia of a plane lamina about any axis is equal to the moment of inertia about a parallel axis through the centre of mass plus the product of the mass of the body and the square of the distance between the two axes.
The moment of inertia of a reference point about axis $XY$ is given by:
$$I = \sum m(R+x)^2$$or,
$$I = \sum m(R^2+2Rx+x^2)$$or,
$$I = \sum mR^2+\sum 2mxR+\sum mx^2$$or,
$$I = \sum mR^2+2R\sum mx+\sum mx^2$$or,
$$I = R^2\sum m+0+I_{cm} \qquad (\because \sum mx=0)$$or,
$$I = MR^2+I_{cm}$$ $$\therefore \boxed{I = I_{cm}+MR^2}$$The sum of moments of all the particles of the lamina about an axis through the centre of mass is $0$.
Calculation of Moment of Inertia
1. M.I. of a Thin Uniform Rod
(a) About an axis passing through its centre and perpendicular to its length
[Diagram: Thin Uniform Rod]
Let us consider a uniform rod $AB$ of length $l$. Let $M$ = mass of rod, $XY$ = axis passing through the centre and perpendicular to length, $dm$ = mass of elemental length, $dx$ = length of element, $x$ = distance of elemental mass from the axis of rotation.
Since the rod is uniform, its mass per unit length is $\dfrac{M}{l}$. The mass of the element is:
$$dm = \frac{M}{l}dx$$Moment of inertia about an axis $XY$ is:
$$dI = dm\,x^2$$or,
$$dI = \frac{M}{l}x^2dx$$The moment of inertia $(I)$ of the whole rod about the axis $XY$ is thus given by the integral of the above expression between the limits $x=-l/2$ and $x=+l/2$, or by twice its integral between the limits $x=0$ and $x=l/2$, i.e.,
$$I = \int dI = \int_{-l/2}^{l/2} \frac{M}{l}x^2dx = \frac{M}{l}\cdot 2\int_0^{l/2}x^2dx$$ $$= \frac{2M}{l}\left[\frac{x^3}{3}\right]_0^{l/2} = \frac{2M}{3l}\left(\frac{l}{2}\right)^3$$ $$\therefore \boxed{I = \frac{Ml^2}{12}}$$(b) About an axis passing through its one end and perpendicular to its length
Let us consider a uniform rod $AB$ of length $l$. Let $M$ = mass of rod, $XY$ = axis passing through one end of the rod and perpendicular to its length, $x$ = distance of elemental mass from axis of rotation, $dm$ = mass of elemental length, $dx$ = length of element.
Since the rod is uniform, its mass per unit length is $\dfrac{M}{l}$. The mass of the element is:
$$dm=\frac{M}{l}dx$$Moment of inertia about an axis $XY$ is:
$$dI = dm\,x^2 \qquad \text{or,}\qquad dI = \frac{M}{l}x^2dx$$Now, the moment of inertia of the whole rod about the axis $XY$ is obtained by integrating between the limits $x=0$ and $x=l$.
$$I = \int_0^l dm = \int_0^l \frac{M}{l}x^2dx = \frac{M}{l}\int_0^l x^2dx = \frac{M}{l}\left[\frac{x^3}{3}\right]_0^l = \frac{M}{3l}(l)^3$$ $$\therefore \boxed{I = \frac{Ml^2}{3}}$$Alternatively, using theorem of parallel axis,
$$I = MR^2+I_{cm} = \frac{Ml^2}{4}+\frac{Ml^2}{12} = \frac{Ml^2}{3}$$2. Moment of Inertia of a Rectangular Lamina (or Bar)
(a) About an axis through its centre and parallel to end and perpendicular to length
Let $ABCD$ = rectangular lamina, $M$ = mass of the lamina, $l$ = length of the lamina, $b$ = breadth of the lamina, $XY$ = axis through its centre and parallel to side $AD$ and $BC$, $dx$ = length of element, $dm$ = mass of elemental length, $x$ = distance of elemental mass from axis of rotation.
The area of the strip (element) $= dx\times b$. And since the mass per unit area of the lamina $=\dfrac{M}{l\times b}$, we have,
$$\text{mass of the element} = \frac{M}{l\times b}\times dx\times b$$or,
$$dm = \frac{M}{l}dx$$Moment of inertia of $dm$ about an axis $XY$ is:
$$dI = dm\,x^2 \qquad \text{or,}\qquad dI = \frac{M}{l}x^2dx$$Now, the M.I. of the whole rectangular lamina is obtained by integrating between the limits $x=-l/2$ and $x=l/2$.
$$I = \int_{-l/2}^{l/2}dI = \int_{-l/2}^{l/2}\frac{M}{l}x^2dx = \frac{2M}{l}\int_0^{l/2}x^2dx = \frac{2M}{l}\left[\frac{x^3}{3}\right]_0^{l/2}=\frac{2M}{3l}\left(\frac{l}{2}\right)^3$$ $$\therefore \boxed{I = \frac{Ml^2}{12}}$$(b) About an axis passing through its one end and parallel to side $AD$ and $BC$
Let $ABCD$ = rectangular lamina, $M$ = mass of the lamina, $l$ = length of the lamina, $b$ = breadth of the lamina, $XY$ = axis passing through its one end and parallel to side $AD$ and $BC$, $dx$ = length of element, $dm$ = mass of elemental length, $x$ = distance of elemental mass from axis of rotation.
The area of the element $=dx\times b$. And since the mass per unit area of the lamina $=\dfrac{M}{l\times b}$, we have,
$$\text{mass of the element}=\frac{M}{l\times b}\times dx\times b$$or,
$$dm=\frac{M}{l}dx$$Moment of inertia of $dm$ about $XY$ is:
$$dI = dm\,x^2 \qquad \text{or,}\qquad dI = \frac{M}{l}x^2dx$$$\therefore$ Total M.I. of the lamina is obtained by integrating between the limits $x=0$ and $x=l$.
$$I = \int_0^l dI = \int_0^l \frac{M}{l}x^2dx = \frac{M}{l}\int_0^l x^2dx = \frac{M}{l}\left[\frac{x^3}{3}\right]_0^l = \frac{M}{3l}l^3$$ $$\therefore \boxed{I = \frac{Ml^2}{3}}$$(c) About an axis passing through its centre and perpendicular to the plane of lamina
From the theorem of perpendicular axes, the M.I. of the lamina about an axis through $O$ and perpendicular to its plane = M.I. of the lamina about an axis through $O$ parallel to $b$ + M.I. of the lamina about an axis through $O$ parallel to $l$, i.e.,
$$I_z = I_x+I_y = \frac{Mb^2}{12}+\frac{Ml^2}{12}$$ $$\therefore \boxed{I = \frac{M(l^2+b^2)}{12}}$$(d) About an axis passing through the midpoint of one side and perpendicular to the plane of lamina
When the axis passes through the midpoint of $AD$ or $BC$, then from the theorem of parallel axis,
$$I = I_{cm}+MR^2$$ $$= \frac{M(l^2+b^2)}{12}+M\left(\frac{l}{2}\right)^2$$ $$= \frac{Ml^2}{12}+\frac{Mb^2}{12}+\frac{Ml^2}{4}$$ $$= \frac{Ml^2}{3}+\frac{Mb^2}{12}$$ $$\therefore \boxed{I = M\left(\frac{l^2}{3}+\frac{b^2}{12}\right)}$$When the axis passes through the midpoint of $AB$ or $CD$, then from the theorem of parallel axis,
$$I = I_{cm}+MR^2 = \frac{M(l^2+b^2)}{12}+M\left(\frac{b}{2}\right)^2$$ $$= \frac{Ml^2}{12}+\frac{Mb^2}{12}+\frac{Mb^2}{4} = \frac{Ml^2}{12}+\frac{Mb^2}{3}$$ $$\therefore \boxed{I = M\left(\frac{l^2}{12}+\frac{b^2}{3}\right)}$$(e) About an axis passing through one corner and perpendicular to the plane of lamina
To find the moment of inertia about an axis passing through corner $D$ and perpendicular to the plane of lamina, from the theorem of parallel axis:
$$I = I_{cm}+Mr^2$$ $$= \frac{M(l^2+b^2)}{12}+M\left\{\left(\frac{l}{2}\right)^2+\left(\frac{b}{2}\right)^2\right\}$$ $$= \frac{Ml^2}{12}+\frac{Mb^2}{12}+\frac{Ml^2}{4}+\frac{Mb^2}{4}$$ $$= \frac{4Ml^2}{12}+\frac{4Mb^2}{12}$$ $$\therefore \boxed{I = \frac{M(l^2+b^2)}{3}}$$3. Moment of Inertia of a Thin Circular Ring
(a) About an axis through the centre and perpendicular to the plane
Let us consider a ring. Let $M$ = total mass of ring, $R$ = radius of ring, $XY$ = axis through the centre and perpendicular to the plane of ring, $dx$ = length of element, $dm$ = mass of elemental length $=\dfrac{M}{2\pi R}dx$.
M.I. of $dm$ about axis $XY$ is:
$$dI = dm\,R^2 = \frac{M}{2\pi R}dx\cdot R^2 = \frac{M}{2\pi}R\,dx$$Total M.I. of the whole ring about $XY$ is:
$$I = \int_0^{2\pi R}dI = \int_0^{2\pi R}\frac{MR}{2\pi}dx = \frac{MR}{2\pi}\int_0^{2\pi R}dx = \frac{MR}{2\pi}\Big[x\Big]_0^{2\pi R}$$ $$= \frac{MR}{2\pi}\times 2\pi R$$ $$\boxed{I=MR^2}$$(b) About an axis through its diameter
Let $I_x$ = moment of inertia along the $x$-axis, $I_y$ = M.I. along the $y$-axis. Then, from the theorem of perpendicular axis,
$$I_z = I_x+I_y$$or,
$$I_z = I+I$$or,
$$I_z = 2I$$or,
$$I = \frac{I_z}{2} = \frac{MR^2}{2}$$ $$\therefore \boxed{I = \frac{MR^2}{2}}$$4. Moment of Inertia of a Circular Lamina or Disc
(a) About an axis through its centre and perpendicular to the plane
Let $M$ = mass of disc or circular lamina, $R$ = radius of disc. $x$ = radius of elemental ring, $dx$ = thickness of elemental ring, $dm$ = mass of elemental ring.
$$\therefore dm = \frac{M}{\pi R^2}\cdot 2\pi x\,dx = \frac{2M}{R^2}x\,dx$$M.I. of the elemental ring about an axis through the centre and perpendicular to the plane is:
$$dI = dm\,x^2 = \frac{2M}{R^2}x\,dx\cdot x^2 = \frac{2M}{R^2}x^3dx$$Total M.I. of the disc about an axis through the centre and perpendicular to its plane is:
$$I = \int_0^R dI = \int_0^R \frac{2M}{R^2}x^3dx = \frac{2M}{R^2}\int_0^R x^3dx = \frac{2M}{R^2}\left[\frac{x^4}{4}\right]_0^R = \frac{M}{2R^2}\cdot R^4$$ $$\boxed{I = \frac{1}{2}MR^2}$$(b) About an axis through its diameter
From the theorem of perpendicular axes,
$$I_z=I_x+I_y \;\Rightarrow\; I_z=I+I \;\Rightarrow\; I_z=2I$$ $$I = \frac{I_z}{2} = \frac{MR^2/2}{2}$$ $$\therefore \boxed{I = \frac{1}{4}MR^2}$$5. M.I. of Annular Ring or Disc
(a) About an axis through the centre and perpendicular to the plane
Let $R$ = external radius, $r$ = internal radius, $M$ = mass. Then, mass per unit area $=\dfrac{M}{\pi R^2-\pi r^2}$.
$\therefore$ Mass of elemental ring $(dm)=\dfrac{M}{\pi(R^2-r^2)}\cdot 2\pi x\,dx = \dfrac{2M}{(R^2-r^2)}x\,dx$
$\therefore$ M.I. of ring about axis through the centre and perpendicular to the plane is:
$$dI = dm\,x^2 = \frac{2M}{(R^2-r^2)}x\,dx\cdot x^2 = \frac{2M}{(R^2-r^2)}x^3dx$$$\therefore$ Total M.I. of annular ring or disc:
$$I = \int_r^R dI = \int_r^R \frac{2M}{(R^2-r^2)}x^3dx = \frac{2M}{(R^2-r^2)}\int_r^R x^3dx$$ $$= \frac{2M}{(R^2-r^2)}\left[\frac{x^4}{4}\right]_r^R = \frac{2M}{2\cdot(R^2-r^2)}(R^4-r^4)$$ $$= \frac{M(R^2+r^2)}{2}$$ $$\therefore \boxed{I = \frac{1}{2}M(R^2+r^2)}$$(b) About an axis through diameter
From the theorem of perpendicular axes,
$$I_z = I_x+I_y \;\Rightarrow\; I_z=I+I \;\Rightarrow\; I_z=2I$$or,
$$I = \frac{I_z}{2} = \frac{M(R^2+r^2)}{2\cdot 2}$$ $$\therefore \boxed{I = \frac{1}{4}M(R^2+r^2)}$$6. M.I. of Solid Cylinder
(a) About an axis through the axis of cylindrical symmetry
M.I. of a solid cylinder about an axis through the axis of cylindrical symmetry is equal to the M.I. of a thick disc of the same mass and radius about an axis through its centre and perpendicular to the plane.
$$\therefore \boxed{I = \frac{1}{2}MR^2}$$(b) About an axis through its centre and perpendicular to the axis of cylindrical symmetry
Let $M$ = mass of the cylinder, $R$ = radius of the cylinder, $l$ = length of the cylinder. Let us consider an elemental disc of thickness $dx$ at distance $x$ from the axis of rotation.
$\therefore$ mass of elemental disc $=\dfrac{M}{l}dx$
Now, M.I. of the elemental disc about its diameter $AB = \left(\dfrac{M}{l}\right)dx\cdot\dfrac{R^2}{4}$
Now, M.I. of the disc about a parallel axis passing through the centre and perpendicular to the axis of the cylinder is:
$$dI = I_{cm}+M x^2 = \left(\frac{M}{l}\right)dx\cdot\frac{R^2}{4}+\left(\frac{M}{l}\right)x^2dx$$$\therefore$ Total M.I. $(I) = \displaystyle\int_{-l/2}^{l/2}dI = 2\int_0^{l/2}dI$
$$= 2\int_0^{l/2}\left(\frac{M}{l}\,dx\cdot\frac{R^2}{4}+\frac{M}{l}x^2\,dx\right)$$ $$= \frac{2M}{l}\left[\frac{R^2}{4}\int_0^{l/2}dx+\int_0^{l/2}x^2dx\right]$$ $$= \frac{2M}{l}\left[\frac{R^2}{4}\Big[x\Big]_0^{l/2}+\left[\frac{x^3}{3}\right]_0^{l/2}\right]$$ $$= \frac{2M}{l}\left[\frac{R^2}{4}\cdot\frac{l}{2}+\frac{l^3}{3\cdot 8}\right]$$ $$\boxed{I = \frac{MR^2}{4}+\frac{Ml^2}{12}}$$7. M.I. of Spherical Shell
(a) About diameter
Let $M$ = mass of spherical shell, $R$ = radius of spherical shell, $AB$ = axis about which M.I. should be determined.
Let us consider a slice $EFHG$ at distance $x$ from the centre $O$. Let us consider $\angle COE=\theta$, $\angle EOG=d\theta$.
Mass per unit area $=\dfrac{M}{4\pi R^2}$
Mass of the slice (ring) $=\dfrac{M}{4\pi R^2}\cdot 2\pi(EP)(EG)$
Here, $OP=R\sin\theta$, and $x=R\sin\theta \;\Rightarrow\; \dfrac{dx}{d\theta}=R\cos\theta$
$$\text{Mass of the slice} = \frac{M}{4\pi R^2}\cdot 2\pi(EP)(EG) = \frac{M}{2R^2}(EP)(EG)$$ $$= \frac{M}{2R^2}\cdot R\cos\theta\cdot R\,d\theta$$$\therefore$ Mass of the slice $=\dfrac{M}{2R^2}\cdot\dfrac{dx}{d\theta}\cdot R\,d\theta = \dfrac{M}{2R}dx$
$\therefore$ Moment of inertia of the ring about axis $AB$ is:
$$dI = \text{mass of ring}\times(\text{radius of ring})^2 = \frac{M}{2R}dx\times(EP)^2 = \frac{M}{2R}dx\,(R^2-x^2)$$$\therefore$ Total moment of inertia of the spherical shell about axis $AB$ is:
$$I = \int_{-R}^{R}dI = 2\int_0^R dI = 2\int_0^R \frac{M}{2R}(R^2-x^2)dx$$ $$= \frac{M}{R}\left[\int_0^R R^2dx-\int_0^R x^2dx\right] = \frac{M}{R}\left[R^2\times R-\left[\frac{x^3}{3}\right]_0^R\right]$$ $$= \frac{M}{R}\left(R^3-\frac{1}{3}R^3\right) = \frac{M}{R}\cdot\frac{2R^3}{3}$$ $$\boxed{I = \frac{2}{3}MR^2}$$(b) About tangent
Using the theorem of parallel axes,
$$I = I_{cm}+MR^2 = \frac{2}{3}MR^2+MR^2$$ $$\therefore \boxed{I = \frac{5}{3}MR^2}$$8. M.I. of Solid Sphere
(a) About diameter
Let us consider a solid sphere. Let $M$ = mass of the solid sphere, $R$ = radius of the sphere, $r$ = radius of the disc.
$\therefore$ Mass per unit volume of sphere $=\dfrac{M}{\frac{4}{3}\pi R^3}$
Let us consider a slice (disc) of radius $r$ and thickness $dx$ at distance $x$ from the centre of the sphere. Then, the surface area of the slice $=\pi r^2=\pi(R^2-x^2)$.
$\therefore$ Volume of slice $=\pi(R^2-x^2)\,dx$
$\therefore$ Mass of slice (disc) $(dm)=\dfrac{M}{\frac{4}{3}\pi R^3}\cdot\pi(R^2-x^2)dx = \dfrac{3M}{4R^3}(R^2-x^2)dx$
Now, M.I. of disc about axis $AB$ is:
$$dI = \text{mass of disc}\times\frac{(\text{radius})^2}{2} = \frac{3M}{4R^3}(R^2-x^2)dx\cdot\frac{(R^2-x^2)}{2}$$ $$\therefore dI = \frac{3}{8}\frac{M}{R^3}(R^2-x^2)^2dx$$$\therefore$ Total M.I. of solid sphere about $AB$ is:
$$I = \int_{-R}^R dI = 2\int_0^R dI = \frac{2\cdot 3}{8}\frac{M}{R^3}\left[\int_0^R(R^4-2R^2x^2+x^4)dx\right]$$ $$= \frac{3}{4}\frac{M}{R^3}\left[\int_0^R R^4dx-2\int_0^R R^2x^2dx+\int_0^R x^4dx\right]$$ $$= \frac{3M}{4R^3}\left[R^5-2R^2\cdot\frac{R^3}{3}+\frac{R^5}{5}\right]$$ $$= \frac{3M}{4R^3}\left[\frac{15R^5-10R^5+3R^5}{15}\right]$$ $$= \frac{3M}{4R^3}\cdot\frac{8R^5}{15}$$ $$\boxed{I = \frac{2}{5}MR^2}$$(b) About tangent
Using the theorem of parallel axis,
$$I = I_{cm}+MR^2 = \frac{2}{5}MR^2+MR^2$$ $$\therefore \boxed{I = \frac{7}{5}MR^2}$$Rotational K.E. (About Axis Through C.M.)
Let $m_1,m_2$ up to $m_n$ be the masses and $r_1,r_2$ up to $r_n$ be the radii from the axis. Let $v_1,v_2,\ldots,v_n$ be their corresponding velocities.
Let linear velocity varies and angular velocity remains constant throughout the motion.
$$v_1 = \omega r_1$$Total,
$$(K.E.)_{rot} = \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\cdots+\frac{1}{2}m_nv_n^2$$ $$= \frac{1}{2}m_1\omega^2r_1^2+\frac{1}{2}m_2\omega^2r_2^2+\cdots+\frac{1}{2}m_n\omega^2r_n^2$$ $$= \frac{1}{2}\omega^2(m_1r_1^2+m_2r_2^2+\cdots+m_nr_n^2)$$ $$\boxed{(K.E.)_{rot} = \frac{1}{2}I\omega^2}$$K.E. of a Rotating Body in which C.M. has Linear Velocity
Consider a wheel of radius $r$ rolling on a horizontal surface. Let $\omega$ = angular velocity of wheel, $v$ = linear velocity of C.M. Then, displacement of wheel in one complete rotation is $x=2\pi r$.
$$(K.E.)_{rolling} = (K.E.)_{rotational}+(K.E.)_{translational}$$ $$= \frac{1}{2}I\omega^2+\frac{1}{2}mv^2$$ $$= \frac{1}{2}mk^2\left(\frac{v}{r}\right)^2+\frac{1}{2}mv^2$$ $$\boxed{(K.E.)_{rolling} = \frac{1}{2}mv^2\left[\frac{k^2}{r^2}+1\right]}$$Acceleration of a Rolling Body on an Inclined Plane
Consider a spherical body rolling on an inclined plane. Let $m$ = mass of the body, $R$ = radius of the body, $s$ = displacement of the body, $v$ = linear velocity of the body.
In this case,
$$P.E._{\text{lost}} = K.E._{\text{gain}}$$or,
$$mgs\sin\theta = \frac{1}{2}mv^2\left(\frac{K^2}{R^2}+1\right)$$or,
$$v^2 = \frac{2gs\sin\theta}{\left(\dfrac{K^2}{R^2}+1\right)}$$Using $v^2=2as$ (starting from rest),
$$2as = \frac{2gs\sin\theta}{\left(\dfrac{K^2}{R^2}+1\right)}$$ $$\boxed{a = \frac{g\sin\theta}{\left(\dfrac{K^2}{R^2}+1\right)}}$$or, in terms of moment of inertia,
$$a = \frac{mg\sin\theta}{\dfrac{mK^2}{R^2}+m}$$ $$\boxed{a = \dfrac{mg\sin\theta}{\dfrac{I}{R^2}+m}} \qquad \text{(in terms of M.I.)}$$M.I. of a Diatomic Molecule
Let us consider a diatomic molecule. Let $r_0$ = bond length (interatomic distance), C.M. = centre of mass, $r_1$ = distance of first atom from C.M., $r_2$ = distance of second atom from C.M., $m_1$ = mass of first atom, $m_2$ = mass of second atom.
When the molecule is in stable equilibrium,
$$\text{clockwise moment} = \text{anticlockwise moment}$$ $$F_1r_1 = F_2r_2$$ $$m_1gr_1 = m_2gr_2$$ $$m_1r_1 = m_2r_2 \qquad \cdots (1)$$From the figure, $r_0=r_1+r_2$, so $r_1=r_0-r_2$.
Replacing $r_1$ in eq.(1), we get,
$$m_1(r_0-r_2)=m_2r_2$$ $$m_1r_0-m_1r_2=m_2r_2$$ $$m_1r_0=(m_1+m_2)r_2$$ $$\Rightarrow r_2=\frac{m_1r_0}{m_1+m_2} \qquad \cdots (2)$$Similarly, we can derive,
$$r_1=\frac{m_2r_0}{m_1+m_2} \qquad \cdots (3)$$$\therefore$ M.I. of molecule about the axis through C.M. and perpendicular to bond length is given by:
$$I=\sum mr^2 = m_1r_1^2+m_2r_2^2$$ $$= m_1\cdot\frac{m_2^2}{(m_1+m_2)^2}r_0^2+m_2\cdot\frac{m_1^2}{(m_1+m_2)^2}r_0^2$$ $$= m_1m_2r_0^2\left[\frac{m_2}{(m_1+m_2)^2}+\frac{m_1}{(m_1+m_2)^2}\right]$$ $$= m_1m_2r_0^2\left[\frac{m_1+m_2}{(m_1+m_2)^2}\right]$$ $$\boxed{I = \frac{m_1m_2}{m_1+m_2}r_0^2=\mu r_0^2}$$where $\mu$ is the reduced mass of the molecule,
$$\mu=\frac{m_1m_2}{m_1+m_2}$$Numericals
Q1.
A uniform sphere of mass 2 kg and radius 10 cm is released from rest on an inclined plane which makes angle $30^\circ$ with horizontal. Find the K.E. as it travels 2 m along the plane.
Solution:
For a uniform sphere, $m=2$ kg, $R=10$ cm, $\theta=30^\circ$, $s=2$ m, $(K.E.)=?$
We have,
$$(K.E.)_{gain} = P.E._{lost}$$ $$= mgs\sin\theta$$ $$= 2\times9.8\times2\times\sin30^\circ$$ $$= 2\times9.8\times2\times\frac{1}{2}$$ $$\boxed{= 19.6\ \text{J}}$$Q2.
A satellite of mass 1000 kg is launched with speed of 10 km/s. It settles into a circular orbit of radius $8.68\times10^3$ km above the centre of earth. What is its speed in this orbit? [Mass of earth $(M)=5.98\times10^{24}$ kg, Radius of earth $(R)=6.38\times10^6$ m]
Solution:
$m=1000$ kg, $r=8.68\times10^3$ km $=8.68\times10^6$ m, $M=5.98\times10^{24}$ kg, $R=6.38\times10^6$ m, launching velocity $(v)=10$ km/s, orbital velocity $(v_0)=?$
Since $g=\dfrac{GM}{R^2}$, so $v_0=R\sqrt{\dfrac{g}{r}}$, or directly:
$$\frac{mv_0^2}{r} = \frac{GMm}{r^2}$$ $$v_0 = \sqrt{\frac{GM}{r}}$$ $$v_0 = \sqrt{\frac{6.67\times10^{-11}\times5.98\times10^{24}}{8.68\times10^6}}$$Q3.
A satellite $A$ is 5 times farther from a planet than satellite $B$. If it takes satellite $A$ 22 weeks to complete a full orbit around the planet, how long will it take satellite $B$ to travel around the planet once?
Solution:
$T_A=22$ weeks, $T_B=?$, $r_A=5r_B$.
We have,
$$v_0=\omega r,\qquad v_0=\frac{2\pi r}{T}$$ $$T = 2\pi\sqrt{\frac{r^3}{GM}}$$since
$$T=\frac{2\pi r}{v_0}=\frac{2\pi r}{\sqrt{GM/r}}=2\pi r\sqrt{\frac{r}{GM}}=2\pi\sqrt{r^3/GM}$$$\therefore$
$$T_A = 2\pi\sqrt{\frac{r_A^3}{GM}}$$ $$T_B = 2\pi\sqrt{\frac{r_B^3}{GM}}$$Q4. (In a Uniform Disc)
A thin uniform disc of radius 25 cm and mass 1 kg has a hole of radius 5 cm in it. If the centre of the hole is at a distance of 10 cm from the centre of the disc, calculate the moment of inertia about an axis perpendicular to its plane and passing through the centre of the hole.
Solution:
$M=1$ kg $=1000$ gm, $R=25$ cm (disc radius), $r=5$ cm (hole radius), distance between centres $=10$ cm.
Total area of disc before hole made:
$$=\pi R^2 = \pi(25)^2 = 625\pi\ \text{cm}^2$$Area of the hole:
$$=\pi r^2 = \pi(5)^2 = 25\pi\ \text{cm}^2$$Area of disc after hole made:
$$= 625\pi-25\pi=600\pi\ \text{cm}^2$$M.I. of the complete disc about the axis through its centre and perpendicular to plane $=\dfrac{1}{2}mR^2$:
$$= \frac{1}{2}\left(\frac{625\pi}{600\pi}\times1000\right)\times(25)^2$$ $$= \frac{1}{2}\times1041.66\times625$$ $$= 325520.83\ \text{gm}\cdot\text{cm}^2$$M.I. of complete disc about axis through the centre of hole and perpendicular to plane $=I_{cm}+MR^2$:
$$= I_{cm}+\frac{625\pi}{600\pi}\times1000\times(10)^2$$ $$= 325520.83+1041.66\times100$$ $$= 325520.83+104166$$ $$= 429687.4967\ \text{gm}\cdot\text{cm}^2$$M.I. of disc (the piece removed as hole) about an axis through its centre and perpendicular to plane $=\dfrac{1}{2}mR^2$:
$$= \frac{1}{2}\left(\frac{25\pi}{600\pi}\times1000\right)\times5^2 = 520.83\ \text{gm}\cdot\text{cm}^2$$$\therefore$ M.I. of disc with hole about the axis through the centre of the hole and perpendicular to the plane:
$$= (429687.49-520.83)\ \text{gm}\cdot\text{cm}^2$$ $$\boxed{= 429166.66\ \text{gm}\cdot\text{cm}^2}$$Q5.
Calculate the self-energy of the Sun taking the mass to be equal to $2\times10^{30}$ kg and radius to be $7\times10^8$ m. If its radius contracts by 1 km/year without affecting its mass, calculate the rate at which it radiates energy.
Solution:
$M=2\times10^{30}$ kg, $R=7\times10^8$ m, $G=7\times10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$
$$\frac{dR}{dt} = \frac{1\ \text{km}}{1\ \text{year}} = \frac{1000}{365\times86400} = 3.17\times10^{-5}\ \text{m/s}$$(i) Gravitational self energy:
$$U_s = -\frac{3}{5}\frac{GM^2}{R}$$ $$= -\frac{3}{5}\cdot\frac{7\times10^{-11}(2\times10^{30})^2}{7\times10^8}$$ $$\boxed{= -2.4\times10^{41}\ \text{J}}$$(ii) $\dfrac{dU_s}{dt}=?$
$$\frac{dU_s}{dt} = \frac{dU_s}{dR}\cdot\frac{dR}{dt}$$ $$= \frac{d}{dR}\left\{-\frac{3}{5}\frac{GM^2}{R}\right\}\cdot\frac{dR}{dt}$$ $$= -\frac{3}{5}GM^2\left(\frac{dR^{-1}}{dR}\right)\cdot\frac{dR}{dt}$$ $$= -\frac{3}{5}GM^2(-1)R^{-1-1}\cdot\frac{dR}{dt}$$ $$= \frac{3}{5}\frac{GM^2}{R^2}\cdot\frac{dR}{dt}$$ $$\boxed{= 1.087\times10^{28}\ \text{J/s}}$$Q6.
The intermolecular distance between two atoms of hydrogen is $0.77\ \text{\AA}$ and mass of proton is $1.67\times10^{-27}$ kg. Find the M.I. of the hydrogen molecule about an axis through the midpoint of the line joining the two atoms and perpendicular to it.
Solution:
$r_0=0.77\ \text{\AA}=0.77\times10^{-10}$ m, $m_p=1.67\times10^{-27}$ kg
Now,
$$I=\mu r_0^2=\left(\frac{m_p\cdot m_p}{m_p+m_p}\right)r_0^2=\frac{m_p}{2}\times r_0^2$$ $$= \frac{1.67\times10^{-27}\times(0.77\times10^{-10})^2}{2}$$ $$\boxed{= 4.95\times10^{-48}\ \text{kg}\cdot\text{m}^2}$$Q7.
Calculate the velocity with which a body is projected upward so as to reach a height $4R_e$ from the surface of earth. [$R_e$ is the radius of earth, $R_e=6.4\times10^6$ m]
Solution:
$R_e=6.4\times10^6$ m, $h=4R_e=25.6\times10^6$ m, velocity of projection $(v)=?$
Here, $r=R_e+h=R_e+4R_e=5R_e=32\times10^6$ m.
Change in K.E. from earth's surface to the given height = change in P.E. from earth's surface to the given height.
$$\frac{1}{2}mv^2-\frac{1}{2}m(0)^2 = -\frac{GMm}{5R_e}-\left(-\frac{GMm}{R_e}\right)$$or,
$$\frac{1}{2}mv^2 = \frac{GMm}{R_e}-\frac{GMm}{5R_e}$$or,
$$\frac{1}{2}mv^2 = \frac{GMm}{R_e}\left[1-\frac{1}{5}\right]$$or,
$$v^2 = \frac{2GM}{R_e}\cdot\frac{4}{5}$$or,
$$v = \sqrt{\frac{8}{5}\frac{GM}{R_e}} = \sqrt{\frac{8}{5}\cdot\frac{GM\cdot R_e}{R_e^2}}$$ $$= \sqrt{\frac{8}{5}\,gR_e}$$ $$= \sqrt{\frac{8}{5}\times9.8\times6.4\times10^6}$$ $$\boxed{= 10047.58\ \text{m/s}}$$