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The rotating blade of a fan has some kinetic energy due to rotational motion which can not be expressed directly as \(K.E = \frac{1}{2}mv^{2}\) since all the points do not have same speed.

To find rotational K.E. we take the fan's blade as a collection of different very small particles called elements. One such element has mass dm and is at distance r from the rotational axis as shown. Its kinetic energy can be given as

\[\begin{matrix} d(K.E.) & \ = \frac{1}{2}(dm)v^{2} = \frac{1}{2}(dm)(r\omega)^{2} \\ & \ = \frac{1}{2}(dm)r^{2}\omega^{2} \end{matrix}\]

The rotational kinetic energy of the body is given by summing
i.e. intergrating the kinetic energy of all the elements of the body.

\[\text{~}\text{K.E.}\text{~} = \int_{}^{}\ d(K.E.) = \int_{}^{}\ \frac{1}{2}(dm)r^{2}\omega^{2}\]

Since \(\omega\) is same for every element of a rigid body so we take \(\omega^{2}\) outside the integral.
\(\therefore\ \) K.E. \(= \frac{1}{2}\omega^{2}\int_{}^{}\ r^{2}dm\)
we may write K.E. \(= \frac{\mathbf{1}}{\mathbf{2}}\mathbf{I}\mathbf{\omega}^{\mathbf{2}}\)
where \(I = \int_{}^{}\ r^{2}dm\) called Moment of Inertia.
The above equation is analogous to the K.E. \(= \frac{1}{2}mv^{2}\) i.e. Kinetic energy of a body having translational motion. Here \(\omega\) is analogous to \(v\). Also \(I\) is analogous to mass \(m\) i.e. I plays the same role in rotational motion as that of mass in translational motion. In other words as the inertia to the translational motion is due to the mass, inertia to the rotational motion is due to the quantity Moment of Inertia.

Moment of Inertia

Definition

It is the property of a rigid body by virtue of which it opposes change in its rotational motion.

• This is always taken w.r.t. a axis of rotation.

• This plays same role in rotational motion as mass plays in translational motion

• Difference between mass & M.I. (Moment of Inertia) is that mass is property of body & is independent of any reference axis choosen but MI depends on the mass as well as its distribution about the given axis of rotation. In other words it depends on
  (i) axis of rotation
  (ii) shape of the body
  (iii) size of the body
  (iv) density of the material of the body \(\rbrack\) mass depends only on these two things.

MI of a point mass :

\(r\) is perpendicular distance from mass to axis of rotation \({yy}^{'}\)

As we have discussed in previous topic for a continuous rigid body, \(I = \int_{}^{}\ r^{2}dm\)
To solve for above equation, we take dm in terms of variable \(r/dr\) or both r and dm in terms of another variable.
To substitute dm, we take linear, areal and volume mass densities as we had taken in previous chapter.

M.I. of a ring

Let mass of the ring is M and radius is R .
\(\therefore\ \) M.I. of small element of mass dm is \(dI = (dm)R^{2}\)
\[\therefore\ I = \int R^{2}(dm) \]but here distance of each element from the axis is same \(( = R)\) and so can be taken out of integral
\[{\therefore\ I = R^{2}\int dm }{\therefore\ I = {MR}^{2} }\]

M.I. of uniform rod

(i) About one of its end
M.I. of the element which is at distance \(r\) from the end is
\[I = \int dI = \int r^{2}(dm) \]Where, mass of the dement, \(dm =\) (mass per unit length) \(\times\) (length of the element)
\[{dm = \left( \frac{M}{L} \right)dr }{\therefore\ I = \int_{0}^{L}\mspace{2mu} r^{2}\left( \frac{M}{L} \right)dm = \frac{M}{L}\int_{0}^{L}\mspace{2mu} r^{2}dr = \frac{M}{L}\left\lbrack \frac{r^{3}}{3} \right\rbrack_{0}^{L} }\]

\[\therefore\ I = \frac{{ML}^{2}}{3} \](ii) About its C.O.M
\[dI = \left( r^{2} \right)(dm) = \frac{M}{L}r^{2}dr \]

here M.I. can be fround out by integrating the above from left end to right end

\[\begin{matrix} \therefore & I = \frac{M}{\text{ }L}\int_{- L/2}^{L/2}\mspace{2mu}\mspace{2mu} r^{2}dr = \left\lbrack \frac{r^{2}}{3} \right\rbrack_{- L/2}^{L/2} = \frac{M}{\text{ }L} \times \frac{1}{3}\left\lbrack \left( \frac{{\text{ }L}^{3}}{8} \right) = \left( \frac{- L^{3}}{8} \right) \right\rbrack \\ \therefore & I = \frac{{ML}^{2}}{12} \end{matrix}\]

Note : Here we can observe M.I. about the axis passing through C.O.M is less then that about end. In fact for various axis which are parallel to each other M.I. is minimum about the axis which passes through C.O.M. This is also proved later on in this chapter only.

M.I. of a uniform disc

Mass-M, Radius R. Lets taken an elementary ring of radius r and thickness dr .

\[\begin{matrix} \therefore & dI = (dm)r^{2}\text{~}\text{(for ring)}\text{~} \\ \text{~}\text{Also}\text{~} & dm = (\text{~}\text{mass per unit area}\text{~}) \times (\text{~}\text{area}\text{ of alimentary ring}\text{~}) \\ \text{~}\text{Where}\text{~} & \text{~}\text{area of elementary ring,}\text{~}dA = \pi\left\lbrack (r + dr)^{2} - r^{2} \right\rbrack \\ & = \pi\left\lbrack 2rdr + (dr)^{2} \right\rbrack \\ \therefore & dA \simeq 2\pi rdr\ \left\lbrack \text{~}\text{neglecting}\text{~}(dr)^{2}\text{~}\text{being a ring small quantity}\text{~} \right\rbrack \\ \therefore & dm = \frac{M}{\pi R^{2}} \times 2\pi rdr = \frac{2M}{R^{2}}rdr \\ \therefore & dI = (dm)r^{2} = \frac{2M}{R^{2}}r^{3}dr \\ \therefore & I = \frac{2M}{R^{2}}\int_{0}^{R}\mspace{2mu}\mspace{2mu} r^{3}dr \\ \therefore & I = \frac{{MR}^{2}}{2} \end{matrix}\]

M.I. of solid sphere (Mass-M, Radius - R)

Let taken \(n\) elemntary disc of radius \(r\) and at distance from the centre. Let its thikness is dy.

\[\begin{matrix} & dm = \frac{M}{\frac{4}{3}\pi R^{3}} \times \pi r^{2}dy \\ & \ = \frac{3M}{4R^{3}}r^{2}dy \end{matrix}\]

\(\therefore\ \) M.I. of elementary disc is \(dI = \frac{1}{2}(dm)r^{2} = \frac{3M}{8R^{2}}r^{4}dy\)
Also \(r^{2} = R^{2} - y^{2} \Rightarrow r^{4} = \left( R^{2} - y^{2} \right)^{2} = R^{4} + y^{4} - 2R^{2}y^{2}\)
\[\therefore\ dI = \frac{3M}{8R^{3}}\left( R^{4} + y^{4} - 2R^{2}y^{2} \right)dy \therefore\ \] M.I. of the sphere is

\[I = \int_{}^{}\ dI = \frac{3M}{8R^{3}}\left\lbrack \int_{- R/2}^{R/2}\mspace{2mu}\mspace{2mu} dy + \int_{- R/2}^{R/2}\mspace{2mu}\mspace{2mu} y^{4} \cdot dy - 2R^{2}\int_{- R/2}^{R/2}\mspace{2mu}\mspace{2mu} y^{2} \cdot dy \right\rbrack\]

\[{\therefore\ I = \frac{3M}{8R^{3}}\left\lbrack R^{4}\lbrack y\rbrack_{- R/2}^{R/2} + \left\lbrack \frac{y^{5}}{5} \right\rbrack_{- R/2}^{R/2} - 2R^{2}\left\lbrack \frac{y^{3}}{3} \right\rbrack_{- R/2}^{R/2} \right\rbrack }{\therefore\ I = \frac{2}{5}{MR}^{2}}\]

M.I. of thin hollow sphere (mass - M; Radius - R)

Here we take an elementary ring of radius \(r\) and thickness \(= Rd\theta\)
\(\therefore\ \) Mass of elementary ring, \(dm = \sigma dA\)
\[\therefore\ dm = \frac{M}{4\pi R^{2}} \times (2\pi r)Rd\theta\]

\[= \frac{M}{2R}rd\theta\]

\[\therefore\ dI = (dm)r^{2} = \frac{M}{2R}r^{3}\text{ }d\theta \]Aslo \(\ r = sin\theta \Rightarrow dI = \frac{M}{2R}\sin^{3}\theta d\theta\)
or \(dI = \frac{M}{2R}\left( 1 - \cos^{2}\theta \right)sin\theta d\theta\)
[Taking \(\sin^{2}\theta = 1 - \cos^{2}\theta\) ]

Let \(cos\theta = x \Rightarrow \frac{dx}{d\theta} = - sin\theta \Rightarrow sin\theta d\theta = - dx\)
\[\therefore\ dI = \frac{M}{2R}\left( 1 - x^{2} \right)( - dx) = \frac{M}{2R}\left( x^{2} - 1 \right)dx \]Also When \(\theta\) varies from 0 to \(\pi\); x i.e. \(cos\theta\) varies from 1 to -1
\[{\therefore\ I = \frac{M}{2R}\int_{1}^{- 1}\mspace{2mu}\left( x^{2} - 1 \right)dx = \frac{M}{2R}\left\lbrack \frac{x^{3}}{3} - x \right\rbrack_{1}^{- 1} }{\therefore\ I = \frac{2}{3}{MR}^{2}}\]

M.I. of some common bodies

Used to find moment of inertia about an axis which parallel to the axis passing through C.M.

C.M. is at origin
\(I_{CM} \rightarrow MI\) of the rigid body about an axis through CM
\(I_{P} \rightarrow MI\) of the rigid body about an axis which is parallel to the above axis through CM & is at distance d from the axis through CM

\[\begin{matrix} & I_{CM} = \sum_{i}^{}\mspace{2mu}\mspace{2mu}{\text{ }m}_{i}\left( x_{i}^{2} + y_{i}^{2} \right) \\ & I_{P} = \sum_{i}^{}\mspace{2mu}\mspace{2mu}{\text{ }m}_{i}\left\lbrack \left( x_{i} - a \right)^{2} + \left( y_{i} - b \right)^{2} \right\rbrack \\ & \ = \Sigma m_{i}\left( x_{i}^{2} + y_{i}^{2} \right) + \Sigma m_{i}\left( a^{2} + b^{2} \right) - \Sigma 2{am}_{i}x_{i} - \Sigma 2{bm}_{i}y_{i} \\ & \ = I_{CM} + \left( a^{2} + b^{2} \right)\Sigma m_{i} - 2a\Sigma{\text{ }m}_{i}x_{i} - 2\text{ }b\Sigma{\text{ }m}_{i}y_{i} \\ & \text{~}\text{of ant be an particles placed on}\text{~}with\text{~}\text{co-ordinate}\text{~}\left( x_{i},y \right) \end{matrix}\]

If M.I. of a body of mass M about an axis passing through its C.O.M. is \(I_{C}\) them M.I. of another axis which is parallel to the above central axis and is parallel to it is given by \(I_{{AA}^{'}} = I_{C} + {Mh}^{2}\)

This theorem is applicable only for the laminar bodies (i.e. plane bodies). (e.g. ring, disc, not sphere) \(I_{x}\& I_{y}\) are MI of body about a common pt. O in two mutually \(\bot\) directions in the plane of body \(I_{z}\) is MI of body about an axis \(\bot\) to X & Y axis & passing through pt. O

\[{I_{x} = \Sigma m_{i}y_{i}\ ^{2} }{I_{y} = \Sigma m_{i}x_{i}\ ^{2} }{I_{z} = \Sigma m_{i}\left( x_{i}\ ^{2} + y_{i}\ ^{2} \right)}\]

\[I_{z} = I_{x} + I_{y}\]

The point of intersection of the three axis need not be centre of mass, it can be any point in the plane of body which lie on the body or even outside it. For relation from perpendicular axis theorem, \(I_{3} = I_{1} + I_{2}\) axis (3) must be perpendicular the plane of the body and axis (1) and axis (2) must be in the plane of the body.

Illustation :

Find M.I. of the ring and the disc about the axis passing through their diameters.
Sol. By perpendicular axis theorem

\[I_{X} + I_{Y} + I_{Z}\]

Also, since mass distrioution of the body about \(x\) and \(y\) axis are similar

\[\therefore\ I_{x} = I_{y}\ \Rightarrow \ I_{x} = I_{y} = I_{z}/2 \](i) For ring, \(I_{Z} = MR^{2} \Rightarrow \ I_{X} = I_{y} = \frac{MR^{2}}{2}\)
(ii) For disc, \(I_{Z} = \frac{MR^{2}}{2} \Rightarrow I_{X} + I_{y} = \frac{MR^{2}}{4}\)

Note :

(i)

(ii)

(iii)

We can observe mass distrbution about the diamenteral axis shown in (i), (ii) and (iii) are similar
\[\therefore\ I_{(i)} = I_{(ii)} = I_{(ii)}\]

Similarly for square plate

(i)

(ii)

(iii)

\[I_{(i)} = I_{(ii)} = I_{(iii)}\frac{Ml^{2}}{6}\]

If M.I. aobut an axis of a system of mass M is I , then we may write, \(I = {MK}^{2}\) where k is called radius of gyration of the body about that axis \(= \sqrt{\frac{I}{M}}\)
In other words M.I. aobut an axis of system is same as M.I. of a particle of same mass placed t the distance equal to K from the axis of a uniform disc of radius R , radius of gyration about an axis passing through its centre & is to its plane is
\(K = \sqrt{\frac{\left( \frac{{MR}^{2}}{2} \right)}{M}} = \frac{R}{\sqrt{2}}\) i.e. the disc will same rotational inertia as that of a particle of same mass placed at distance \(\frac{R}{\sqrt{2}}\) from the axis

If M.I. aobut an axis of a system of mass M is I , then we may write, \(I = {MK}^{2}\) where k is called radius of gyration of the body about that axis \(= \sqrt{\frac{I}{M}}\)
In other words M.I. aobut an axis of system is same as M.I. of a particle of same mass placed t the distance equal to K from the axis of a uniform disc of radius R , radius of gyration about an axis passing through its centre & is to its plane is
\(K = \sqrt{\frac{\left( \frac{{MR}^{2}}{2} \right)}{M}} = \frac{R}{\sqrt{2}}\) i.e. the disc will same rotational inertia as that of a particle of same mass placed at distance \(\frac{R}{\sqrt{2}}\) from the axis

Note : In case of system of combitionation of various bodies \(k = \sqrt{\frac{I_{\text{total}\text{~}}}{M_{\text{total}\text{~}}}}\)

MI of Plate:

\[{I_{x} = \int\frac{1}{12}dma^{2} = \frac{1}{12}ma^{2} }{\therefore I_{y} = \frac{1}{12}mb^{2} }{\therefore I_{z} = I_{x} + I_{y} = \frac{1}{12}m\left( a^{2} + b^{2} \right) }\]

1. What would have happened if it was a square plate?

2. Prove that MI of square plate of mass \(m\) and side \(a\) about any axis passing through COM along the surface of plate is \(ma^{2}/12\)

Practice Exercise

Q.1. Find the moment of inertia of a pair of spheres, each having a mass \(m\) and radius \(r\), kept in contact about the tangent passing through the point of contact.
Q. 2 The moment of inertia of a uniform rod of mass \(m = 0.50\text{ }kg\) and length \(l = 1\text{ }m\) is \(I = 0.10\text{ }kg - m^{2}\) about a line perpendicular to the rod. Find the distance of this line from the middle point of the rod.
Q. 3 Find the moment of inertia of a uniform square plate of mass \(m\) and edge a about one of its diagonals.
Q. 4 The radius of gyration of a uniform disc about a line perpendicular to the disc equals its radius. Find the distance of the line form the centre.
Q. 5 Calculate the moment of inertia of a uniform rod of mass \(m\) & length \(l\) about an axis passing through one end & making angle \(\theta = 45^{\circ}\) with its length.
Q. 6 The surface density (mass/area) of a circular disc of radius a depends on the distance from the centre of \(\rho(r) = A + Br\). Find its moment of inertia about the line perpendicular to the plane of the disc through its centre.

Answers

Q.1. \(14{mr}^{2}\)
Q. \(2\sqrt{\frac{1}{2}\left( 1 - \frac{ml^{2}}{12} \right)} = \sqrt{\frac{7}{60}} = 0.34\text{ }m\)
Q. \(3\ {ma}^{2}/12\)
Q. \(4\ r/\sqrt{2}\)
Q. \(5\frac{\text{ }ml^{2}}{6}\)
Q. \(62\pi\left( \frac{{Aa}^{4}}{4} + \frac{{Ba}^{5}}{5} \right)\)

The quantitative measure of the tendency of a force to cause or change the rotational motion of a body is called torque. Consider an example to understand this.

In the figure below, the wrench is trying to open the nut. Now the ability of wrernch to open the nut will depend not only on the applied force, but the distance at which force is applied. This gives birth to new physical quantity called torque.

If only radial force \(F_{r}\) were present, the nut could not be turned. Thus the force causing the rotation is tangential force \(F_{T}\) only. The magnitude of the torque about an axis due to a force is given by
\(\tau = (\) Force causing the rotation \() \times (\) distance of point of application of force from the axis \()\)
i.e., \(\ = (Fsin\phi)r\)
we may also write \(\tau = F(rsin\phi) = F\left( r_{\bot} \right)\)
\[\therefore\tau = rFsin\phi = (rsin\phi)F\]

\[= rF_{r} = r_{\bot}F\]

\(r_{\bot}(d)\) : moment arm, lever arm
\(\overrightarrow{\tau} = \overrightarrow{r}x\overrightarrow{F}\ \) Direction of torque is found by sliding the force vector at the axis of rotation and using right hand thumb rule.
Torque also follows superposition principle. \(\overrightarrow{\tau} = {\overrightarrow{\tau}}_{1} + {\overrightarrow{\tau}}_{2} + \ldots + {\overrightarrow{\tau}}_{n}\)
\(\overrightarrow{OA} = \overrightarrow{r} =\) P.V. of pt. of application of force wrt fixed axis.(centre of rotation)

Note :

1. Torque & force are entirely diff. quantities. As torque is always defined with reference to point about which body is rotating, while force does not depend on it. Like torque of \(F_{3}\) about O is zero, while about \(A\) or \(B\) is not zero.

2. When equal & opposite force acts on a body having different line of action is called couple.

Rotational analog of Newton's second law
Consider a particle of mass \(m\) rotating in a circle of radius \(r\) under the influence of a tangential force \(F_{\tau}\) and a radial force \(F_{r}\), as shown in figure.
\[F_{t} = ma_{t} \]Magnitude of torque about the center of circle is:
\[{\tau = F_{t}r = ma_{t}r = m(\alpha r)r = \alpha\left( mr^{2} \right) = I\alpha }{\therefore\tau = I\alpha }\]

That is, the torque acting on the particle is proportional to its angular acceleration.
We can also understand that since torque is written about O , we should write \(\ ^{'}T\) also about O .

Torque on rigid body:

Proof: Consider a rigid body shown.
Torque acting on \(i\) th particle will be:
\[{{\overrightarrow{\tau}}_{i} = \left( m_{i}r_{i}^{2} \right)\overrightarrow{\alpha} \Rightarrow {\overrightarrow{\tau}}_{\text{net}\text{~}} = \sum{\overrightarrow{\tau}}_{i} = \sum\left( m_{i}r_{i}^{2} \right)\overrightarrow{\alpha} }{\therefore{\overrightarrow{\tau}}_{\text{net}\text{~}} = I\overrightarrow{\alpha} }\]

\(\overrightarrow{\tau} = I\overrightarrow{\alpha} = \overrightarrow{r} \times \overrightarrow{F}\) : This is the general relation that we are going to use in this chapter.
Note:
a) Torque due to internal forces is alwys zero.
b) Torque due to gravity is found by showing the gravitational force at the COM of the rigid body.

Proof (a): Consider two particles as shown in figure. From Newton's third law, the forces exerted by these particles are equal and opposite.

\[{\overrightarrow{F}}_{2,1} = {\overrightarrow{F}}_{1,2}\]

The sum of torques of these forces about origin O is

\[\begin{matrix} {\overrightarrow{\tau}}_{1} + {\overrightarrow{\tau}}_{2} & \ = {\overrightarrow{r}}_{1} \times {\overrightarrow{F}}_{2,1} + {\overrightarrow{r}}_{2} \times {\overrightarrow{F}}_{1,2} \\ & \ = {\overrightarrow{r}}_{1} \times {\overrightarrow{F}}_{2,1} + {\overrightarrow{r}}_{2} \times \left( - {\overrightarrow{F}}_{2,1} \right) \\ & \ = \left( {\overrightarrow{r}}_{1} - {\overrightarrow{r}}_{2} \right) \times {\overrightarrow{F}}_{2,1} \end{matrix}\]

The vector \({\overrightarrow{r}}_{1} - {\overrightarrow{r}}_{2}\) is along the line joining the two particles, so \({\overrightarrow{F}}_{2,1}\) is either parallel or antiparallel to ( \({\overrightarrow{r}}_{1} - {\overrightarrow{r}}_{2}\) ) thus

\[\left( {\overrightarrow{r}}_{1} - {\overrightarrow{r}}_{2} \right) \times {\overrightarrow{F}}_{2,1} = 0\]

So the internal forces (torques) cancel in pairs.

Proof (b):
\(\tau =\) torque of gravity about O
\[{\overrightarrow{\tau} = {\overrightarrow{r}}_{1} \times m_{1}\overrightarrow{g} + {\overrightarrow{r}}_{2} \times m_{2}\overrightarrow{g} + \ldots\ldots\ldots{\overrightarrow{r}}_{n} \times m_{n}\overrightarrow{g} = \left( m_{1}{\overrightarrow{r}}_{1} + m_{2}{\overrightarrow{r}}_{2} + m_{3}{\overrightarrow{r}}_{3} + \ldots\ldots + m_{n}{\overrightarrow{r}}_{n} \right) \times \overrightarrow{g} }{\overrightarrow{\tau} = \frac{\left( m_{1}{\overrightarrow{r}}_{1} + m_{2}{\overrightarrow{r}}_{2} + m_{3}{\overrightarrow{r}}_{3} + \ldots\ldots + m_{n}{\overrightarrow{r}}_{n} \right)}{\left( m_{1} + m_{2} + m_{3} + \ldots\ldots + m_{n} \right)} \times (M\overrightarrow{g}) }{\overrightarrow{\tau} = {\overrightarrow{r}}_{cm} \times (M\overrightarrow{g}) }\]

Where \(r_{cm}\) is PV of C.M. wrt. point O.

Practice Exercise

Q. 1 A force \(F = A\widehat{i} + Bj\) is applied to a point whose radius vector relative to the origin of coordinates O is equal to \(r = a\widehat{i} + b\widehat{j}\), where \(a,b\& A,B\) are constants, and \(\widehat{i},\widehat{j}\) are the unit vectors of the \(x\) and \(y\) axes. Find the Torque due to force.

Answers

Q. \(1Z = (aB - bA)\widehat{k}\)

If net external torque acting on the body is zero, then the body is said to be in rotational equilibrium.

The centre of mass of a body remains in equilibrium if the total external force acting on the body is zero. Similarly, a body remains in rotational equilibrium if the total external torque acting on the body is zero.
For translational equilibrium.
and

\[\begin{matrix} & \Sigma F_{x} = 0 \\ & \Sigma{\text{ }F}_{y} = 0 \\ & \Sigma{\text{ }F}_{z} = 0 \end{matrix}\]

The condition of rotational equilibrium is

\[\Sigma Z_{\text{ext}\text{~}} = 0\]

The equilibrium of a body is called stable if the body tries to regain its equilibrium position after being slightly displaced and released. It is called unstable if it gets further displaced after being slightly dis-
placed and released. If it can say in equilibrium even after being slightly displaced and released, it is said to be is neutral equilibrium.

Practice Exercise

Q. 1 Assuming frictionless contacts, determine the magnitude of external horizontal force P applied at the lower end of equilibrium of the rod. The rod is uniform and its mass ' \(m\) '

Q. 2 A uniform metre stick of mass 200 g is suspended from the ceiling through two vertical strings of equal lengths fixed at the ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tensions in the two strings.

Answers

Q. \(1P = \frac{W}{2}cot\theta\) or \(P = \frac{mg}{2}cot\theta\)
Q. 21.04 N in the left string and 1.12 N in the right.

Since torque is a rotational analong of force, therefore, Newton's second law for rotational motion is given by

\[\begin{array}{r} \tau_{net} = I\alpha\#(i) \end{array}\]

Note that the above equation (i) is not a vector equation.
It is valid in two situation:
(i) The axis is fixed in position and direction.
(ii) The axis passes through the center of mass and is fixed in direction only the equation

\[\begin{array}{r} \tau_{cm} = I_{cm}\alpha_{cm}\#(ii) \end{array}\]

is valid even if the center of mass is accelerating.

Practice Exercise

Q. 1 A light rod of length 1 m is pivoted at its centre and two masses of 5 kg and 2 kg are hung from the ends as shown in figure
(a) Find the initial angular acceleration of the rod assuming that it was horizontal in the beginning.
(b) If the rod has a mass of 1 kg distributed uniformly over its length.

(i) Find the initial angular acceleration of the rod
(ii) Find the tension in the supports to the blocks of mass 2 kg and 5 kg .
Q. 2 A meter stick is held vertically with one end on a rough horizontal floor. It is gently allowed to fall on the floor. Assuming that the end the floor does not slip, find the angular speed of the rod when it hits the floor.

Answers

Q. \(1\ \) (a) \(\frac{2\text{ }g\left( {\text{ }m}_{1} - m_{2} \right)}{\mathcal{l}\left( m_{1} + m_{2} \right)} = \frac{60}{7} = 8.4rad/s^{2}\)
(b) (i) \(\frac{2\text{ }g\left( {\text{ }m}_{1} - m_{2} \right)}{\mathcal{l}\left( m_{1} + m_{2} + m_{3}/3 \right)} = \frac{90}{22} = 8.4rad/s^{2}\), (ii) \(\left( m_{1}\text{ }g - m_{1}\alpha\frac{\mathcal{l}}{2} \right) = 29\text{ }N;\left( m_{2}\text{ }g + m_{2}\alpha\frac{\mathcal{l}}{2} \right) = 27.6\text{ }N\)
Q. \(2\sqrt{\frac{3\text{ }g}{\mathcal{l}}} = 5.4rad/s\)

The orbital angular momentum : Irrespective of the path or trajectory of the particle, be it a straight line, curved path or a closed orbital path, the orbital angular momentum \(\overrightarrow{L}\) of the particle at any position w.r.t. a reference point is

\[\begin{matrix} & \overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{P} \\ & \ |\overrightarrow{\text{ }L}| = rpsin\phi \\ & \ = r_{\bot} \times mv \end{matrix}\]

The \(rsinf\) is know as the moment arm, or lever arm designated as \(r_{\bot}\). The orbital angular momentum of particle in circular motion is expressed as

\[\overrightarrow{L} = {mr}^{2}\overrightarrow{\omega}\]

Note that direction of angular momentum vector \(\overrightarrow{L}\) is parallel to
angular velocity \(\overrightarrow{\omega}\).Figure shows the right hand thumb rule for determining direction of angular momentum. Curl your finger in rotrational sense from \(\overrightarrow{r}\) vector to \(\overrightarrow{p}\) vector, then the thumb points in the direction of angular momentum.

Spin angular momentum of a rigid body

We consider two cases:
(i) Axis of rotation passes through centre of mass of the body, referred to as centroidal roation.

(according to right hand thumb rule)
(ii) Axis of rotation is shifted from centre of mass, but passes through the body, referred to as noncentroidal rotation.

For non-centroidal rotation. \(\ \overrightarrow{L} = I_{0}\overrightarrow{\omega}\)
For non-centroidal rotation. \(\ \overrightarrow{L} = I\overrightarrow{\omega}\)
Where \(I_{0}\) is moment of inertia about centre of mass and I is moment of inertia about rotational axis, to be calculated with the help of parallel axis theorem.

Simultaneous spin and orbital motion

The total angular momentum is the vector sum of the spin and orbital angular momentum

\[\begin{matrix} & {\overrightarrow{L}}_{\text{total}\text{~}} = {\overrightarrow{L}}_{\text{spin}\text{~}} + {\overrightarrow{L}}_{\text{orbit}\text{~}} \\ & \ = I_{CM}{\overrightarrow{\omega}}_{\text{spin}\text{~}} + {mr}_{\bot}^{2}{\overrightarrow{\omega}}_{\text{orbit}\text{~}} \end{matrix}\]

Angular momentum of an inverted conical pendulum

Angular momentum about O,

\[\begin{matrix} & \overrightarrow{r} = lsin\theta\widehat{i} + lcos\theta\widehat{k} \\ & \overrightarrow{u} = (lsin\theta)\omega\widehat{j} \\ & \overrightarrow{\text{ }L} = \overrightarrow{r} \times m\overrightarrow{v} = ml^{2}\sin^{2}\theta\omega\widehat{k} - ml\omega^{2}sin\theta cos\theta\widehat{j} \end{matrix}\]

Concept : Angular momentum vector \(\overrightarrow{L}\) is perpendicular to position vector \(\overrightarrow{r}\) as well as momentum vector \(\overrightarrow{p}\). The magnitude of \(\overrightarrow{L}\) is constant but its direction is continuously varying. As the particle swings, \(\overrightarrow{L}\) vector sweeps out a cone. The z-component of \(\overrightarrow{L}\) is constant but the horizontal component travels around the circle with the particle.

Torque and Angular Momentum

When a number of force act on a particle, the net torque about origin 0 is sum of the torques due to each force.

\[\tau_{\text{net}\text{~}} = \overrightarrow{r} \times {\overrightarrow{F}}_{1} + \overrightarrow{r} \times {\overrightarrow{F}}_{2} + \ldots\ldots.. = \overrightarrow{r} \times \sum_{i}^{}\mspace{2mu}{\overrightarrow{\text{ }F}}_{1} = \overrightarrow{r} \times {\overrightarrow{F}}_{\text{net}\text{~}}\]

From Newton's second Law the net force is equal to rate of change of linear momentum. So we have

\[\begin{array}{r} \tau_{net} = \overrightarrow{r} \times {\overrightarrow{F}}_{net} = \overrightarrow{r} \times \frac{d\overrightarrow{p}}{dt}\#(i) \end{array}\]

As rate of change of angular momentum

\[\begin{array}{r} \frac{d\overrightarrow{\text{ }L}}{dt} = \frac{d}{dt}(\overrightarrow{r} \times \overrightarrow{p}) = \frac{d\overrightarrow{r}}{dt} \times \overrightarrow{p} + \overrightarrow{r} \times \frac{d\overrightarrow{p}}{dt}\#(ii) \\ \frac{\text{ }d\overrightarrow{r}}{dt} \times \overrightarrow{p} = \overrightarrow{v} \times m\overrightarrow{v} = 0\#(ii) \end{array}\]

As

\[\begin{array}{r} \text{~}\text{ecomes}\text{~}\#(iii) \\ \frac{d\overrightarrow{\text{ }L}}{dt} = \overrightarrow{r} \times \frac{d\overrightarrow{p}}{dt}\#(iii) \end{array}\]

On comparing eqns. (i) and (iii), we get

\[\begin{array}{r} {\overrightarrow{\tau}}_{\text{net}\text{~}} = \frac{d\overrightarrow{\text{ }L}}{dt}\#(iv) \end{array}\]

If the next externa torque on a systme is zero, the total angular momentum is constant in magnitude and direction.

That is, if \(\tau_{\text{ext}\text{~}} = 0\ \frac{dL}{dt} = 0\)
Thus, \(L =\) constant
For rigid body rotating about a fixed axis.

\[\begin{matrix} & L_{f} = L_{i} \\ \text{~}\text{or}\text{~} & L_{f}\omega_{f} = I_{i}\omega_{i} \end{matrix}\]

Angular Impulse

In complete analogy with the linear momentum, angular impulse is defined as

\[\tau = \int_{}^{}\ \tau_{ext}dt\]

Using Newton's second law for rotation motion,

\[\begin{matrix} t_{ext} & \ = \frac{dL}{dt} \\ \therefore\ \overrightarrow{\tau}t & \ = \Delta{\overrightarrow{L}}_{f} - {\overrightarrow{L}}_{i} \end{matrix}\]

The net angular impulse acting on a rigid body is equal to the change in angular momentum of the body. This is called the impulse-momentum theorem for rotational dynamics.

Practice Exercise

Q. 1 A wooden long of mass M and length L is hinged by a frictionless nail at O . A bullet of mass \(m\) strikes with velocity \(v\) and sticks to it. Find angular velocity of the system, immediately after the collision, about O .

Q. 2 A disc of mass \(M\) and radius \(r\) is rotating about its axis with angular velocity \(\omega\). Now if a mass \(m\) falls vertically on its rim and sticks to it. Then find final angular velocity of the combined system.
Q. 3 A man of mass 100 kg stands at the rim of a turn-table of radius 2 m , moment of inertia \(4000\text{ }kg - m^{2}\) mounted on a vertical frictionless shaft at its centre. The whole system is initially at rest. The man walks along the outer edge of the turn-table with a velocity of \(1\text{ }m/s\) relative to the earth.
(a) With what angular velocity and in what direction does the turn-table rotate?
(b) Through what angle will it have rotated when the man reaches his initial position on the turn-table?
(c) Through what angle will it have rotated when the man reaches his initial position relative to earth?
Q. 4 In horizontal smooth plane. If particle sticks after collision to the rod, find

(a) Final angular velocity
(b) The impulse on particle
Q. 5 A particle having mass 2 kg is moving along straight line \(3x + 4y = 5\) with speed \(8\text{ }m/s\). Find angular momentum of the particle about origin. \(x\) and \(y\) are in meters.
Q. 6 A particle having mass 2 kg is moving with velocity \((2\widehat{i} + 3\widehat{j})m/s\). Find angular momentum of the particle about origin when it is at \((1,1,0)\).
Q. 7 A wheel of moment of inertia \(0.500\text{ }kg - m^{2}\) and radius 20.0 cm is rotating about its axis at an angular speed of \(20.0rad/s\). It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel.

Answers

Q. \(1\ \omega = \frac{3mv}{L(M + 3\text{ }m)}\)
Q. \(2\ \omega^{'} = \left( \frac{M}{M + 2m} \right)\omega\)
Q. \(3\ \) (a) \(- \frac{1}{20}rad/s\) (b) \(- \frac{2\pi}{11}radian(\) c \()\theta_{t} = - \frac{\pi}{5}radianQ.4\ \) (a) \(\omega = \frac{3{\text{ }m}_{1}u}{\left( m_{2} + 3{\text{ }m}_{1} \right)}\ \) (b) \(\frac{m_{1}{\text{ }m}_{2}u}{m_{2} + 3{\text{ }m}_{1}}\)
Q. \(516\text{ }kg{\text{ }m}^{2}/s\ \) Q. \(6\ 2\widehat{kkgm}/s\ \) Q. \(7\ 19.7rad/s\)

It is the axis about which the motion of a rigid body undergoing plane motion is assumed to be pure rotational motion. It is always perpendicular to the plane of motion of rigid body and instantaneously remains at rest. The point of intersection of instantaneous axis of rotation with the plane of motion of the remains at rest. The point of intersection of instantaneous axis of rotation with the plane of motion of the rigid body is called instantaneous centre of rotation (ICR) about which all points of the rigid body are assumed to be going in circles of different radii equal to their respective distances from ICR with the same \(\omega\) and \(\alpha\) as that about CM of the rigid body at that instant.
\(\therefore\) Velocity of IAR \(= 0\)
i.e. \(v_{A} = 0\)

By finding the position of IAR, we can easily find the velocity of any point of rigid body at that instant. \({\overrightarrow{v}}_{p} = \overrightarrow{\omega} \times {\overrightarrow{r}}_{P,A}\)
We can also find the acceleration of any point P of rigid body at that instant provided the acceleration of IAR should be know.

\[{\overrightarrow{a}}_{P} = {\overrightarrow{a}}_{P,A} + {\overrightarrow{a}}_{A}\]

(i) Kinetic energy of the rigid body is \(K = \frac{1}{2}I_{A}\omega^{2}\)
(ii) Angular momentum of rigid body about IAR is \(L_{A} = I_{A}\omega\)
(iii) \(\ \tau_{A} = I_{A}\alpha\), where \(\tau_{A}\) includes the torque of pseudo force acting on the CM about IAR also.

How to find the position of ICR

The position of ICR can be fund out in the following cases.

Case-I :

If the velocity of a point of the body and angular velocity are given.

Draw a line perpendicular to \(\overrightarrow{v}\), the instantaneous centre must be
lying on this line a distance ' \(r\) ' given by \(r = v/\omega\)

Case II :

If the lines of action of two non-parallel velocities of two points of the rigid boy are given.
Draw the normals on the two non-parallel velocities \({\overrightarrow{v}}_{P}\) and \({\overrightarrow{v}}_{Q}\) at points P and Q , respectively. The point of intersection of these normals is the instantaneous centre at that instant.

Case-III :

If magnitudes and direction of two parallel velocities are given.
In figure (i), if the two velocities \({\overrightarrow{v}}_{P}\) and \({\overrightarrow{v}}_{Q}\) are in the opposite direction, then IC must be lying in between P and Q.
\(\frac{v_{P}}{v_{Q}} = \frac{r_{P}}{r_{Q}}\ \) and \(r_{P} + r_{Q} = d\)

(i)

(ii)

In figure (ii), if the two velocities \({\overrightarrow{v}}_{P}\) and \({\overrightarrow{v}}_{Q}\) are in the same direction, the IC must be lying outside PQ (near P if \(v_{P} < v_{Q}\) and near Q if \(v_{p} > v_{Q}\) )
\(\frac{V_{P}}{r_{Q}} = \frac{r_{P}}{r_{Q}}\) and \(r_{P} - r_{Q} = d\)

Rolling Motion

Pure rolling means no sliding. Now, the motion of any body can be divided into pure translation & pure rotation. And we can see rotation about any axis. So; for a wheel rolling on a horizontal surface, I m taking its COM as reference point to study its motion. If C is the reference point, then the wheel can be considered rotating about C . And the point C would be translating with velocity vc.
Sliding refers to the condition under which two bodies in contact have relative velocity. And under pure rolling, the relative velocity at point of contact should be zero.

Now for the wheel shown, point P is in contact with the ground. The point P on the ground has zero velocity. Thus, the point P on the wheel should also have zero velocity.

Trajectory of a point on a periphery of the wheel is a cycloid

\[\begin{matrix} \overrightarrow{v} & \ = \frac{d\overrightarrow{r}}{dt} = v\lbrack(1 + cos\omega t)\widehat{i} + sin\omega t( - \widehat{j})\rbrack \\ \overrightarrow{r} & \ = \int_{}^{}\ d\overrightarrow{r} = v\int_{}^{}\ \lbrack(1 + cos\omega t)\widehat{i} + sin\omega t( - \widehat{j})\rbrack dt \\ & \ = \omega R\left( t + \frac{1}{\omega}sin\omega t \right)\widehat{i} + \frac{1}{\omega}cos\omega t(\widehat{j}) \\ & \ = R\lbrack(\omega t + sin\omega t)\widehat{i} + cos\omega t(\widehat{j})\rbrack \end{matrix}\]

\[{\frac{d\overrightarrow{v}}{dt} = \frac{dv}{dt}\lbrack(1 + cos\omega t)\widehat{i} + sin\omega t( - \widehat{j})\rbrack + \omega v\lbrack - sin\omega t(\widehat{i}) - cos\omega t(\widehat{j})\rbrack }{= \frac{dv}{dt}(\widehat{i}) + \frac{dv}{dt}\lbrack(cos\omega t)\widehat{i} + sin\omega t( - \widehat{j})\rbrack + \frac{v^{2}}{R}\lbrack - \widehat{r}\rbrack }{= a_{c}(\widehat{i}) + \alpha R(\widehat{\tau}) + \frac{v^{2}}{R}( - \widehat{r})}\]

Kinematics of Boby in pure Rolling

1. Velocity

wrt centre

2. Acceleration wrt centre

Constraint Equation:

Written at contact points where no slipping takes place.
\(\left\lbrack \begin{matrix} \text{~}\text{velocity of a contact}\text{~} \\ \text{~}\text{point on}\text{~}1^{\text{st}\text{~}} \end{matrix} \right.\ \) rigid body \(\rbrack = \begin{bmatrix} \text{~}\text{velocity of same contact}\text{~} \\ \text{~}\text{point on}\text{~}2^{\text{nd}\text{~}}\text{~}\text{rigid body}\text{~} \end{bmatrix}\)

Write contraint equation for following exmaples:

To get the direction of friction in pure rolling, we set two criteria's:
(a) Acceleration : Its direction should be such that vector sum all forces comply with it.
(b) Angular acceleration: Its direction should be such that vector sum all torques comply with it.

So, from above point we understand that if we show wrong direction of friction in our free body diagram, we will get a negative answer. So, direction of friction force in pure rolling should not be cause of concern.

Here it is also to be noted that in case of sliding, (kinetic friction) we can not take any arbitrary direction of friction and solve that question. In case of rolling static friction is unknown so after sloving, its value comes out negative if taken wrongly, but in case of kinetic friction its value is known beforehand \(( = \mu N)\), so there is no case of value of a known quantity coming negative after solving.

Here are few examples wherein, one might try to guess the direction of friction force.

Rolling with slipping

Case-I \(\ V_{cm} > \omega R\)

Note : The friction will be kinetic in nature & its magnitude can be determined using \(f_{k} = \mu_{k}N\). Direction will be opposite to \(V_{cm}\) (because pt. of contact A is moving forward w.r.t. ground)
(This is how barkes stop a car)

Case -2 \(\ V_{cm} < \omega R\)

(This is how a car accelerate because of friction)

Case -3 \(\ V_{cm} = \omega r\)
Known as perfect rolling
Maximum problems we come across this situation
Note : 1 . The friction will be be static in nature. Its magnitude be determined, it will very from O to \(\mu_{s}\text{ }N\).
2. Even its direction can not be perdicted
3. The total work done by static friction is zero. Thus mechanical cnergy of the system will remain conserved.

Application of Newton's Second Law in Rolling Motion

1. Write \(F_{net} = {Ma}_{cm}\) for the object as if it were a point-mass, that is, ignoring ratation.

2. Write \(\tau = I_{cm}\alpha\) as if the object were only rotating about the centre of mass, that is ignoring translation.

3. Use of no-slip condition

4. Solve the resulting equations simultaneously for any unknown.

Caution:

• In general, it is not the case that \(f = \mu N\)

• Be certain that the sign convention of forces and torques are consistent.

Applying Newon's Second Law

\[\begin{array}{r} \begin{matrix} \Sigma F_{x} = ma & Mgsin\theta - f = Ma \\ \Sigma\tau = I\alpha & fR = I\alpha = \frac{2}{5}MR^{2}\alpha \end{matrix}\#(ii) \end{array}\]

Since the sphere rolls without slipping, the speed of the centre is \(v = \omega R\).
By differentiating it with respect to time, we get

\[a = \alpha R.\]

Solving equation (i), (ii) & (iii)

A shpere rolls down an incline. Since the sphere is not driven, the force of friction is direction up the incline.

\[\begin{array}{r} \text{~}\text{we get}\text{~}\ f = \frac{2}{5}Ma\#(iv) \end{array}\]

\[\begin{array}{r} \text{~}\text{and}\text{~}\ a = \frac{5}{7}gsin\theta\#(v) \end{array}\]

(b) Substituting (v) into (iv) yields

\[\begin{array}{r} f = \frac{2}{7}Mgsin\theta\#(vi) \end{array}\]

We use equation (vi) to find the minimum coefficient of friction required for the sphere to roll without slipping.
By definition, \(f = \mu N\) where \(N = Mgcos\theta\). Combining this with equation (vi) we have

\[\mu = \frac{2}{7}tan\theta\]

If the coefficient of static friction is less than this value, the sphere will slip as it rolls down the incline.

Planck:

Retardation \(a = \frac{f}{M} = \frac{\mu mg}{M}\)

Instantaneous velocity \(v = v_{0} - \frac{\mu mg}{M}t\)
Condition of pure rolling

\[v = \frac{7}{2}\mu gt = v_{0} - \frac{\mu mg}{M}t\]

or \(\ t = \frac{v_{0}}{\left\lbrack \frac{7}{5} + \frac{m}{M} \right\rbrack\mu g}\)

Since the rolling motion is a combination of linear velocity of the center and rotational motion about the center. Therefore, the total kinetic energy of a rolling body is given by

\[\begin{array}{r} K = \frac{1}{2}{mv}_{c}^{2} + \frac{1}{2}I_{c}\omega^{2}\#(i) \end{array}\]

Where \(\frac{1}{2}mv_{c}^{2}\) is the translational kinetic energy and
\(\frac{1}{2}I_{c}\omega^{2}\) is the rotational kinetic energy about the center of mass
In pure rolling motion, \(v_{c} = \omega R\)
\[\therefore\ K = \frac{1}{2}\text{ }m(\omega R)^{2} + \frac{1}{2}I_{c}\omega^{2} \]or \(\ K = \left( I_{c} + {mR}^{2} \right)\omega^{2}\)
Using parallel axes therem, the term \(I_{c} + mR^{2}\) given the moment of inertia about the point of contact, therefore,

\[\begin{array}{r} I_{0} = I_{c} + {mR}^{2}\#(ii) \end{array}\]

and \(K = \frac{1}{2}I_{0}\omega^{2}\)

Note that equation (ii) give the rotational kinetic energy of the wheel aobut the point of contact.

\[\begin{matrix} {\overrightarrow{l}}_{i} & \ = \sum_{i = 1}^{n}\mspace{2mu}\mspace{2mu} m_{i}\left( {\overrightarrow{r}}_{i} \times {\overrightarrow{v}}_{i} \right) \\ & \ = \sum_{i = 1}^{n}\mspace{2mu}\mspace{2mu} m_{i}\left\lbrack \left( {\overrightarrow{r}}_{o} + {\overrightarrow{r}}_{ic} \right) \times \left( {\overrightarrow{v}}_{c} + \overrightarrow{\omega} \times {\overrightarrow{r}}_{ic} \right) \right\rbrack \\ & \ = \sum_{i = 1}^{n}\mspace{2mu}\mspace{2mu} m_{i}\left( {\overrightarrow{r}}_{o} \times {\overrightarrow{v}}_{c} \right) + {\overrightarrow{r}}_{o} \times \sum_{i = 1}^{n}\mspace{2mu}\mspace{2mu} m_{i}\left( \overrightarrow{\omega} \times {\overrightarrow{r}}_{ic} \right) + \sum_{i = 1}^{n}\mspace{2mu}\mspace{2mu} m_{i}\left( {\overrightarrow{r}}_{ic} \times {\overrightarrow{v}}_{c} \right) + \sum_{i = 1}^{n}\mspace{2mu}\mspace{2mu} m_{i}\left\lbrack {\overrightarrow{r}}_{ic} \times \left( \overrightarrow{\omega} \times {\overrightarrow{r}}_{ic} \right) \right\rbrack \\ & \ = \left\lbrack \sum_{i = 1}^{n}\mspace{2mu}\mspace{2mu} m_{i} \right\rbrack\left( {\overrightarrow{r}}_{o} \times {\overrightarrow{v}}_{c} \right) + {\overrightarrow{r}}_{o} \times \sum_{i = 1}^{n}\mspace{2mu}\mspace{2mu} m_{i}{\overrightarrow{v}}_{ic} + \left\lbrack \sum_{i = 1}^{n}\mspace{2mu}\mspace{2mu} m_{i}{\overrightarrow{r}}_{ic} \right\rbrack \times {\overrightarrow{v}}_{c} + \sum_{i = 1}^{n}\mspace{2mu}\mspace{2mu} m_{i}\left\lbrack {\overrightarrow{r}}_{ic}^{2}\overrightarrow{\omega} - \left( {\overrightarrow{r}}_{ic} \cdot \overrightarrow{\omega} \right){\overrightarrow{r}}_{ic} \right\rbrack \\ & \ = M\left( {\overrightarrow{r}}_{o} \times {\overrightarrow{v}}_{c} \right) + \overrightarrow{\omega}\sum_{i = 1}^{n}\mspace{2mu}\mspace{2mu} m_{i}{\overrightarrow{r}}_{ic}^{2} = M\left( {\overrightarrow{r}}_{o} \times {\overrightarrow{v}}_{c} \right) + I_{COM}\overrightarrow{\omega} \end{matrix}\]

Angular momentum of a rigid body in planar motion about \(O:\overrightarrow{l} = M\left( {\overrightarrow{r}}_{o} \times {\overrightarrow{v}}_{c} \right) + I_{COM}\overrightarrow{\omega}\)

Spin and Orbital Angular Momentum

For a rigid body undergoing linear and rotational motion, the total angular momentum may be split into two pats-the orbital angular momentum and the spin angular momentum. The orbital angular momentum \(L_{0}\) is the angular momentum relative to the center of mass.
The orbital term treats the system as a point particle at the center of mass, where as the spin term is the sum of the angular momenta of the particles relative to the center of mass. The total angular momentum relative to the origin O in an inertial frame is the sum:

\[L = L_{0} + L_{cm}\]

Practice Exercise

Q. 1 A disc of mass and radius R is rolling with slipping on a rough horizontal surface which has coefficient of friction \(\mu\). At some instant the velocity of its center of mass is v and angular speed about center of mass is \(\omega\). The torque of the frictional force about the point which is instantaneously at rest at this moment is?

Q. 2 A uniform circular disc of radius r placed on a rough horizontal surface has initially a velocity \(v_{0}\) and angular velocity \(\omega_{0}\) as shown in the figure. The disc comes to rest (neither translates nor rotates) after moving some

distance. Then \(\frac{v_{0}}{r\omega_{0}}\) is?
Q. 3 A ball of radius \(R = 10.0\text{ }cm\) rolls without slipping down an inclined plane so that its centre moves with constat acceleration \(a = 2.50\text{ }cm/s^{2};t = 2.00\text{ }s\) after the beginning of motion its position corresponds to that shown in figure. Find:

(A) the velocities of the points \(A,B\) and O ;
(B) the acceleration of these points
Q. 4 A cylinder rolls without slipping over a horizotal plane. The radius of the cylinder is equal to r. Find the curvature radii of trajectories traced out by the points A and B in figure.

Q. 5 A sphere kept on a rough inclined plane is in equilibrium by a string wrapped over it. If the angle of inclination is \(\theta\), the tension in the string will be equal to

Q. 6 A wheel of radius R is rolling without slipping on a stationary horizontal surface. Find the acceleration of point of contact P at the instant shown.

Q. 7 A uniform circular disc of radius R rolls without slipping with a constant velocity on a stationary horizontal surface. Find the distance covered by a point P on its circumference in one full rotation.

Q. 8 A uniform cylinder rolls from rest from \(A\) down the side of a trough whose vertical dimension y si given by the equation \(y = kx^{2}\). The cylinder does not slip form \(A\) to \(B\) but the surface of trough is frictionless from \(B\) to \(C\). Then height of ascent of cylinder towards \(C\) is

Q. 9 A sphere of mass M and radius r slips on a rough horiozntal plane. At some instant it has translational velocity \(v_{0}\) and rotational velocity about the center \(v_{0}/2r\). The rotational velocity when the sphere starts pure rolling is
Q. 10 A solid cylinder of mass M and radius R lying on a rough horizontal surface for which coefficient of friction is \(\mu\) is pushed by applying a horizontal force P at a distance \(R/2\) from the center as shown in figure. The frictional force acting at the contact is

Answers

Q. \(1\mu mg\left( R - \frac{v}{\omega} \right)\ \) Q. \(2\ \frac{1}{2}\)
Q. 3 (A) \(v_{A} = 10\text{ }cm/s,v_{B} = 7.1\text{ }cm/sv_{0} = 0\) (B) \(a_{A} = 5.6\text{ }cm/s^{2},a_{B} = 2.5\text{ }cm/s^{2},a_{0} = 2.5\text{ }cm/s^{2}\)
Q. \(4\ R_{A} = 4r,R_{B} = 2\sqrt{2}r\)
Q. \(5\frac{mgsin\theta}{2}\)
Q. \(6{\overrightarrow{a}}_{p} = \omega^{2}R\widehat{j}\)
Q. \(7\ 8R\)
Q. \(8\ \frac{2\text{ }h}{3}\)
Q. \(9\ \frac{6}{7}v_{0}\)
Q. 10 zero