Rotational Motion

It is the movement of an object around a fixed axis or point, where every point on the object follows a circular path. It is commonly seen in spinning wheels, rotating fans, or the Earth's rotation. Key quantities involved in rotational motion include angular displacement, angular velocity, and angular acceleration, which describe how much and how fast an object rotates. Torque is the force that causes rotation, and moment of inertia determines how resistant an object is to changes in its rotational motion. Rotational motion is an essential concept in physics, closely related to linear motion but involving circular paths.

1.0Definition

Rigid Body: A rigid body is an assemblage of a large number of material particles which do not change their mutual distance under any circumstances.

Rotational Motion of a Rigid Body: Any kind of motion is identified by change in position or change in orientation or change in both. If a body changes its orientation during its motion it is said to be in rotational motion.

2.0Axis of Rotation

An imaginary line perpendicular to the plane of circular paths of particles of a rigid body in rotation and containing the centres of all these circular paths is known as the axis of rotation.

It is not necessary that the axis of rotation should pass through the body.

3.0Kinematics of Rotational Motion

S.No	Topic	Diagram	Formula
1	Moment of Inertia For Discrete System of Particles		$I=m{r}^2$ m=mass, r=Perpendicular distance from AOR $\Rightarrow$I depends on mass, mass distribution, position of AOR. $\Rightarrow$I does not depend on angular velocity, angular acceleration, torque, angular momentum. $\Rightarrow$As the distance of mass increases from the rotational axis, MOI increases $\Rightarrow$SI Unit- $Kg-m^2$
2	Moment Of Inertia(MOI) for Continuous Mass Distribution		$I_{AB}=\int r^{2}dm$

4.0Moments of inertia of some regular shaped bodies about specific axes

S.No.	MOI of Body	Geometrical Figure	Axis
1.	Rod about an axis passing through the centre		$I=\frac{M{L}^2}{12}$
2.	Rod about an axis passing through its end and perpendicular to its Length		$I=\frac{M{L}^2}{3}$
3.	Rod Inclined at some angle		$I=\frac{M{L}^2}{3}Si{n}^2\theta$
4.	Ring		$I=M{R}^2$
5.	Disc		$I=\frac{1}{2}M{R}^2$
6.	Hollow Cylinder		$I=M{R}^2$
7.	Solid Cylinder		$I=\frac{M{R}^2}{2}$
8.	Rectangular Plate about an axis that lies in the plane and parallel to shorter sides		$I=\frac{M{L}^2}{3}$
9.	Rectangular Plate about the axis AA’		$I_{A{A}^{\prime }}=\frac{M{L}^2}{12}$
10.	Rectangular Plate about the axis AA’		$I_{A{A}^{\prime }}=\frac{M{b}^2}{3}$
11.	Rectangular Plate about the axis AA’		$I_{A{A}^{\prime }}=\frac{M{b}^2}{12}$
12.	Solid Sphere about its diametric axis		$I=\frac{2}{5}M{R}^2$
13.	Hollow Sphere about its diametric axis		$I=\frac{2}{3}M{R}^2$

5.0Perpendicular Axis Theorem and Parallel Axis Theorem

S.No	Theorem	Diagram
1.	Perpendicular Axis Theorem-The moment of inertia (MI) of a plane lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about any two mutually perpendicular axes in its own plane intersecting each other at the point through which the perpendicular axis passes. $I_Z=I_X+I_Y$ Applicable only for two dimensional bodies or plane lamina
2.	Parallel Axis Theorem-The moment of inertia of a body about any axis is the sum of its moment of inertia about a parallel axis through the center of mass and the product of the body's mass and the square of the distance between the two axes. $I=I_{CM}+M{d}^2$

6.0Radius of Gyration

It is the distance from the axis of rotation, the square of this distance when multiplied by the mass of the body then it gives the moment of inertia of the body (I = MK²) about the same axis of rotation.

• Radius of Gyration $K=\sqrt{\frac{I}{M}}$
• The radius of gyration depends on both the axis of rotation and the body's mass distribution.
• Radius of Gyration does not depend on mass of the body,angular quantities.

7.0Moment of Force or Torque

• It is the physical agency which is responsible for change in state of rotation. Torque is essential for producing turning or toppling phenomena.
• For producing torque the force is required & it is product of force and perpendicular distance of line of action of force (lever arm from axis).

• $\tau =FrSin\theta$
• In vector form $\vec{\tau }=\vec{r}\times \vec{F}$
• Units of Torque: N-m
• If $\theta =0°,{\tau }_{min}=0$ and $\theta =90°,{\tau }_{max}=Fr$

Couple of Forces

• When two forces of equal magnitude act on different points and in opposite directions with distinct lines of action these forces form a couple. This couple tries to rotate their bodies
• Moment of couple $\tau =FrSin\theta =Fb$

S.no	Conceptual Points	Diagram
1.	If the line of action of a force passes through point O, then the Torque of this force about point O will be zero.
2.	If a force is acting parallel to the axis of rotation, then its rotational effect about that axis will be zero.
3.	If the line of action of a force passes through the axis of rotation, then its rotational effect (Torque) about that axis of rotation will be zero.

8.0Equilibrium of Rigid Bodies

A system is said to be in mechanical equilibrium, if it is in translational as well as in rotational equilibrium i.e. $\overrightarrow{F_{net}}=0and\overrightarrow{{\tau }_{net}}=0$ (about every point)

9.0Rotational Equilibrium

A rigid body is in rotational equilibrium if its angular acceleration is zero, meaning it is either at rest or rotating with constant angular velocity. When subject to several coplanar forces, the resultant torque about any axis perpendicular to the plane must be zero.

$\sum \overrightarrow{{\tau }_p}=0$

Note:

• A body cannot be in rotational equilibrium under the action of a single force unless the line of action passes through the axis of rotation.
•    If a body is in rotational equilibrium under the action of three forces, the lines of action of the three forces must be either concurrent or parallel.

10.0Relation Between Torque and Angular Acceleration

${\tau }_{net}=I\alpha$

This equation is valid for fixed axis.

I→ moment of inertia of the body about the axis of rotation

$\alpha$= angular acceleration

11.0Rotating Pulley

Angular acceleration for heavy pulley

Here friction is present between the string and pulley i.e. There is no relative slipping between string and pulley.

$m_{1}g-T_1=m_{1}a$…..(1)

$T_2-m_{2}g=m_{2}a$……(2)

${\tau }_{\text{net}/P}=I_P\alpha$

$0+0+\left(T_{1}R-T_{2}R\right)=I\alpha$

No slipping,

$\alpha =\frac{a}{R}$

$a=\frac{m_{1}g-m_{2}g}{m_1+m_2+\frac{I}{R^2}}\Rightarrow a=\frac{\text{Net force along string}}{\text{Total mass of blocks}+\frac{I}{R^2}}$

12.0Toppling

In some cases, an external force is applied to a body to make it slide along a surface. However, in certain situations, the body may tip over before it begins to slide, a phenomenon known as toppling.

Conditions for Toppling

• If $\mu \ge \frac{a}{b}$, in this case body will topple if $F>mg\frac{a}{b}$
• If $\mu <\frac{a}{b}$, body will not topple for any value of F, applied at COM
• In case of Toppling Normal Reaction must pass through end points.

13.0Angular Momentum of a Particle

The angular momentum of a body about a specific axis is the product of its linear momentum and the perpendicular distance from the axis of rotation to the line of action of the linear momentum.

$L=mv\times rSin\theta$

$\vec{L}=\vec{r}\times \vec{p}$

S.No	Topic	Diagram	Formula
1.	Angular Momentum of a Rigid Body Rotating about Fixed Axis		$L=I\omega$
2.	Angular Momentum of Combined Rotation and Translation		${\vec{L}}_P=I_{\text{CM}}\vec{\omega }+\vec{r}\times m{\vec{v}}_{\text{CM}}$
3.	Total Kinetic Energy in Combined Translation and Rotation		$K_{\text{Total}}=\frac{1}{2}M{V}_C^2+\frac{1}{2}{I}_C{\omega }^2$

Newton 2nd Law of Rotation

It states that net external Torque acting on a system about any point O (or axis) is equal to the rate of change of angular momentum of the system about any point O (or axis).

${\left({\vec{\tau }}_{\text{ext}}\right)}_o=\frac{d{\vec{L}}_o}{dt},\text{also}{\vec{L}}_o=I_o\vec{\omega }$

${\left({\vec{\tau }}_{\text{ext}}\right)}_o=\frac{d(I_o\vec{\omega })}{dt}=I_o\frac{d\vec{\omega }}{dt}=I_o\vec{\alpha }$

${\vec{\tau }}_{\text{avg}}=\frac{\Delta \vec{L}}{\Delta t}=\frac{{\vec{L}}_F-{\vec{L}}_I}{t}$

14.0Conservation of Angular Momentum(COAM)

• If the resultant external torque acting on a system is zero then the total angular momentum of the system remains constant.

$\overrightarrow{{\tau }_{ext}}=0\Rightarrow \frac{\overrightarrow{dL}}{dt}=0\Rightarrow \vec{L}$$\text{ is constant}{L}_i=L_f$

• If a system is isolated from its surroundings any internal interaction between its different parts cannot alter its total angular momentum.

15.0Rotational Kinetic Energy

The energy possessed due to rotational motion of a body is known as rotational kinetic energy.

• $K.E_r=\frac{1}{2}I{\omega }^2$
• $K.E_{Rotation}=\frac{L^2}{2I}$
• $K.E_{Rotation}\propto \frac{1}{I}\Rightarrow$If I increases ,then K.E rotation decreases.

16.0Rotational Power

The power associated with the work done by torque acting on a rotating body.

$P_r=\vec{\tau }.\vec{\omega }$

17.0Work Energy Theorem in Rotational Motion

Work done by the torque = Change in kinetic energy of rotation $W=\frac{1}{2}I{\omega }_2^2-\frac{1}{2}I{\omega }_1^2$

The change in a rigid body's rotational kinetic energy equals the work done by external torques, similar to the work-energy theorem for linear motion.

18.0Rolling Motion

• When a body performs translatory motion as well as rotatory motion combinedly then it is said to undergo rolling motion

Pure Rolling

• If the relative velocity of the point of contact of the rolling body with the surface is zero then it is known as pure rolling, $v_{CM}=R\omega$
• If a body is performing rolling then the velocity of any point of the body with respect to the surface is given by $\vec{v}=\overrightarrow{v_{CM}}+\vec{\omega }\times R$

Total kinetic energy of a purely rolling objects

• $K.E_{\text{total}}=\frac{1}{2}M{V}_C^2\left[1+\frac{K^2}{R^2}\right]$
• $\frac{K.E_{\text{total}}}{K.E_{\text{trans}}}=\frac{K^2}{R^2}$
• So value of $\frac{K^2}{R^2}$ is maximum for ring or hollow cylinder=1 and minimum for solid sphere=$\frac{2}{5}=0.4$

19.0Rolling With Slipping

When a body rolls on a surface under external force, the frictional force on the body (if any) will be static in nature, less than its limiting value. But if the object rolls with slipping the nature of friction should be kinetic in nature

S.No	General Conditions	Diagrams
1.	No Friction Pure Rolling, $V_A=V_B$ $V_C-\omega R=V_{Surface}$
2.	$V_C>\omega R$, Contact point slips forward so kinetic friction will act in backward direction, so $V_C$ decreases and $\omega$Increases.
3.	$V_C<\omega R$, Contact point slips in backward direction, so kinetic friction will act in forward direction, so $V_C$ increases and decreases.
4.	$V=\omega R$(Initial), Static Friction whose values can lies between zero and $\mu N$ will act in backward direction.

20.0Accelerated Pure Rolling

S.no	Conditions	Diagram
1.	Accelerated Pure Rolling, $\Rightarrow a_{COM}=\alpha R$
2.	If Surface is also Moving, $\Rightarrow a_{COM}-\alpha R=a_{Plank}$

21.0Rolling Motion on an Inclined Plane

Note: If different bodies are allowed to roll down on an inclined plane than the body with

22.0Sample Questions on Rotational Motion

Q-1. A body of mass M and radius r, rolling with velocity v on a smooth horizontal floor, rolls up a rough irregular inclined plane up to a vertical height $(\frac{3v^2}{4g})$. Compute the moment of inertia of the body and comment on its shape?

Solution:

$K{E}_{Total}=K{E}_{Tr}+K{E}_{Rot}=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$

$E=\frac{1}{2}M{v}^2[1+(\frac{I}{M{r}^2})][\text{as}v=r\omega ]$

When it rolls up on an irregular inclined plane of height $h=(\frac{3v^2}{4g})$, its KE is fully converted into PE, So, by conservation of mechanical energy

$E=\frac{1}{2}M{v}^2[1+(\frac{I}{M{r}^2})]=Mg[\frac{3v^2}{4g}]$

which on simplification gives

$I=\frac{1}{2}M{r}^2$

This result clearly indicates that the body is either a disc or a cylinder.

Q-2. A disc placed on a rough surface such that its initial angular velocity is zero. Find the velocity of COM when pure rolling starts?

Solution:

$\omega =0$

Applying COAM about point O

$(\overrightarrow{L_i}{)}_0=(\overrightarrow{L_f}{)}_0$

$[-4M{V}_{O}R]+[0]=[-4M{V}_{c}R]+[-12\times 4M{R}^2\omega ]$

$\Rightarrow -4M{V}_{O}R=-4M{V}_{c}R-\frac{1}{2}\times 4M{V}_{c}R$

$\Rightarrow -4M{V}_{O}R=-6M{V}_{c}R\Rightarrow V_c=\frac{2V_O}{3}$

Q-3. Find velocity of COM of hollow sphere on reaching ground, if sphere maintains pure rolling throughout motion ? [Assume R<<H]. Also find the direction of friction on the sphere during rolling ?

Solution:

Apply conservation of mechanical energy :

$K_1+U_1=K_2+U_2$

$0+MgH=\frac{1}{2}M{V}_{COM}^2[1+\frac{2}{3}]+0\Rightarrow MgH=\frac{5}{6}M{V}_{COM}^2$

$V_{COM}=\sqrt{\frac{6gH}{5}}$

For pure rolling, $V_{COM}=\omega R$

So, as $V_{COM}$ increases during rolling so  must also increase to maintain pure rolling, so torque of

friction should be clockwise, and friction must be upward along the plane.

Q-4. If the rotational kinetic energy accounts for 50% of the total kinetic energy, what can be inferred about the body?

Solution:

$\frac{K{E}_{Rot}}{K{E}_{Tot}}\times 100=50$

$\frac{\frac{K^2}{R^2}}{1+\frac{K^2}{R^2}}\times 100=50$$\Rightarrow \frac{\frac{K^2}{R^2}}{1+\frac{K^2}{R^2}}=\frac{1}{2}$$\Rightarrow \frac{K^2}{R^2}=1$

Body can be ring or hollow cylinder

Q-5.An automobile engine is rated at 200 horsepower when operating at 6000 revolutions per minute (rpm). What is the torque produced by the engine at this speed?

Solution:

$P=200\times 746W=1.49\times 10^{5}W$

$\omega =6000rev/min=6000\times \frac{2\pi }{60}=628rad/s$

$\tau =\frac{P}{\omega }=\frac{1.49\times 10^5}{628}=237.5N-m$