Dynamics of Rotational Motion and Rolling

Kinematics Of Rotational Motion About A Fixed Axis

Kinematics is the study of motion without considering the forces that cause it. For a rigid body rotating about a fixed axis, we can describe the motion using angular variables, analogous to the linear variables used for translational motion.

Let the fixed axis of rotation be the z-axis. Consider a rigid body rotating about this axis. Every particle in the body moves in a circle parallel to the xy-plane, centered on the z-axis.

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Angular Position, Displacement, Velocity, and Acceleration

• Angular Position ($\theta$): The angle of a reference line on the rigid body with respect to a fixed reference direction (e.g., the positive x-axis). Measured in radians.
• Angular Displacement ($\Delta\theta$): The change in angular position, $\Delta\theta = \theta_{final} - \theta_{initial}$.
• Angular Velocity ($\omega$): The rate of change of angular position. For a fixed axis, we often use the scalar angular speed, $\omega = \frac{d\theta}{dt}$. As a vector, $\vec{\omega} = \omega \hat{k}$ if the axis is z and rotation is counterclockwise. Units: rad/s.
• Angular Acceleration ($\alpha$): The rate of change of angular velocity. For a fixed axis, $\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}$. As a vector, $\vec{\alpha} = \alpha \hat{k}$ if angular speed is increasing (and $\vec{\omega}$ is in +z direction). Units: rad/s$^2$.

These angular variables are related to the linear variables of a particle at a perpendicular distance $r$ from the axis by $s = r\theta$ (arc length), $v = r\omega$ (tangential speed), and $a_t = r\alpha$ (tangential acceleration).

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Kinematic Equations for Constant Angular Acceleration

If a rigid body rotates about a fixed axis with constant angular acceleration $\alpha$, the following kinematic equations hold, analogous to the linear kinematic equations:

Linear Motion (Constant a)	Rotational Motion (Constant $\alpha$)
$v = u + at$	$ \omega = \omega_0 + \alpha t $
$s = ut + \frac{1}{2}at^2$	$ \theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2 $ (using $\Delta\theta = \theta - \theta_0$)
$v^2 = u^2 + 2as$	$ \omega^2 = \omega_0^2 + 2\alpha (\theta - \theta_0) $
$s = \frac{1}{2}(u+v)t$	$ \theta - \theta_0 = \frac{1}{2}(\omega_0+\omega)t $

Here, $\omega_0$ is the initial angular velocity, $\omega$ is the final angular velocity, $\alpha$ is the constant angular acceleration, $t$ is the time interval, and $\theta - \theta_0$ is the angular displacement during time $t$. These equations are valid only for rotational motion about a fixed axis with constant angular acceleration.

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Dynamics Of Rotational Motion About A Fixed Axis

Dynamics deals with the forces and torques that cause motion. For a rigid body rotating about a fixed axis, the net external torque causes angular acceleration, and the moment of inertia quantifies the body's resistance to this change in rotational motion.

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Rotational Analogue of Newton's Second Law

The relationship between the net external torque ($\vec{\tau}_{net, ext}$) acting on a rigid body and its angular acceleration ($\vec{\alpha}$) about a fixed axis is given by:

$ \vec{\tau}_{net, ext} = I \vec{\alpha} $

where $I$ is the moment of inertia of the body about the fixed axis of rotation.

This equation is the fundamental equation of rotational dynamics for a rigid body about a fixed axis. It directly mirrors Newton's Second Law for translational motion ($\vec{F}_{net} = m\vec{a}$), with torque ($\vec{\tau}$) replacing force ($\vec{F}$), moment of inertia ($I$) replacing mass ($m$), and angular acceleration ($\vec{\alpha}$) replacing linear acceleration ($\vec{a}$).

In many fixed-axis problems, the net external torque vector $\vec{\tau}_{net, ext}$ and the angular acceleration vector $\vec{\alpha}$ are both directed along the axis of rotation. In such cases, we can work with their scalar components along the axis:

$ \tau_{net, ext} = I \alpha $

where $\tau_{net, ext}$ is the component of the net external torque along the axis of rotation and $\alpha$ is the component of the angular acceleration along the axis. We typically assign a sign (e.g., positive for counterclockwise rotation/acceleration when the axis is pointing out of the page) to these components.

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Work Done by Torque in Rotational Motion

When a torque acts on a rigid body and causes it to rotate through an angular displacement, the torque does work. For a constant torque $\tau$ about a fixed axis and an angular displacement $\Delta\theta$ about the same axis, the work done ($W$) is:

$ W = \tau \Delta\theta $

In general, for a variable torque, the work done during an infinitesimal angular displacement $d\theta$ is $dW = \tau \, d\theta$. The total work done during a finite angular displacement from $\theta_1$ to $\theta_2$ is:

$ W = \int_{\theta_1}^{\theta_2} \tau \, d\theta $

The units of work are Joules (J).

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Rotational Kinetic Energy

A rotating rigid body possesses kinetic energy due to the motion of its constituent particles. This is called rotational kinetic energy ($KE_{rot}$).

Consider a rigid body rotating about a fixed axis with angular velocity $\omega$. A particle of mass $m_i$ at a perpendicular distance $r_i$ from the axis moves with linear speed $v_i = r_i\omega$. The kinetic energy of this particle is $\frac{1}{2}m_i v_i^2 = \frac{1}{2}m_i (r_i\omega)^2 = \frac{1}{2}m_i r_i^2 \omega^2$.

The total rotational kinetic energy of the rigid body is the sum of the kinetic energies of all its particles:

$ KE_{rot} = \sum_i \frac{1}{2} m_i r_i^2 \omega^2 $

$ KE_{rot} = \frac{1}{2} \left(\sum_i m_i r_i^2\right) \omega^2 $

The term in the parenthesis is the moment of inertia $I$ about the axis of rotation ($\sum_i m_i r_i^2 = I$).

So, the rotational kinetic energy of a rigid body rotating about a fixed axis with angular velocity $\omega$ is:

$ KE_{rot} = \frac{1}{2} I \omega^2 $

This is the rotational analogue of translational kinetic energy $KE_{trans} = \frac{1}{2} m v^2$, with $I$ replacing $m$ and $\omega$ replacing $v$.

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Work-Energy Theorem for Rotational Motion

The net work done by external torques on a rigid body rotating about a fixed axis is equal to the change in its rotational kinetic energy.

$ W_{net, ext} = \Delta KE_{rot} = \frac{1}{2} I \omega_{final}^2 - \frac{1}{2} I \omega_{initial}^2 $

This is the rotational version of the Work-Energy Theorem.

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Angular Momentum In Case Of Rotation About A Fixed Axis

We defined the angular momentum of a single particle relative to an origin as $\vec{L} = \vec{r} \times \vec{p}$. For a rigid body rotating about a fixed axis, the total angular momentum is the sum of the angular momenta of all its constituent particles.

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Total Angular Momentum of a Rigid Body about a Fixed Axis

Consider a rigid body rotating about a fixed axis (say, the z-axis) with angular velocity $\vec{\omega} = \omega \hat{k}$. A particle of mass $m_i$ is located at position $\vec{r}_i$ relative to the origin. The linear velocity of this particle is $\vec{v}_i = \vec{\omega} \times \vec{r}_i$. The angular momentum of this particle relative to the origin is $\vec{l}_i = \vec{r}_i \times \vec{p}_i = \vec{r}_i \times (m_i \vec{v}_i) = m_i (\vec{r}_i \times \vec{v}_i)$.

Substituting $\vec{v}_i = \vec{\omega} \times \vec{r}_i$, we get $\vec{l}_i = m_i [\vec{r}_i \times (\vec{\omega} \times \vec{r}_i)]$. This expression involves a vector triple product. For a particle at a perpendicular distance $r_i$ from the z-axis, the angular momentum vector $\vec{l}_i$ is generally not parallel to the angular velocity vector $\vec{\omega}$ (which is along the axis), unless the particle is on the axis or all particles are on the axis (which is not a rigid body).

However, for rotation about a fixed axis (like the z-axis), the component of the angular momentum of each particle along the axis of rotation adds up simply. Let $r_{i,perp}$ be the perpendicular distance of $m_i$ from the axis. The component of $\vec{l}_i$ along the z-axis is $(l_i)_z$. Summing the z-components for all particles:

$ L_z = \sum_i (l_i)_z = \sum_i m_i r_{i,perp}^2 \omega $

This sum $\sum_i m_i r_{i,perp}^2$ is the moment of inertia $I$ of the rigid body about the fixed axis of rotation.

So, the component of the total angular momentum of the rigid body along the fixed axis of rotation is:

$ L_z = I \omega $

This scalar equation relates the component of angular momentum along the axis to the moment of inertia about that axis and the angular speed. If the fixed axis is a symmetry axis for the body, then the total angular momentum vector $\vec{L}$ is parallel to the angular velocity vector $\vec{\omega}$, and we can write $\vec{L} = I \vec{\omega}$. In general, however, $\vec{L}$ and $\vec{\omega}$ are not parallel unless the axis of rotation is a principal axis of inertia.

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Rotational Analogue of Newton's Second Law (in terms of Angular Momentum)

We previously established that the net external torque on a rigid body is equal to the rate of change of its total angular momentum: $\vec{\tau}_{net, ext} = \frac{d\vec{L}}{dt}$.

For rotation about a fixed axis (say, the z-axis), the component of the net external torque along the axis is equal to the rate of change of the component of angular momentum along that axis:

$ (\tau_{net, ext})_z = \frac{d L_z}{dt} $

Substituting $L_z = I\omega$ (where $I$ is constant for a fixed axis):

$ (\tau_{net, ext})_z = \frac{d (I\omega)}{dt} = I \frac{d\omega}{dt} = I \alpha $

This recovers the fundamental equation of rotational dynamics $\tau = I\alpha$ for the component along the fixed axis.

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Conservation Of Angular Momentum

Just as the total linear momentum of a system is conserved when the net external force is zero, the total angular momentum of a system is conserved when the net external torque is zero.

From the rotational analogue of Newton's Second Law for a system:

$ \vec{\tau}_{net, ext} = \frac{d\vec{L}_{total}}{dt} $

where $\vec{L}_{total}$ is the total angular momentum of the system (sum of angular momenta of all particles or parts) relative to a chosen origin, and $\vec{\tau}_{net, ext}$ is the net external torque on the system relative to the same origin.

Statement of Conservation of Angular Momentum: If the net external torque acting on a system is zero, the total angular momentum of the system remains constant (conserved).

$ \text{If } \vec{\tau}_{net, ext} = 0 \text{, then } \vec{L}_{total} = \text{constant} $

This is a vector conservation law. If the net torque about a specific axis is zero, the component of angular momentum about that axis is conserved.

Applications of Conservation of Angular Momentum:

• Spinning Ice Skater: When an ice skater pulls her arms and legs inwards, she decreases her moment of inertia about the vertical axis. Since there is negligible external torque about this axis, her angular momentum ($L = I\omega$) is conserved. As $I$ decreases, her angular speed $\omega$ must increase, causing her to spin faster.
• Diver or Gymnast: A diver or gymnast can control their rate of rotation in the air by changing their body shape, thus changing their moment of inertia about their axis of rotation. Tucking the body inwards decreases $I$, increasing $\omega$; stretching the body outwards increases $I$, decreasing $\omega$.
• Planetary Motion: Kepler's Second Law (Law of Areas) is a direct consequence of the conservation of angular momentum of a planet orbiting the Sun. The gravitational force acts along the line joining the planet and the Sun, so the torque about the Sun is zero ($\vec{\tau} = \vec{r} \times \vec{F}_g = \vec{r} \times (-G\frac{M_S m_p}{r^2}\hat{r}) = 0$). Since the net torque is zero, the angular momentum of the planet about the Sun is conserved.

The conservation of angular momentum is a fundamental principle, arising from the rotational symmetry of space (i.e., the laws of physics are the same regardless of orientation in space).

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Rolling Motion

Rolling motion is a common type of motion exhibited by objects like wheels, spheres, and cylinders on a surface. It is a combination of translational motion (the object as a whole moves from one place to another) and rotational motion (the object spins about an axis).

The special characteristic of pure rolling motion is that there is no slipping between the object and the surface it is rolling on. At the point of contact between the rolling object and the surface, the instantaneous velocity of the point of contact on the object is zero relative to the surface.

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Condition for Pure Rolling

Consider a rigid body (like a wheel or cylinder) of radius $R$ rolling on a horizontal surface. Let $\vec{v}_{CM}$ be the linear velocity of its centre of mass, and $\omega$ be its angular speed of rotation about the axis passing through the centre of mass. The axis of rotation for rolling is generally the axis passing through the centre of mass and perpendicular to the direction of motion.

For a point P on the circumference of the object at the lowest point, touching the surface:

The velocity of point P relative to the centre of mass is $\vec{v}_{P,CM} = \vec{\omega} \times \vec{r}$, where $\vec{r}$ is the vector from CM to P. If CM velocity is $\vec{v}_{CM}$ (say, in +x direction), and rotation is such that point P moves backward relative to CM (clockwise rotation if rolling in +x direction), the velocity of P relative to CM has magnitude $v_{P,CM} = R\omega$ and is in the -x direction.

The absolute velocity of point P relative to the ground is $\vec{v}_P = \vec{v}_{CM} + \vec{v}_{P,CM}$.

For pure rolling (no slipping), the point of contact P has zero instantaneous velocity relative to the surface. If the surface is stationary, the velocity of P relative to the ground must be zero.

$ v_P = v_{CM} - R\omega $ (magnitudes, assuming $\vec{v}_{CM}$ and the linear velocity component $R\omega$ due to rotation at the contact point are opposite)

For no slipping, $v_P = 0$:

$ v_{CM} - R\omega = 0 $

$ v_{CM} = R\omega $

This is the condition for pure rolling: the linear speed of the centre of mass is equal to the product of the radius and the angular speed about the CM axis. If $v_{CM} > R\omega$, there is forward slipping. If $v_{CM} < R\omega$, there is backward slipping.

Taking the time derivative, the condition for pure rolling with acceleration is $a_{CM} = R\alpha$, where $a_{CM}$ is the linear acceleration of the CM and $\alpha$ is the angular acceleration about the CM axis.

(Image Placeholder: A wheel on a surface. Arrow shows v_CM of the center. Curved arrow shows omega. Point at the bottom contact is labelled P. Arrow v_P,CM points left with magnitude R*omega. Arrow v_CM points right. Label v_P = v_CM - R*omega = 0 for rolling.)

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Kinetic Energy Of Rolling Motion

Since rolling motion is a combination of translation and rotation, the total kinetic energy of a rigid body undergoing rolling motion is the sum of its translational kinetic energy and its rotational kinetic energy about an axis through its centre of mass.

$ KE_{total} = KE_{trans} + KE_{rot} $

The translational kinetic energy is due to the motion of the centre of mass:

$ KE_{trans} = \frac{1}{2} M v_{CM}^2 $

where $M$ is the total mass of the body and $v_{CM}$ is the speed of the centre of mass.

The rotational kinetic energy is due to the rotation about the axis passing through the centre of mass (which is the axis of rotation in the CM frame):

$ KE_{rot} = \frac{1}{2} I_{CM} \omega^2 $

where $I_{CM}$ is the moment of inertia about the axis through the centre of mass, and $\omega$ is the angular speed of rotation about this axis.

So, the total kinetic energy of a rolling body is:

$ KE_{total} = \frac{1}{2} M v_{CM}^2 + \frac{1}{2} I_{CM} \omega^2 $

For pure rolling, we have the relation $v_{CM} = R\omega$, or $\omega = v_{CM}/R$. We can substitute this into the formula for $KE_{rot}$:

$ KE_{total} = \frac{1}{2} M v_{CM}^2 + \frac{1}{2} I_{CM} \left(\frac{v_{CM}}{R}\right)^2 $

$ KE_{total} = \frac{1}{2} M v_{CM}^2 + \frac{1}{2} I_{CM} \frac{v_{CM}^2}{R^2} $

$ KE_{total} = \frac{1}{2} v_{CM}^2 \left(M + \frac{I_{CM}}{R^2}\right) $

This formula shows that the total kinetic energy of a rolling body is the sum of the kinetic energy of a particle of mass M moving with velocity $v_{CM}$ and an additional term related to its rotational inertia.

Alternatively, we can express $I_{CM}$ using the radius of gyration $k_{CM}$ about the CM axis: $I_{CM} = M k_{CM}^2$.

$ KE_{total} = \frac{1}{2} M v_{CM}^2 + \frac{1}{2} (M k_{CM}^2) \frac{v_{CM}^2}{R^2} $

$ KE_{total} = \frac{1}{2} M v_{CM}^2 \left(1 + \frac{k_{CM}^2}{R^2}\right) $

The ratio $\frac{k_{CM}^2}{R^2}$ depends on the shape of the object (e.g., for a solid sphere, $I_{CM} = \frac{2}{5}MR^2$, so $k_{CM}^2 = \frac{2}{5}R^2$, and $\frac{k_{CM}^2}{R^2} = \frac{2}{5}$).

This formula is useful for solving problems involving rolling motion using energy conservation, such as a body rolling down an inclined plane. As the body rolls down, its potential energy ($PE_g = Mgh$) is converted into translational and rotational kinetic energy.

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