11.1 Uniform Circular Motion, Centripetal Force & Moment of Inertia

Uniform Circular Motion, Centripetal Force & Moment of Inertia

In the Intermediate Physics Part I syllabus and the University of Engineering and Technology (UET) ECAT engineering entrance examination, circular motion and rotational dynamics represent core problem-solving domains. Unlike pure rectilinear motion, rotational mechanics requires analyzing angular kinematics, vector cross products, distributed mass distributions, and energy partitioning between translational and rotational degrees of freedom. Mastering the relations between linear and angular parameters, the mechanics of banked curves, moments of inertia of symmetric bodies, and the law of conservation of angular momentum provides essential problem-solving speed without reliance on electronic calculators.

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1. Angular Kinematics and Linear Relationships

When a particle traces a curved or circular trajectory of radius $r$, its spatial position is most conveniently described using polar angular variables referenced to a central rotation axis.

Angular Displacement ($\theta$)

Angular displacement is defined as the angle subtended at the center of curvature by an arc of length $s$ traced along the circumference:

$\theta = \frac{s}{r}$

Where:

• $s$ is the arc length traversed (meters, $\text{m}$).
• $r$ is the radius of the circular path (meters, $\text{m}$).
• $\theta$ is the angular displacement in radians ($\text{rad}$).

Although the radian is formally a dimensionless SI unit (listed as supplementary in older texts) (ratio of two lengths, $[L][L]^{-1} = [M^0 L^0 T^0]$), it serves as the required unit in all physical kinematic formulas. Angular displacement conversions frequently tested on the ECAT include:

$1\text{ revolution (rev)} = 2\pi\text{ rad} = 360^\circ$

$1\text{ rad} = \left(\frac{180}{\pi}\right)^\circ \approx 57.3^\circ$

Directionality: Infinitesimal angular displacement $d\boldsymbol{\theta}$ is a true vector whose direction is given by the right-hand grip rule: curl the fingers of the right hand along the direction of rotation, and the outstretched thumb points along the vector axis of $d\boldsymbol{\theta}$. Finite angular displacements do not commute under vector addition ($\boldsymbol{\theta}_1 + \boldsymbol{\theta}_2 \ne \boldsymbol{\theta}_2 + \boldsymbol{\theta}_1$) and are therefore treated as scalar quantities.

Angular Velocity ($\omega$)

The time rate of change of angular displacement is defined as angular velocity:

$\omega_{\text{avg}} = \frac{\Delta\theta}{\Delta t}, \qquad \boldsymbol{\omega} = \lim_{\Delta t \to 0} \frac{\Delta\boldsymbol{\theta}}{\Delta t} = \frac{d\boldsymbol{\theta}}{dt}$

• SI Unit: radians per second ($\text{rad}\cdot\text{s}^{-1}$), with dimensional formula $[T^{-1}]$.
• Rotational speed is frequently expressed in revolutions per minute ($\text{rpm}$):

$1\text{ rpm} = \frac{1\text{ rev}}{60\text{ s}} = \frac{2\pi\text{ rad}}{60\text{ s}} = \frac{\pi}{30}\text{ rad}\cdot\text{s}^{-1} \approx 0.1047\text{ rad}\cdot\text{s}^{-1}$

Angular Acceleration ($\alpha$)

The time rate of change of angular velocity is angular acceleration:

$\alpha_{\text{avg}} = \frac{\Delta\omega}{\Delta t}, \qquad \boldsymbol{\alpha} = \lim_{\Delta t \to 0} \frac{\Delta\boldsymbol{\omega}}{\Delta t} = \frac{d\boldsymbol{\omega}}{dt} = \frac{d^2\boldsymbol{\theta}}{dt^2}$

• SI Unit: radians per second squared ($\text{rad}\cdot\text{s}^{-2}$), dimensions $[T^{-2}]$.

Equations of Constant Angular Acceleration

For a rigid body rotating about a fixed axis with uniform angular acceleration $\alpha = \text{constant}$, the rotational equations of motion directly mirror rectilinear kinematic equations:

Coupling Linear and Angular Variables

For a point at distance $r$ from the fixed axis of rotation:

1. Linear Arc Displacement: $s = r\theta$
2. Tangential Linear Velocity: In vector notation, $\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$. Because $\boldsymbol{\omega} \perp \mathbf{r}$ in planar circular motion, the magnitude is:

$v = r\omega$

1. Tangential Linear Acceleration ($a_t$): Responsible for altering the linear speed of the particle:

$\mathbf{a}_t = \boldsymbol{\alpha} \times \mathbf{r} \implies a_t = r\alpha$

1. Centripetal / Radial Acceleration ($a_c$): Responsible for continuously altering the direction of the velocity vector, directed perpendicular to $\mathbf{v}$ and radially inward toward the rotation center:

$a_c = \frac{v^2}{r} = \frac{(r\omega)^2}{r} = r\omega^2 = v\omega$

In non-uniform circular motion (where $\alpha \ne 0$), the particle experiences both tangential acceleration and radial centripetal acceleration simultaneously. Because $\mathbf{a}_t \perp \mathbf{a}_c$, the magnitude of the total linear acceleration vector is:

$a_{\text{total}} = \sqrt{a_t^2 + a_c^2} = \sqrt{(r\alpha)^2 + (r\omega^2)^2}$

The angle $\phi$ of total acceleration relative to the radial inward vector is $\tan\phi = \frac{a_t}{a_c}$. In uniform circular motion (constant angular speed $\omega$), $\alpha = 0 \implies a_t = 0$, so $a_{\text{total}} = a_c$.

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2. Centripetal Force and Banking of Roads

The Centripetal Force Law

By Newton's second law of motion, any body of mass $m$ traversing a circular arc of radius $r$ at tangential speed $v$ must experience a net resultant force directed toward the instantaneous center of curvature, termed the centripetal force ($F_c$):

$F_c = m a_c = \frac{m v^2}{r} = m r \omega^2 = m v \omega$

Critical Conceptual Fundamentals for ECAT:

• Centripetal force is not an autonomous new type of physical interaction; it is an orbital requirement provided by existing real physical forces, such as string tension for a whirling bob, gravitational pull for planetary orbits ($G \frac{M m}{r^2} = \frac{m v^2}{r}$), Coulomb attraction for orbital electrons, or tire-road friction for turning vehicles.
• Work Done by Centripetal Force is Strictly Zero: Because the instantaneous velocity vector $\mathbf{v}$ (and instantaneous displacement $d\mathbf{s}$) is tangent to the circle and $\mathbf{F}_c$ is strictly radial, $\mathbf{F}_c \cdot d\mathbf{s} = F_c ds \cos(90^\circ) = 0$. Centripetal force alters direction without changing kinetic energy or speed.

Banking of Curved Roads and Tracks

When an automobile rounds an unbanked flat horizontal curve of radius $r$, the centripetal force is supplied entirely by lateral static friction between the road and tires:

$f_s \le \mu_s N = \mu_s m g \implies \frac{m v^2}{r} \le \mu_s m g \implies v_{\text{max}} = \sqrt{\mu_s g r}$

If the road is icy or wet ($\mu_s \to 0$), high-speed turning causes skidding. To eliminate dependence on friction, engineers bank the roadway by elevating the outer edge at an angle $\theta$ relative to the horizontal.

On a frictionless banked incline of angle $\theta$, the normal contact force $\mathbf{N}$ exerted by the road surface acts perpendicular to the incline. Resolving $\mathbf{N}$ into orthogonal components:

1. Vertical equilibrium: The vertical component balances gravitational weight:

$N \cos\theta = m g$

1. Horizontal radial acceleration: The horizontal component points directly toward the center of curvature, providing the required centripetal acceleration:

$N \sin\theta = \frac{m v^2}{r}$

Dividing the horizontal equation by the vertical equation eliminates the mass $m$ and normal force $N$:

$\frac{N \sin\theta}{N \cos\theta} = \frac{m v^2 / r}{m g} \implies \tan\theta = \frac{v^2}{r g}$

Solving for the ideal rated design speed $v_{\text{ideal}}$ (at which zero lateral friction is required):

$v_{\text{ideal}} = \sqrt{r g \tan\theta}$

ECAT Analysis for Deviations from Design Speed:

• If $v > v_{\text{ideal}}$, the vehicle tends to skid up the embankment; static friction acts down the incline along the road surface to prevent outward sliding.
• If $v < v_{\text{ideal}}$, the vehicle tends to slide down the embankment; static friction acts up the incline to prevent inward slipping.

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3. Moment of Inertia (Rotational Inertia)

In linear mechanics, mass $m$ represents inertia—the resistance of an object to changes in rectilinear velocity. In rotational mechanics, the resistance of a rigid body to changes in its rotational state (angular acceleration $\alpha$) depends not only on total mass, but critically on how that mass is distributed relative to the axis of rotation.

Mathematical Formulation

For a system of $N$ discrete particles of masses $m_1, m_2, \dots, m_N$ positioned at perpendicular distances $r_1, r_2, \dots, r_N$ from the rotation axis, the moment of inertia $I$ is:

$I = \sum_{i=1}^N m_i r_i^2$

For continuous rigid bodies of mass density $\rho$, the summation generalizes to the integral $I = \int r^2 dm$.

• SI Unit: kilogram meter squared ($\text{kg}\cdot\text{m}^2$).
• Dimensional Formula: $[M L^2]$.
• Torque relation: $\tau_{\text{net}} = I \alpha$, directly analogous to Newton's second law $F_{\text{net}} = m a$.

Radius of Gyration ($k$)

The radius of gyration $k$ is the radial distance from the axis of rotation at which the entire mass $M$ of a rigid body could be concentrated as a point mass without altering its moment of inertia:

$I = M k^2 \implies k = \sqrt{\frac{I}{M}}$

Standard Moments of Inertia of Symmetric Bodies

The following table summarizes moments of inertia about central principal axes of symmetry for standard geometric bodies tested in ECAT:

ECAT Rapid Recognition: For identical mass $M$ and exterior radius $r$, the moments of inertia rank strictly in order of mass concentration away from the central axis: $I_{\text{hoop}} > I_{\text{shell}} > I_{\text{disc}} > I_{\text{sphere}}$.

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4. Rotational Kinetic Energy and Rolling Down an Incline

Rotational Kinetic Energy

A rigid body rotating with angular velocity $\omega$ possesses rotational kinetic energy equal to the sum of kinetic energies of its constituent particles:

$K_{\text{rot}} = \sum \frac{1}{2} m_i v_i^2 = \sum \frac{1}{2} m_i (r_i \omega)^2 = \frac{1}{2} \left(\sum m_i r_i^2\right) \omega^2 = \frac{1}{2} I \omega^2$

Pure Rolling Motion Without Slipping

When a circular body of mass $M$ and radius $r$ rolls without slipping along a surface, its contact point with the surface is instantaneously at rest. The linear velocity of the center of mass $v_{\text{cm}}$ and the angular velocity $\omega$ are constrained by the rolling condition:

$v = v_{\text{cm}} = r\omega \implies \omega = \frac{v}{r}$

The total kinetic energy $K_{\text{total}}$ of the rolling body is the sum of translational and rotational kinetic energy:

$K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} M v^2 + \frac{1}{2} I \left(\frac{v}{r}\right)^2 = \frac{1}{2} M v^2 \left(1 + \frac{I}{M r^2}\right)$

Defining the dimensionless shape factor $\beta = \frac{I}{M r^2}$:

$K_{\text{total}} = \frac{1}{2} M v^2 (1 + \beta)$

Energy Partition Ratios:

• Hoop ($\beta = 1$): $K_{\text{total}} = M v^2 \implies K_{\text{trans}} = 50\%, \quad K_{\text{rot}} = 50\%$.
• Solid Disc ($\beta = 1/2$): $K_{\text{total}} = \frac{3}{4} M v^2 \implies K_{\text{trans}} = \frac{2}{3} K_{\text{total}} \approx 66.7\%, \quad K_{\text{rot}} = \frac{1}{3} K_{\text{total}} \approx 33.3\%$.
• Solid Sphere ($\beta = 2/5$): $K_{\text{total}} = \frac{7}{10} M v^2 \implies K_{\text{trans}} = \frac{5}{7} K_{\text{total}} \approx 71.4\%, \quad K_{\text{rot}} = \frac{2}{7} K_{\text{total}} \approx 28.6\%$.

Speed and Acceleration Down an Inclined Plane

When a body of mass $M$ and radius $r$ rolls from rest without slipping down a rough incline of height $h$ and inclination angle $\theta$, gravitational potential energy converts into total rolling kinetic energy (static friction supplies the torque for rolling but does no work, because the contact point is momentarily at rest):

$M g h = \frac{1}{2} M v^2 \left(1 + \frac{I}{M r^2}\right)$

Solving for the linear velocity $v$ at the bottom of the incline:

$v = \sqrt{\frac{2 g h}{1 + \frac{I}{M r^2}}} = \sqrt{\frac{2 g h}{1 + \beta}}$

Using kinematics ($v^2 = 2 a s = 2 a \frac{h}{\sin\theta}$), the linear acceleration $a$ along the incline is:

$a = \frac{g \sin\theta}{1 + \frac{I}{M r^2}} = \frac{g \sin\theta}{1 + \beta}$

Exam Quantitative Comparison:

• Solid Sphere ($\beta = 0.4$): $v = \sqrt{\frac{2 g h}{1.4}} = \sqrt{\frac{10}{7} g h} \approx 1.195 \sqrt{g h}, \quad a = \frac{5}{7} g \sin\theta \approx 0.714 g \sin\theta$.
• Solid Disc ($\beta = 0.5$): $v = \sqrt{\frac{2 g h}{1.5}} = \sqrt{\frac{4}{3} g h} \approx 1.155 \sqrt{g h}, \quad a = \frac{2}{3} g \sin\theta \approx 0.667 g \sin\theta$.
• Spherical Shell ($\beta = 0.67$): $v = \sqrt{\frac{6}{5} g h} \approx 1.095 \sqrt{g h}, \quad a = \frac{3}{5} g \sin\theta = 0.600 g \sin\theta$.
• Hoop ($\beta = 1.0$): $v = \sqrt{g h} = 1.000 \sqrt{g h}, \quad a = \frac{1}{2} g \sin\theta = 0.500 g \sin\theta$.

ECAT Incline Race Rule: The speed and acceleration of a rolling body down an inclined plane are completely independent of its mass $M$ and radius $r$. They depend exclusively on the shape factor $\beta = \frac{I}{M r^2}$. Because $\beta_{\text{sphere}} < \beta_{\text{disc}} < \beta_{\text{shell}} < \beta_{\text{hoop}}$, the bodies reach the bottom in order: Sphere first (fastest) $\to$ Disc second $\to$ Spherical Shell third $\to$ Hoop last (slowest).

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5. Angular Momentum and Conservation Law

Vector Angular Momentum ($\mathbf{L}$)

The angular momentum $\mathbf{L}$ of a particle of linear momentum $\mathbf{p} = m\mathbf{v}$ relative to an origin is defined by the cross product:

$\mathbf{L} = \mathbf{r} \times \mathbf{p} = m (\mathbf{r} \times \mathbf{v})$

• Magnitude: $L = r p \sin\theta = m v r \sin\theta$, where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{p}$.
• For a rigid body rotating about a principal symmetry axis with angular velocity $\boldsymbol{\omega}$:

$\mathbf{L} = I \boldsymbol{\omega}$

• SI Unit: Joule-second ($\text{J}\cdot\text{s}$) or kilogram meter squared per second ($\text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}$).
• Dimensional Formula: $[M L^2 T^{-1}]$ — notably identical to Planck's constant $h$.

Torque and Rate of Change of Angular Momentum

Differentiating $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ with respect to time:

$\frac{d\mathbf{L}}{dt} = \left(\frac{d\mathbf{r}}{dt} \times \mathbf{p}\right) + \left(\mathbf{r} \times \frac{d\mathbf{p}}{dt}\right) = (\mathbf{v} \times m\mathbf{v}) + (\mathbf{r} \times \mathbf{F}_{\text{net}})$

Because the cross product of any vector with itself is zero ($\mathbf{v} \times \mathbf{v} = \mathbf{0}$):

$\boldsymbol{\tau}_{\text{net}} = \frac{d\mathbf{L}}{dt}$

This represents the rotational formulation of Newton's second law: the net external torque acting on a body equals the time rate of change of its angular momentum.

Law of Conservation of Angular Momentum

If the net external torque acting on an isolated rotating system is zero ($\boldsymbol{\tau}_{\text{ext}} = \mathbf{0}$), the total angular momentum of the system remains constant in both magnitude and direction over time:

$\frac{d\mathbf{L}}{dt} = \mathbf{0} \implies \mathbf{L} = \text{constant} \implies I_1 \boldsymbol{\omega}_1 = I_2 \boldsymbol{\omega}_2$

Physical Applications Tested on ECAT:

1. Acrobat / Ice Skater: A spinning skater pulls their outstretched arms inward toward the chest. Mass moves closer to the rotation axis, causing the moment of inertia to decrease ($I_2 < I_1$). By conservation of angular momentum, the spin speed increases dramatically ($\omega_2 = \frac{I_1}{I_2} \omega_1 > \omega_1$).
   • Kinetic Energy Increase: $K_{\text{rot}, 2} = \frac{L^2}{2 I_2} > \frac{L^2}{2 I_1} = K_{\text{rot}, 1}$. The additional kinetic energy comes from positive internal chemical work performed by the skater's arm muscles overcoming centrifugal forces.
2. Planetary Keplerian Motion: Gravitational force exerted by the Sun on an orbiting planet is central, passing directly through the Sun's center. Hence, $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}_g = \mathbf{0}$. Therefore, orbital angular momentum is conserved:

$m r_p v_p = m r_a v_a \implies r_p v_p = r_a v_a$

At perihelion (closest distance $r_p$), the planet moves fastest ($v_p$ maximum); at aphelion (farthest distance $r_a$), it moves slowest ($v_a$ minimum).