Rotational Dynamics - Comprehensive Study Notes

Kinematics and Characteristics of Circular Motion

• Foundational Conceptual Inquiries:

  • Circular motion is defined as the motion of an object or particle along the circumference of a circle.
  • The center of mass of a body is a point at which the entire mass of the body can be supposed to be concentrated to investigate its translational motion.
  • Kinematical equations of motion describe the relations between displacement, velocity, acceleration, and time under constant acceleration.
  • Real forces originate from known physical interactions (gravitational, electromagnetic, strong, weak), whereas pseudo forces are fictitious forces introduced in non-inertial (accelerated) frames of reference to apply Newton's laws of motion.

• Circular Motion in Rigid Bodies:

  • Circular motion is an essential part of daily mechanics encountered in revolving and rotating rigid objects.
  • Revolution: During revolution, every particle in the object undergoes circular motion about an external axis or point located outside the object.
  • Rotation: During rotation, the motion takes place about an internal axis of rotation passing through the object itself.

• Primary Characteristics of Circular Motion:

  • Accelerated Motion: The direction of the linear velocity changes continuously at every instant along the tangent to the trajectory; hence, it is intrinsically an accelerated motion even if speed is constant.
  • Periodic Motion: During its travel, the particle repeatedly traverses the identical spatial path at regular intervals, making the motion periodic in space.

• Kinematical Parameters:

  • Translational quantities have direct angular analogues:
  • Angular displacement $\vec{\theta}$ is analogous to linear displacement $\vec{s}$.
  • Angular velocity $\vec{\omega} = \frac{d\vec{\theta}}{dt}$ is analogous to linear velocity $\vec{v} = \frac{d\vec{s}}{dt}$.
  • Angular acceleration $\vec{\alpha} = \frac{d\vec{\omega}}{dt}$ is analogous to linear acceleration $\vec{a} = \frac{d\vec{v}}{dt}$.
  • Relation Between Linear and Angular Velocity:$\vec{v} = \vec{\omega} \times \vec{r}$     where $\vec{r}$ is the position (radius) vector drawn outward from the center of the circular path to the particle. In magnitude:     $v = \omega r$
  • Right-Hand Thumb Rule for $\vec{\omega}$:
  • The direction of the angular velocity vector $\vec{\omega}$ lies permanently along the axis of rotation.
  • Curl the fingers of the right hand along the sense of rotation with the thumb outstretched along the axis; the outstretched thumb points in the vector direction of $\vec{\omega}$.

• Period and Frequency:

  • If $T$ is the period of revolution (periodic time) and $n$ is the frequency of revolution:     $\omega = 2\pi n = \frac{2\pi}{T}$

• Uniform Circular Motion (UCM):

  • When the linear speed of a particle remains constant throughout its circular trajectory, the motion is termed Uniform Circular Motion.
  • Only the direction of the velocity vector changes at every instant, remaining strictly tangential.
  • The acceleration causing this directional change is the radial or centripetal acceleration:     $\vec{a}_r = -\omega^2 \vec{r}$
  • The scalar magnitude of the centripetal acceleration in UCM is constant:     $a = \omega^2 r = \frac{v^2}{r} = v\omega$
  • The direction is always radially inward along $-\vec{r}$ toward the center of curvature.
  • Physical Illustration: Any point on the blade of an electric fan rotating at a constant, steady speed.

• Non-Uniform Circular Motion:
  • When the linear speed of the particle varies (such as when an electric fan is switched ON or OFF), the motion becomes non-uniform.
  • The radial acceleration $\vec{a}_r$ persists because the velocity direction changes, but its magnitude is no longer constant.
  • A tangential acceleration component $\vec{a}_T$ arises along or opposite to the tangential velocity $\vec{v}$ to modify its scalar magnitude:     $\vec{a}_{total} = \vec{a}_r + \vec{a}_T$
  • The changing linear velocity implies that angular velocity $\vec{\omega}$ changes over time via the angular acceleration $\vec{\alpha} = \frac{d\vec{\omega}}{dt}$.
  • For planar rotation, $\vec{\alpha}$ remains axial:
  • Directed parallel to $\vec{\omega}$ if rotational speed is increasing.
  • Directed antiparallel to $\vec{\omega}$ if rotational speed is decreasing.

• Off-Axis Angular Acceleration Dynamics:

  • If $\vec{\alpha}$ has any component perpendicular to the axis of rotation, it alters the direction of $\vec{\omega}$.
  • Because $\vec{\omega}$ is perpendicular to the plane of rotation, altering the direction of $\vec{\omega}$ tilts the instantaneous plane of rotation.
  • If $\vec{\alpha}$ is constant in magnitude and perpetually perpendicular to $\vec{\omega}$, it changes only the direction of $\vec{\omega}$ without altering its magnitude, creating continuous precession of the plane of rotation (analogous to linear centripetal acceleration acting perpendicularly to linear velocity).
  • When $\vec{\alpha}$ is constant and directed purely along the axis of rotation, scalar kinematic equations are valid.

• Worked Example 1.1:

  • Problem Statement: An electric fan rotates at $90\,\text{rpm}$. It is switched OFF and comes to a complete rest after completing $21$ rotations. Calculate the time taken to stop assuming constant frictional torque.
  • Given Data:
  • Initial frequency: $n_0 = 90\,\text{rpm} = \frac{90}{60}\,\text{rev/s} = 1.5\,\text{rps}$
  • Initial angular velocity: $\omega_0 = 2\pi n_0 = 2\pi(1.5) = 3\pi\,\text{rad/s}$
  • Total angular displacement: $\theta = 2\pi N = 2\pi(21) = 42\pi\,\text{rad}$
  • Final angular velocity: $\omega = 0\,\text{rad/s}$
  • Kinematical Solution:$\alpha = \frac{\omega - \omega_0}{t} = \frac{\omega^2 - \omega_0^2}{2\theta}$$\frac{0 - 3\pi}{t} = \frac{0 - (3\pi)^2}{2(42\pi)}$$\frac{-3\pi}{t} = \frac{-9\pi^2}{84\pi} = \frac{-3\pi}{28}$$t = 28\,\text{s}$
  • Alternative Unit Method: Measuring displacement directly in revolutions without $\pi$ yields $\omega_0 = 1.5\,\text{rev/s}$, $\theta = 21\,\text{rev}$, giving identical result $t = 28\,\text{s}$.

Dynamics of Circular Motion: Centripetal and Centrifugal Forces

• Centripetal Force (CPF):

  • Centripetal force is the net real force responsible for maintaining circular motion by supplying centripetal acceleration:     $\vec{F}_{CPF} = -m\omega^2 \vec{r} = -\frac{mv^2}{r^2}\vec{r}$
  • In scalar magnitude:     $F_{CPF} = mr\omega^2 = \frac{mv^2}{r} = mv\omega$
  • The term "centripetal" is an adjective describing the direction of the force (center-seeking along the radius); it is not a distinct fundamental interaction of nature (such as gravity or electromagnetism).
  • Any motion directed toward a central point is not necessarily circular; for example, simple harmonic motion also has a central restoring force. A precise, matching tangential velocity is necessary to produce a circular path.
  • Analyzed exclusively from an inertial frame of reference, centripetal force represents the true vector resultant of all physical forces acting on the particle.

• Centrifugal Force (c.f.f.):

  • Centrifugal force is a pseudo (fictitious or non-real) force that arises strictly when analyzing motion within a non-inertial, rotating reference frame rotating with angular velocity $\vec{\omega}$.
  • It has magnitude equal to the centripetal force but acts radially outward away from the center:     $\vec{F}_{cff} = +m\omega^2 \vec{r}$
  • Centrifugal force is not imaginary; an observer within the rotating frame experiences an outward pull, and physical measuring instruments (e.g., spring balances) register this force within that frame.
  • In the rotating frame of reference, the particle is statically at rest because the centrifugal pseudo force balances the resultant of all real forces.

• Dynamic Equations in Different Reference Frames:

  • Inertial Frame Formulation:$\sum \vec{F}_{real} = -m\omega^2 \vec{r}$
  • Non-Inertial (Rotating) Frame Formulation:$m\omega^2 \vec{r} + \sum \vec{F}_{real} = 0$

• Experimental Observation (Spring Balance Whirling Activity):

  • A mass suspended from a spring balance causes an initial static extension governed by weight $mg$.
  • When whirled in a horizontal circle, the spring balance extends further, reading an apparent mass higher than the static value.
  • In the non-inertial frame of the rotating mass, the outward centrifugal force stretches the spring further to provide the necessary inward reaction.

Applications of Circular Motion on Flat Roads and Vertical Walls

• Vehicle Along an Unbanked Horizontal Circular Track:
  • Consider a vehicle modeled as a point mass $m$ executing a horizontal turn of radius $r$.
  • Forces Acting on the Vehicle:
  • Weight $mg$ directed vertically downwards.
  • Normal reaction $N$ directed vertically upwards from the road surface.
  • Force of static friction $f_s$ acting horizontally inward between the road and tyres, directed toward the track center $C$.

• Equilibrium and Motion Equations:

  • Vertical balance:       $N = mg$
  • Centripetal requirement provided by static friction:       $f_s = mr\omega^2 = \frac{mv^2}{r}$
  • Force ratio:       $\frac{f_s}{N} = \frac{v^2}{rg}$

• Maximum Safe Speed Calculation:

  • Static friction adjusts with speed up to a maximum threshold:       $(f_s)_{max} = \mu_s N$

  • Substituting into the force ratio:       $\frac{(f_s)_{max}}{N} = \mu_s = \frac{v_{max}^2}{rg}$$v_{max} = \sqrt{\mu_s rg}$

  • Exceeding $v_{max}$ causes the vehicle to skid outwards.

  • Vehicle as an Extended Body and Toppling Mechanics:

• For a real four-wheeled vehicle, the normal reaction is distributed across four tyres and is unequal during a turn.

• The center of mass of the vehicle is situated at an elevation $h$ above ground level.

• The inward frictional force acts at the ground level, while the outward centrifugal force acts through the center of mass.

• These two equal and opposite forces form an unbalance couple (torque) that tends to overturn (topple) the vehicle outwards.

• To avoid toppling, the torque produced by unequal normal reactions must counter this overturning torque.

• When toppling occurs during a turn, the inner wheels lift off the road first.

• Loading the vehicle lower down lowers the center of mass and widens the base, reducing the risk of toppling.

• For a two-wheeler, the rider cannot shift normal reactions across separated tyres; thus, the rider must lean inward at an angle $\theta$ to the vertical so the resultant contact force passes through the center of mass.

  • Well of Death (Wall of Death / मौत का कुआँ):

• A vertical cylindrical chamber of inner radius $r$ inside which a stunt vehicle performs horizontal circular loops along the wall.

• Forces Acting on the Vehicle:

  • Normal reaction $N$ directed horizontally inwards toward the central axis, providing centripetal acceleration.
  • Weight $mg$ acting vertically downwards.
  • Force of static friction $f_s$ acting vertically upwards between the vertical wall and tyres to prevent downward slippage.

• Derivation of Minimum Safe Speed:

  • Horizontal centripetal relation:       $N = mr\omega^2 = \frac{mv^2}{r}$
  • Vertical equilibrium:       $f_s = mg$
  • Condition preventing downward slippage:       $f_s \le \mu_s N$$mg \le \mu_s \left(\frac{mv^2}{r}\right)$$g \le \frac{\mu_s v^2}{r}$$v^2 \ge \frac{rg}{\mu_s}$$v_{min} = \sqrt{\frac{rg}{\mu_s}}$

• Key Operating Conditions:

  • The normal force is:       $N = \frac{mv^2}{r}$$N_{min} = \frac{mg}{\mu_s}$

  • The relation $f_s = \mu_s N$ is strictly satisfied only at $v = v_{min}$; for higher speeds, $f_s$ remains constant at $mg$, while $N$ increases.

  • Real riders must tilt relative to the horizontal wall surface to prevent overturning caused by the torque of their weight.

  • A vertical wall corresponds to a banking angle of $90^\circ$, establishing a strict lower speed limit without an upper limit.

  • Worked Example 1.2:

• Problem Statement: A motorcyclist performs horizontal circles inside a cylindrical well of inner radius $4\,\text{m}$. The coefficient of static friction is $0.4$. Determine the minimum speed and revolution frequency required. (Take $g = 10\,\text{m/s}^2$).

• Calculation:$v_{min} = \sqrt{\frac{rg}{\mu_s}} = \sqrt{\frac{4 \times 10}{0.4}} = \sqrt{100} = 10\,\text{m/s}$$n_{min} = \frac{v_{min}}{2\pi r} = \frac{10}{2 \times \pi \times 4} = \frac{10}{8\pi} \cong 0.4\,\text{rev/s}$

Vehicle on a Banked Road

• The Principle of Banking:

  • Relying on friction alone to negotiate curved roads is unsafe because $\mu_s$ fluctuates with rain, oil, and wear, and road friction causes tyre degradation.
  • To provide centripetal force independently of friction, curved road surfaces are tilted inward at an angle $\theta$ relative to the horizontal; this design is termed the banking of roads.

• Ideal Case: Frictionless Banked Road (Most Safe Speed):

  • Consider a road of radius of curvature $r$ banked at angle $\theta$.
  • Forces acting on the vehicle: weight $mg$ vertically downwards, normal reaction $N$ perpendicular to the inclined road surface.
  • Resolving $N$ into orthogonal components:
  • Vertical component balances weight:       $N\cos(\theta) = mg$
  • Horizontal component provides centripetal acceleration:       $N\sin(\theta) = \frac{mv^2}{r}$

• Dividing the horizontal equation by the vertical equation:     $\tan(\theta) = \frac{v^2}{rg}$

• Most Safe Speed ($v_s$):$v_s = \sqrt{rg\tan(\theta)}$

• Banking Angle Formula:$\theta = \tan^{-1}\left(\frac{v^2}{rg}\right)$

  • Speed Limits on Rough Banked Roads:

• When a vehicle travels at speeds other than $v_s$, static friction along the tyre-road interface prevents inward or outward sliding.

• Lower Speed Limit ($v_1 < v_s$):

  • When speed is below $v_s$, the required centripetal force is less than $N\sin(\theta)$:       $\frac{mv_1^2}{r} < N\sin(\theta)$
  • The vehicle tends to slip down the incline, mobilizing static friction $f_s$ directed upwards along the incline.

* Resolving forces:

      $N\cos(\theta) + f_s\sin(\theta) = mg$$N\sin(\theta) - f_s\cos(\theta) = \frac{mv_1^2}{r}$ * Setting $f_s = (f_s)_{max} = \mu_s N$ for minimum speed:       $N(\cos(\theta) + \mu_s\sin(\theta)) = mg$$N(\sin(\theta) - \mu_s\cos(\theta)) = \frac{mv_{min}^2}{r}$ * Dividing equations:       $\frac{v_{min}^2}{rg} = \frac{\sin(\theta) - \mu_s\cos(\theta)}{\cos(\theta) + \mu_s\sin(\theta)} = \frac{\tan(\theta) - \mu_s}{1 + \mu_s\tan(\theta)}$$v_{min} = \sqrt{rg\left(\frac{\tan(\theta) - \mu_s}{1 + \mu_s\tan(\theta)}\right)}$ * If $\mu_s \ge \tan(\theta)$, the minimum speed becomes zero ($v_{min} = 0$), meaning the vehicle will not slip down even when brought to a complete stop.

• Upper Speed Limit ($v_2 > v_s$):
  • When speed exceeds $v_s$, the required centripetal force exceeds $N\sin(\theta)$:       $\frac{mv_2^2}{r} > N\sin(\theta)$
  • The vehicle tends to skid outward and upward, mobilizing static friction $f_s$ directed downwards along the incline.

* Resolving forces:

      $N\cos(\theta) - f_s\sin(\theta) = mg$$N\sin(\theta) + f_s\cos(\theta) = \frac{mv_2^2}{r}$ * Setting $f_s = (f_s)_{max} = \mu_s N$ for maximum speed:       $N(\cos(\theta) - \mu_s\sin(\theta)) = mg$$N(\sin(\theta) + \mu_s\cos(\theta)) = \frac{mv_{max}^2}{r}$ * Dividing equations:       $\frac{v_{max}^2}{rg} = \frac{\sin(\theta) + \mu_s\cos(\theta)}{\cos(\theta) - \mu_s\sin(\theta)} = \frac{\tan(\theta) + \mu_s}{1 - \mu_s\tan(\theta)}$$v_{max} = \sqrt{rg\left(\frac{\tan(\theta) + \mu_s}{1 - \mu_s\tan(\theta)}\right)}$ * If $\mu_s = \cot(\theta)$, the denominator becomes zero, yielding $v_{max} = \infty$. * Because the maximum physical coefficient of static friction is $(\mu_s)_{max} = 1$, any banking angle $\theta \ge 45^\circ$ results in $v_{max} = \infty$. * If friction vanishes ($\mu_s = 0$), both limit formulas collapse back to the safe speed $v = \sqrt{rg\tan(\theta)}$.

• Worked Example 1.3:
  • Problem Statement: A racing track of radius of curvature $9.9\,\text{m}$ is banked at $\theta = \tan^{-1}(0.5)$. The coefficient of static friction between track and tyres is $0.2$. Determine the operational speed limits with a $10\%$ safety margin. ($g = 10\,\text{m/s}^2$
  • Theoretical Limits:$v_{min} = \sqrt{rg\left(\frac{\tan(\theta) - \mu_s}{1 + \mu_s\tan(\theta)}\right)} = \sqrt{(9.9)(10)\left(\frac{0.5 - 0.2}{1 + (0.2)(0.5)}\right)} = \sqrt{99\left(\frac{0.3}{1.1}\right)} = \sqrt{27} \approx 5.196\,\text{m/s}$$v_{max} = \sqrt{rg\left(\frac{\tan(\theta) + \mu_s}{1 - \mu_s\tan(\theta)}\right)} = \sqrt{(9.9)(10)\left(\frac{0.5 + 0.2}{1 - (0.2)(0.5)}\right)} = \sqrt{99\left(\frac{0.7}{0.9}\right)} = \sqrt{77} \approx 8.775\,\text{m/s}$
  • Applying $10\%$ Margins:
  • Minimum allowed speed ($10\%$ higher than theoretical minimum):       $(v_{min})_{allowed} = 5.196 \times \frac{110}{100} = 5.716\,\text{m/s}$
  • Maximum allowed speed ($10\%$ lower than theoretical maximum):       $(v_{max})_{allowed} = 8.775 \times \frac{90}{100} = 7.896\,\text{m/s}$

Conical Pendulum and Rotating Systems

• Physical System:
  • A conical pendulum consists of a small point mass (bob) of mass $m$ attached to a light, flexible, inextensible string of length $L$, suspended from a rigid support $A$.
  • The bob revolves in a horizontal circle of radius $r$ with constant speed, while the string sweeps out the lateral surface of a right circular cone of semi-vertical angle $\theta$.

• Force Derivations:

  • Inertial Frame Resolution:
  • Tension $T_0$ acts along the string toward support $A$.
  • Vertical equilibrium:       $T_0\cos(\theta) = mg$
  • Horizontal centripetal force:       $T_0\sin(\theta) = mr\omega^2 = \frac{mv^2}{r}$
  • Non-Inertial Frame Resolution:
  • Within the bob's rest frame, the outward centrifugal force balances the horizontal component of tension:       $T_0\sin(\theta) = mr\omega^2$
  • Geometry:$r = L\sin(\theta)$
  • Angular Velocity and Period:
  • Dividing the horizontal equation by the vertical equation:       $\tan(\theta) = \frac{r\omega^2}{g} = \frac{L\sin(\theta)\omega^2}{g}$$\frac{\sin(\theta)}{\cos(\theta)} = \frac{L\sin(\theta)\omega^2}{g}$$\omega = \sqrt{\frac{g}{L\cos(\theta)}}$
  • Since $\omega = \frac{2\pi}{T}$:       $T = 2\pi\sqrt{\frac{L\cos(\theta)}{g}}$
  • Frequency of revolution:       $n = \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{g}{L\cos(\theta)}}$

• Constraints on String Angle:

  • For a given string length $L$ and local gravity $g$, the period and frequency depend solely on the semi-vertical angle $\theta$.
  • The string can never become completely horizontal ($\theta = 90^\circ$):
  • Dynamical proof: If $\theta = 90^\circ$, tension is purely horizontal; its vertical component $T_0\cos(90^\circ) = 0$, leaving the downward gravitational force $mg$ uncountered.
  • Mathematical proof: Setting $\theta = 90^\circ$ yields $\cos(90^\circ) = 0$, resulting in $T = 0$, $n = \infty$, tension $T_0 = \frac{mg}{\cos(\theta)} = \infty$, and kinetic energy $K.E. = \frac{1}{2}mr^2\omega^2 = \infty$, all of which are physically impossible.

• Worked Example 1.4:

  • Problem Statement: A merry-go-round features a central vertical pillar with horizontal arms of length $H = 2.1\,\text{m}$. Attached to the end of each horizontal arm is an elbow joint supporting an initially vertical rod of length $V = 1.5\,\text{m}$. During rotation, these rods swing outward to an angle $\theta = 37^\circ$ to the vertical. Calculate the frequency of revolution. (Take $g = \pi^2\,\text{m/s}^2$, $\sin(37^\circ) = 0.6$, $\cos(37^\circ) = 0.8$
  • Radius of Path:$r = H + V\sin(37^\circ) = 2.1 + 1.5(0.6) = 2.1 + 0.9 = 3.0\,\text{m}$
  • Dynamic Equations:$T\cos(\theta) = mg$$T\sin(\theta) = mr\omega^2 = 4\pi^2 mrn^2$$\tan(\theta) = \frac{4\pi^2 r n^2}{g}$
  • Frequency Calculation:$\tan(37^\circ) = \frac{\sin(37^\circ)}{\cos(37^\circ)} = \frac{0.6}{0.8} = 0.75 = \frac{3}{4}$$n = \sqrt{\frac{g\tan(37^\circ)}{4\pi^2 r}} = \sqrt{\frac{\pi^2 (0.75)}{4\pi^2 (3.0)}} = \sqrt{\frac{0.75}{12}} = \sqrt{\frac{1}{16}} = \frac{1}{4} = 0.25\,\text{rev/s}$

• Worked Example 1.5 & Funnel Activity:

  • Problem Statement: The semi-vertical angle of a conical funnel is $37^\circ$. A small ball inside rotates along the inner wall. The maximum speed the ball can maintain while remaining in the funnel is $2\,\text{m/s}$. Determine the inner radius at the brim and calculate the operational frequency limits. ($g = 10\,\text{m/s}^2$, $\sin(37^\circ) = 0.6$
  • Forces:
  • Resolving normal reaction $N$:       $N\sin(\theta) = mg$$N\cos(\theta) = \frac{mv^2}{r}$
  • Dividing equations:       $\tan(\theta) = \frac{rg}{v^2} \implies r = \frac{v^2\tan(\theta)}{g}$
  • Maximum Radius at Brim:$r_{max} = \frac{v_{max}^2 \tan(37^\circ)}{g} = \frac{(2)^2 (0.75)}{10} = \frac{3}{10} = 0.3\,\text{m}$
  • Frequency Analysis:
  • Linear and angular velocity relation: $v = r\omega = 2\pi r n$.
  • As a particle approaches the apex/base ($r \to 0$), $v \to 0$, but $n \propto \frac{v}{r} \to \infty$. Hence, no upper limit exists for frequency.
  • The lower frequency limit occurs at the outer brim ($r = r_{max}$, $v = v_{max}$):       $n_{min} = \frac{v_{max}}{2\pi r_{max}} = \frac{2}{2\pi(0.3)} = \frac{1}{0.3\pi} \cong 1\,\text{rev/s}$

Vertical Circular Motion

• Classification of Vertical Circular Motion:

  • Controlled Vertical Circular Motion: The linear speed is either kept constant or maintained via an external drive system (e.g., giant ferris wheel).
  • Gravity-Controlled Vertical Circular Motion: Initial energy is imparted at the lowest point, after which kinetic energy and gravitational potential energy interconvert under the conservative gravitational field.

• Point Mass Tied to a String (Gravity-Controlled):

  • Consider a mass $m$ attached to an inextensible, massless string of radius $r$.
  • The motion is non-uniform because gravity continuously exerts a tangential component that retards upward motion and accelerates downward motion.

• Uppermost Position (A):

  • Both tension $T_A$ and weight $mg$ act downward toward the center:       $T_A + mg = \frac{mv_A^2}{r}$
  • To complete the loop without the string going slack, the minimum condition is $T_A \ge 0$. At the threshold of slacking, $T_A = 0$:       $mg = \frac{m(v_A)_{min}^2}{r}$$(v_A)_{min} = \sqrt{rg}$

• Lowermost Position (B):

  • Tension $T_B$ acts upward while weight $mg$ acts downward:       $T_B - mg = \frac{mv_B^2}{r}$
  • From energy conservation between position A and position B, the decrease in potential energy over vertical displacement $2r$ equals the gain in kinetic energy:       $mg(2r) = \frac{1}{2}m v_B^2 - \frac{1}{2}m v_A^2$$v_B^2 - v_A^2 = 4rg$
  • Substituting $(v_A)_{min}^2 = rg$:       $(v_B)_{min}^2 = rg + 4rg = 5rg$$(v_B)_{min} = \sqrt{5rg}$

• Tension Difference Across Extremes:

  • Subtracting the equation at A from the equation at B:       $(T_B - mg) - (T_A + mg) = \frac{mv_B^2}{r} - \frac{mv_A^2}{r}$$T_B - T_A - 2mg = \frac{m}{r}(v_B^2 - v_A^2)$
  • Using $v_B^2 - v_A^2 = 4rg$:       $T_B - T_A - 2mg = \frac{m}{r}(4rg) = 4mg$$T_B - T_A = 6mg$
  • This tension difference depends strictly on the body's weight and is entirely independent of trajectory radius $r$ and velocities $v_A, v_B$.
  • Because $T_A > 0$ for a non-slack string, the bottom tension must satisfy $T_B > 6mg$.

• Horizontal String Positions (C and D):

  • At positions C and D, the string is horizontal, so weight $mg$ acts perpendicularly to the string and provides no radial component.

  • The centripetal force is supplied entirely by tension:       $T_C = T_D = \frac{mv_C^2}{r}$

  • Potential energy lost from position A to position C is $mgr$:       $mgr = \frac{1}{2}mv_C^2 - \frac{1}{2}mv_A^2$$v_C^2 - v_A^2 = 2rg$$(v_C)_{min}^2 = rg + 2rg = 3rg$$(v_C)_{min} = (v_D)_{min} = \sqrt{3rg}$

  • Evaluating tensions relative to point A:       $T_C - T_A = \frac{m(3rg)}{r} - 0 = 3mg$$T_C = T_D = 3mg + T_A$

  • Point Mass Tied to a Rigid Rod:

• Unlike a flexible string, a rigid massless rod cannot go slack and can support compressive stress.

• Zero velocity is physically permissible at the uppermost position without collapsing:     $(v_A)_{min} = 0$

• Applying conservation of energy to find the corresponding minimum speeds at other points:

  • Lowermost position (B):       $\frac{1}{2}m(v_B)_{min}^2 - 0 = mg(2r)$$(v_B)_{min} = \sqrt{4rg} = 2\sqrt{rg}$

  • Horizontal positions (C and D):       $\frac{1}{2}m(v_C)_{min}^2 - 0 = mgr$$(v_C)_{min} = \sqrt{2rg}$

  • The tension difference between lowermost and uppermost points remains unchanged:       $T_{lowermost} - T_{uppermost} = 6mg$

  • Sphere of Death (मृत्‍यु गोल):

• A circus stunt where motorcycles loop vertically inside a hollow metal sphere.

• Dynamics match vertical circular motion of a mass on a string, with normal reaction $N$ of the spherical wall replacing string tension $T$.

• The motorcycle starts with nearly horizontal circles, where angular speed is high for small radii. As it transitions to large vertical loops, linear speed increases while frequency adjusts.

  • Vehicle at the Crest of a Convex Over-Bridge:

• Consider a vehicle traversing the convex crest of a bridge of radius of curvature $R$.

• Radial Force Balance at the Crest:$mg - N = \frac{mv^2}{R}$$N = mg - \frac{mv^2}{R}$

• As speed $v$ increases, the normal reaction $N$ decreases.

• The vehicle maintains contact with the road as long as $N \ge 0$.

• The maximum speed to maintain road contact occurs when $N = 0$:     $v_{max} = \sqrt{Rg}$

• Exceeding $v_{max}$ causes the vehicle to launch off the surface into projectile flight.

  • Worked Example 1.6:

• Problem Statement: A stone of mass $20\,\text{g}$ is tied to an inextensible string of length $1.8\,\text{m}$ and whirled in a vertical circle. Its speed is $8\,\text{m/s}$ when the string is horizontal (where centripetal force equals tension). Determine:

  • Minimum and maximum kinetic energies over the circle.
  • Angular position $\theta$ (measured from the lowest point, $\theta = 0$) where string tension numerically equals the weight of the stone. (Take $g = 10\,\text{m/s}^2$

• Kinetic Energy Extremes:

  • Mass: $m = 20\,\text{g} = 0.02\,\text{kg}$, $r = 1.8\,\text{m}$, $v_{horiz} = 8\,\text{m/s}$.
  • From horizontal to lowermost point, vertical drop is $r$:       $K.E._{max} - \frac{1}{2}m(8)^2 = mgr$$K.E._{max} = \frac{1}{2}(0.02)(64) + (0.02)(10)(1.8) = 0.64 + 0.36 = 1.0\,\text{J}$
  • Maximum speed:       $v_{max} = \sqrt{\frac{2 K.E._{max}}{m}} = \sqrt{\frac{2(1.0)}{0.02}} = 10\,\text{m/s}$
  • From horizontal to uppermost point, vertical rise is $r$:       $\frac{1}{2}m(8)^2 - K.E._{min} = mgr$$K.E._{min} = 0.64 - 0.36 = 0.28\,\text{J}$

• Angular Position Where Tension Equals Weight ($T = mg$):

  • Dynamic equation at arbitrary angle $\theta$ from bottom:       $T - mg\cos(\theta) = \frac{mv^2}{r}$
  • Setting $T = mg$:       $mg(1 - \cos(\theta)) = \frac{mv^2}{r} \implies v^2 = rg(1 - \cos(\theta))$
  • Energy conservation from lowermost point (where $v_0 = 10\,\text{m/s}$) to position $\theta$ with vertical rise $h = r(1 - \cos(\theta))$:       $\frac{1}{2}m(10)^2 - \frac{1}{2}mv^2 = mg[r(1 - \cos(\theta))]$$50 - \frac{1}{2}rg(1 - \cos(\theta)) = rg(1 - \cos(\theta))$$50 = \frac{3}{2}rg(1 - \cos(\theta))$
  • Substituting $r = 1.8\,\text{m}$ and $g = 10\,\text{m/s}^2$:       $50 = \frac{3}{2}(1.8)(10)(1 - \cos(\theta)) = 27(1 - \cos(\theta))$$1 - \cos(\theta) = \frac{50}{27}$$\cos(\theta) = 1 - \frac{50}{27} = -\frac{23}{27} \approx -0.8519$$\theta = 148^\circ 25'$

Rotational Inertia and Moment of Inertia

• Concept of Moment of Inertia:

  • In linear mechanics, mass $m$ quantifies a body's inertia (resistance to changes in translational motion).
  • In rotational mechanics, the resistance of a rigid body to rotational acceleration depends not only on total mass, but also on how that mass is distributed relative to the axis of rotation.
  • This rotational analogue of mass is termed the Moment of Inertia (M.I.) or Rotational Inertia, denoted by $I$.

• Derivation of Rotational Kinetic Energy:

  • Consider a rigid body rotating with constant angular velocity $\omega$ about an axis perpendicular to the plane of motion.
  • Let the body comprise $N$ particles with masses $m_1, m_2, \dots, m_N$ at perpendicular distances $r_1, r_2, \dots, r_N$ from the axis.
  • Every particle rotates with the identical angular speed $\omega$, but with distinct linear speeds:     $v_1 = r_1\omega, \quad v_2 = r_2\omega, \quad \dots, \quad v_N = r_N\omega$
  • Total rotational kinetic energy is the sum of the individual translational kinetic energies:     $K.E._{rot} = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 + \dots + \frac{1}{2}m_N v_N^2$$K.E._{rot} = \frac{1}{2}m_1 (r_1\omega)^2 + \frac{1}{2}m_2 (r_2\omega)^2 + \dots + \frac{1}{2}m_N (r_N\omega)^2$$K.E._{rot} = \frac{1}{2}\left(\sum_{i=1}^N m_i r_i^2\right)\omega^2$
  • Comparing this expression with translational kinetic energy $K.E._{trans} = \frac{1}{2}mv^2$, the bracketed quantity defines the Moment of Inertia $I$:     $I = \sum_{i=1}^N m_i r_i^2$$K.E._{rot} = \frac{1}{2}I\omega^2$
  • For a continuous mass distribution:     $I = \int r^2\,dm$

• Moment of Inertia of a Uniform Thin Ring:

  • A uniform ring of mass $M$ and radius $R$ has all its mass distributed uniformly along its circumference.
  • For rotation about its central transverse axis (perpendicular to its plane, passing through center), every mass element $dm$ is at a constant perpendicular distance $r = R$ from the axis:     $I = \int R^2\,dm = R^2\int dm = MR^2$

• Moment of Inertia of a Uniform Disc:

  • A uniform thin disc has mass $M$, radius $R$, and uniform surface mass density:     $\sigma = \frac{M}{\pi R^2}$
  • Divide the disc into concentric elemental rings of radius $r$ and radial width $dr$.
  • Surface area of this elemental ring:     $dA = 2\pi r\,dr$
  • Mass of the elemental ring:     $dm = \sigma dA = 2\pi\sigma r\,dr$
  • The moment of inertia of this elemental ring about the central transverse axis is:     $dI = dm\,r^2 = (2\pi\sigma r\,dr)r^2 = 2\pi\sigma r^3\,dr$
  • Integrating from $r = 0$ to $r = R$:     $I = \int_0^R 2\pi\sigma r^3\,dr = 2\pi\sigma \left[\frac{r^4}{4}\right]_0^R = 2\pi\sigma \frac{R^4}{4} = \frac{1}{2}\pi\sigma R^4$
  • Substituting $\sigma = \frac{M}{\pi R^2}$:     $I = \frac{1}{2}\pi\left(\frac{M}{\pi R^2}\right)R^4 = \frac{1}{2}MR^2$

• Radius of Gyration ($K$):

  • Definition: The radius of gyration of a body about a given axis of rotation is the effective perpendicular distance from the axis at which the entire mass $M$ of the body could be concentrated without altering its moment of inertia:     $I = MK^2 \implies K = \sqrt{\frac{I}{M}}$
  • Physical Significance: $K$ quantifies how spread out the mass is relative to the axis of rotation; a larger $K$ indicates that mass is distributed farther from the axis.
  • Comparison Between Ring and Disc:
  • For a uniform ring: $I_r = MR^2 = MK_r^2 \implies K_r = R$
  • For a uniform disc: $I_d = \frac{1}{2}MR^2 = MK_d^2 \implies K_d = \frac{R}{\sqrt{2}}$
  • Thus, $K_d < K_r$. Among all axisymmetric planar objects of identical mass and radius, the ring has the largest radius of gyration.

Fundamental Theorems on Moment of Inertia

• Theorem of Parallel Axes:

  • Statement: The moment of inertia ($I_O$) of a body about any axis is equal to the sum of its moment of inertia ($I_C$) about a parallel axis passing through its center of mass, plus the product of the total mass of the body ($M$) and the square of the perpendicular distance ($h$) between the two parallel axes:     $I_O = I_C + Mh^2$
  • Derivation:
  • Consider a rigid body of mass $M$.
  • Let the axis passing through point $O$ be parallel to the axis passing through the center of mass $C$, separated by distance $h = CO$.
  • Consider a mass element $dm$ at point $D$. Draw a perpendicular $DN$ to the line extending $OC$.
  • Moment of inertia about axis through $C$:       $I_C = \int (DC)^2\,dm$
  • Moment of inertia about axis through $O$:       $I_O = \int (DO)^2\,dm$
  • In right-angled triangle $\triangle DNO$:       $DO^2 = DN^2 + NO^2 = DN^2 + (NC + CO)^2$$DO^2 = DN^2 + NC^2 + 2(NC)(CO) + CO^2$
  • In right-angled triangle $\triangle DNC$:       $DN^2 + NC^2 = DC^2$
  • Since $CO = h$:       $DO^2 = DC^2 + 2h(NC) + h^2$
  • Integrating to obtain $I_O$:       $I_O = \int [DC^2 + 2h(NC) + h^2]\,dm = \int DC^2\,dm + 2h\int NC\,dm + h^2\int dm$
  • By definition, $\int DC^2\,dm = I_C$ and $\int dm = M$.
  • Because $C$ is the center of mass, the first moment of mass about the center of mass vanishes identically:       $\int NC\,dm = 0$
  • Therefore:       $I_O = I_C + Mh^2$

• Theorem of Perpendicular Axes:

  • Statement: For a planar laminar body, the moment of inertia ($I_z$) about an axis ($z$) perpendicular to its plane equals the sum of its moments of inertia about two mutually perpendicular axes ($x$ and $y$) lying in its plane, with all three axes intersecting at a common origin:     $I_z = I_x + I_y$
  • Derivation:
  • Consider a flat laminar body lying in the $xy$-plane, rotating about the $z$-axis.
  • Consider a mass element $dm$ at point $P(x, y)$.
  • The perpendicular distance of $dm$ from the $x$-axis is $y$, and from the $y$-axis is $x$.
  • Perpendicular distance of $dm$ from the $z$-axis is:       $r = \sqrt{x^2 + y^2}$
  • The moments of inertia about the coordinate axes are:       $I_x = \int y^2\,dm$$I_y = \int x^2\,dm$$I_z = \int r^2\,dm = \int (x^2 + y^2)\,dm$
  • Expanding the integral:       $I_z = \int x^2\,dm + \int y^2\,dm = I_y + I_x$$I_z = I_x + I_y$

• Worked Example 1.7:

  • Problem Statement: A flywheel is a uniform disc of mass $20\,\text{kg}$ and diameter $50\,\text{cm}$ ($R = 0.25\,\text{m}$) rotating about its central transverse axis at $1200\,\text{rpm}$. Calculate its kinetic energy (use $\pi^2 \approx 10$) and determine its moment of inertia about a tangent lying in its plane.
  • Rotational Kinetic Energy:
  • Frequency: $n = \frac{1200}{60} = 20\,\text{rev/s}$
  • Angular speed: $\omega = 2\pi n = 2\pi(20) = 40\pi\,\text{rad/s}$
  • Central moment of inertia:       $I_z = \frac{1}{2}MR^2 = \frac{1}{2}(20)(0.25)^2 = 10(0.0625) = 0.625\,\text{kg\,m}^2$
  • Kinetic energy:       $K.E._{rot} = \frac{1}{2}I_z\omega^2 = \frac{1}{2}(0.625)(40\pi)^2 = \frac{1}{2}(0.625)(1600\pi^2) = (0.625)(800)(10) = 5000\,\text{J}$
  • Moment of Inertia About a Coplanar Tangent:
  • By the perpendicular axes theorem, the diametrical moment of inertia $I_d$ satisfies:       $I_z = I_x + I_y = 2I_d \implies I_d = \frac{I_z}{2} = \frac{1}{4}MR^2$
  • A tangent in the plane of the disc runs parallel to a diameter at distance $h = R$.
  • Applying the parallel axes theorem:       $I_{tangent} = I_d + Mh^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$$I_{tangent} = \frac{5}{4}(20)(0.25)^2 = 25(0.0625) = 1.5625\,\text{kg\,m}^2$