Comprehensive Study Notes on Kinematics and Dynamics of Circular Motion

Translational Motion and Angular Variables

• Position Vector and Polar Coordinates:

  • The position of a particle $P$ in a plane can be completely specified by its position vector $\vec{r}$.
  • The position vector is defined by its magnitude $r = OP$ (the distance from origin $O$ to particle $P$) and its orientation relative to a fixed reference direction, given by the coordinate angle $\theta$ measured from the positive x-axis.
  • The ordered pair $(r, \theta)$ defines the polar coordinates of the particle.

• Types of Particle Motion in Polar Coordinates:

  • Radial Motion Outward: The particle moves directly away from the origin; magnitude $r$ increases while angle $\theta$ remains constant.
  • Radial Motion Inward: The particle moves directly toward the origin; magnitude $r$ decreases while angle $\theta$ remains constant.
  • Circular Motion Centered at Origin: The particle moves along a circular arc of constant radius $r$; only the coordinate angle $\theta$ changes with time $t$
  • General Motion: If a particle moves on any path other than a radial line or a circle centered at the origin, both coordinates $r$ and $\theta$ vary with time $t$

• Definitions of Angular Variables:

  • Angular Motion: Defined as any change in the direction of position vector $\vec{r}$. This occurs whenever a particle moves on a curvilinear path or a straight-line path that does not pass through the origin.
  • Angular Position ($\theta$): The coordinate angle $\theta$ made by the position vector $\vec{r}$ with a reference line at a specific instant.
  • Angular Displacement ($\Delta \theta$): The net change in angular position during a specified time interval $\Delta t = t_2 - t_1$:     $\Delta \theta = \theta_2 - \theta_1$
  • Angular Velocity ($\omega$): The instantaneous rate of change of angular position $\theta$ with respect to time $t$:     $\omega = \frac{d\theta}{dt}$
  • Angular Acceleration ($\alpha$): The instantaneous rate of change of angular velocity $\omega$ with respect to time $t$:     $\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2} = \omega \frac{d\omega}{d\theta}$

• Sign Convention and Analogy to Rectilinear Motion:

  • In planar motion, the position vector turns either clockwise or anticlockwise.
  • Assigning one rotational sense as positive and the opposite as negative allows angular kinematic problems involving $\theta$, $\omega$, and $\alpha$ to be solved using equations directly analogous to 1D rectilinear kinematic problems involving position $x$, velocity $v$, and acceleration $a$

Kinematics of Circular Motion

• Radius Vector and Path Geometry:

  • In circular translational motion, a particle $P$ moves along a circular path of constant radius $r$.
  • The position vector $\vec{r} = \vec{OP}$ originating from the center of the circle $O$ is termed the radius vector.
  • The radius vector is always normal (perpendicular) to the circular path, maintains a constant magnitude $r$, and rotates such that its angle $\theta(t)$ varies continuously.

• Classification of Angular Motion:

  • Motion with Uniform Angular Velocity:
  • Characterized by constant angular velocity $\omega$ and zero angular acceleration ($\alpha = 0$).
  • Analogy to uniform rectilinear motion:       $\theta = \theta_0 + \omega t$
  • Motion with Uniform Angular Acceleration:
  • Characterized by constant angular acceleration ($\alpha = \text{constant}$).
  • Kinematic equations of motion:       $\omega = \omega_0 + \alpha t$$\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2$$\theta = \theta_0 + \left(\frac{\omega_0 + \omega}{2}\right) t$$\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$
  • Motion with Variable Angular Acceleration:
  • Angular acceleration $\alpha$ is specified as a function of time $\alpha(t)$, position $\alpha(\theta)$, or angular velocity $\alpha(\omega)$.
  • Solutions require direct integration or differentiation calculus techniques analogous to variable acceleration in straight-line motion.

• Linear Velocity and Acceleration Components:

  • Arc Length: Arc length $s$ covered along the circle is related to angular displacement $\theta$ by:     $s = r \theta$
  • Linear Speed ($v$): Differentiating arc length $s$ with respect to time $t$ gives the linear speed:     $v = \frac{ds}{dt} = r \frac{d\theta}{dt} = \omega r$
  • Direction of Velocity: Linear velocity $\vec{v}$ is directed tangentially along the path at every instant.
  • Uniform Circular Motion (UCM):
  • Speed $v$ is constant, but velocity vector $\vec{v}$ continuously changes direction.
  • The change in velocity over an infinitesimal interval $dt$ is perpendicular to $\vec{v}$ and points directly toward the center of the path.
  • Infinitesimal velocity magnitude change: $|d\vec{v}| = v \, d\theta$
  • Centripetal (Normal) Acceleration ($a_c$ or $a_n$):       $a_n = a_c = \frac{|d\vec{v}|}{dt} = v \frac{d\theta}{dt} = v \omega = \frac{v^2}{r} = \omega^2 r$
  • Non-Uniform Circular Motion:
  • Speed $v$ changes with time, giving rise to two perpendicular acceleration components:
    1. Tangential Acceleration ($a_t$): Accounts for the rate of change of speed:          $a_t = \frac{dv}{dt} = r \frac{d\omega}{dt} = r \alpha$
    2. Normal/Centripetal Acceleration ($a_n$): Accounts for the rate of change of velocity direction:          $a_n = \frac{v^2}{r} = \omega^2 r$
  • Total Linear Acceleration ($a$):       $a = \sqrt{a_t^2 + a_n^2}$
  • Angle $\beta$ made by the total acceleration vector $\vec{a}$ with the radius vector:       $\tan(\beta) = \frac{a_t}{a_n}$

• Worked Examples and Solutions:

  • Illustration 1: Angular position varies as $\theta = t^3 - 3t^2 + 4t - 2$ (in radians, $t$ in seconds). Find average angular acceleration $\alpha_{avg}$ between $t = 2\,s$ and $t = 4\,s$
  • Instantaneous angular velocity: $\omega(t) = \frac{d\theta}{dt} = 3t^2 - 6t + 4$
  • At $t = 2\,s$: $\omega_2 = 3(2)^2 - 6(2) + 4 = 4\,rad/s$
  • At $t = 4\,s$: $\omega_4 = 3(4)^2 - 6(4) + 4 = 28\,rad/s$
  • Average angular acceleration:       $\alpha_{avg} = \frac{\Delta \omega}{\Delta t} = \frac{\omega_4 - \omega_2}{4 - 2} = \frac{28 - 4}{2} = 12\,rad/s^2$
  • Illustration 2: A particle starts from rest ($\omega_0 = 0$) with constant $\alpha = 3.0\,rad/s^2$. An observer starts a stopwatch at $t_1$ and notes an angular displacement of $\Delta \theta = 120\,rad$ in the 4-second interval ($\Delta t = 4\,s$) up to $t_2$. How long had the particle been moving before the stopwatch started?
  • Angular displacement equation: $\Delta \theta = \omega_1 \Delta t + \frac{1}{2} \alpha (\Delta t)^2$
  • $120 = \omega_1 (4) + \frac{1}{2} (3.0)(4)^2 \implies 120 = 4\omega_1 + 24 \implies \omega_1 = 24\,rad/s$
  • Using $\omega_1 = \omega_0 + \alpha t_1$:       $24 = 0 + 3.0 t_1 \implies t_1 = 8.0\,s$
  • Illustration 3: Particle moves on a circular path of radius $r = 8\,m$ with distance $s = \frac{2}{3}t^3 + 3$. Find its speed when tangential and normal accelerations are equal in magnitude.
  • Speed: $v = \frac{ds}{dt} = 2t^2$
  • Tangential acceleration: $a_t = \frac{dv}{dt} = 4t$
  • Normal acceleration: $a_n = \frac{v^2}{r} = \frac{(2t^2)^2}{8} = \frac{4t^4}{8} = \frac{t^4}{2}$
  • Setting $a_t = a_n$:       $4t = \frac{t^4}{2} \implies t^3 = 8 \implies t = 2\,s$
  • Speed at $t = 2\,s$: $v = 2(2)^2 = 8\,m/s$
  • Illustration 4: Particle moves on a circle of radius $r = 1.5\,m$ with constant $\alpha = 2\,rad/s^2$. Initial angular speed at $t = 0$ is $\omega_0 = \frac{60}{\pi}\,rpm = \frac{60}{\pi} \times \frac{2\pi}{60} = 2\,rad/s$. At $t = 2\,s$, find values:
  • Angular speed: $\omega_2 = \omega_0 + \alpha t = 2 + 2(2) = 6\,rad/s$
  • Angular displacement: $\theta_2 = \omega_0 t + \frac{1}{2}\alpha t^2 = 2(2) + \frac{1}{2}(2)(2)^2 = 8\,rad$
  • Linear velocity: $v_2 = r \omega_2 = 1.5 \times 6 = 9\,m/s$
  • Tangential acceleration: $a_t = r \alpha = 1.5 \times 2 = 3\,m/s^2$
  • Normal acceleration: $a_n = r \omega_2^2 = 1.5 \times 6^2 = 54\,m/s^2$
  • Illustration 5: A particle moves in a circle of radius $R$ starting from rest ($\omega_0 = 0$) with constant tangential acceleration. After $t = 2\,s$, the angle between total acceleration $\vec{a}$ and radius $R$ is $45^\circ$. Find angular acceleration $\alpha$
  • Angle $45^\circ$ implies $a_t = a_n$
  • $\alpha R = \omega^2 R \implies \alpha = \omega^2$
  • Since $\omega = \omega_0 + \alpha t = 2\alpha$:       $\alpha = (2\alpha)^2 = 4\alpha^2 \implies \alpha = \frac{1}{4} = 0.25\,rad/s^2$
  • Illustration 6: Particle moves in a circle of radius $R$ such that $a_n = a_t$ at all times. Initial speed at $t = 0$ is $v_0$. Time taken to complete the first revolution is given as $t = \frac{\alpha R}{v_0}(1 - e^{-\beta \pi})$. Find $\alpha + \beta$
  • Differential equation: $\frac{dv}{dt} = \frac{v^2}{R} \implies \frac{dv}{v^2} = \frac{dt}{R}$
  • Integrating from $v_0$ to $v$:       $-\frac{1}{v} \Big|_{v_0}^v = \frac{t}{R} \implies \frac{1}{v_0} - \frac{1}{v} = \frac{t}{R} \implies v = \frac{v_0}{1 - \frac{v_0 t}{R}}$
  • Integrating $ds = v \, dt$ for one revolution ($s = 2\pi R$):       $2\pi R = \int_0^t \frac{v_0 \, dt}{1 - \frac{v_0 t}{R}} = -R \ln\left(1 - \frac{v_0 t}{R}\right)$$-2\pi = \ln\left(1 - \frac{v_0 t}{R}\right) \implies 1 - \frac{v_0 t}{R} = e^{-2\pi} \implies t = \frac{R}{v_0}(1 - e^{-2\pi})$
  • Comparing with given expression: $\alpha = 1$, $\beta = 2 \implies \alpha + \beta = 3$

Relative Angular Velocity

• Relativity of Angular Velocity:
  • Angular velocity is inherently a relative quantity defined with respect to an origin or reference point from which position vectors are drawn; absolute angular velocity does not exist.
  • For a particle $P$ observed from origin $O$ and reference point $A$:
  • Angular velocity w.r.t. $O$: $\omega_{PO} = \frac{d\alpha}{dt}$
  • Angular velocity w.r.t. $A$: $\omega_{PA} = \frac{d\beta}{dt}$

• Mathematical Definition of Relative Angular Velocity:

  • The angular velocity of particle $A$ with respect to another moving particle $B$ ($\omega_{AB}$) is defined as the rate at which the position vector of $A$ relative to $B$ rotates at that instant:     $\omega_{AB} = \frac{\text{Component of relative velocity of } A \text{ w.r.t. } B \text{ perpendicular to line } AB}{\text{Separation distance between } A \text{ and } B} = \frac{(v_{AB})_\perp}{r_{AB}}$

• Special Motion Cases:

  • Two Particles on Concentric Circles (Closest Approach):
  • For particles $A$ and $B$ on concentric circles of radii $r_A$ and $r_B$ moving in the same direction at closest separation:       $(v_{AB})_\perp = v_B - v_A$$r_{AB} = r_B - r_A$$\omega_{AB} = \frac{v_B - v_A}{r_B - r_A}$
  • Two Particles on Same or Coplanar Concentric Circles (Uniform Speeds):
  • For particles moving in the same direction with uniform angular speeds $\omega_A$ and $\omega_B$:
    • Rate of change of line-of-sight angle between radii $OA$ and $OB$:         $\frac{d\theta}{dt} = \omega_B - \omega_A$
    • Time $T$ taken for one particle to complete one full relative revolution around $O$ w.r.t. the other:         $T = \frac{2\pi}{\omega_{rel}} = \frac{2\pi}{|\omega_2 - \omega_1|} = \frac{T_1 T_2}{|T_1 - T_2|}$
    • Note: $\omega_B - \omega_A$ represents the rate of change of angle between lines $OA$ and $OB$, which is distinct from the angular velocity of $B$ w.r.t. $A$ (the rate of rotation of line $AB$).

• Illustration 7:

  • Particles $P$ and $Q$ are separated by $r = 10\,m$. Velocity of $P$ is $8\,m/s$ at $30^\circ$ to line $PQ$. Velocity of $Q$ is $6\,m/s$ at $30^\circ$ to line $PQ$ in opposite transverse direction.
  • Relative perpendicular velocity component:     $(v_{PQ})_\perp = 8 \sin(30^\circ) - (-6 \sin(30^\circ)) = 8(0.5) + 6(0.5) = 7\,m/s$
  • Relative angular velocity:     $\omega_{PQ} = \frac{(v_{PQ})_\perp}{r} = \frac{7}{10} = 0.7\,rad/s$

Radius of Curvature

• Concept of Radius of Curvature:

  • Any continuous curved path can be modeled as consisting of an infinite series of infinitesimal circular arcs.
  • The radius of curvature $R$ at a specific point on a trajectory is the radius of the circular arc that matches the curve's profile at that location.

• Physical Dynamics Method:

  • Relates particle linear speed $v$ and normal centripetal force $F_c = F_\perp$ or normal acceleration component $a_\perp$:     $F_\perp = \frac{m v^2}{R} \implies R = \frac{m v^2}{F_\perp} = \frac{v^2}{a_\perp}$

• Calculus Trajectory Method:

  • Given explicit trajectory equation $y(x)$, radius of curvature $R$ is evaluated as:     $R = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\left|\frac{d^2y}{dx^2}\right|}$

• Worked Examples:

  • Illustration 8: Particle projected with speed $u$ at angle $\theta$ with horizontal. Find radius of curvature at point of projection and at peak of trajectory.
  • At projection point:
    • Speed $v = u$
    • Normal acceleration perpendicular to velocity: $a_\perp = g \cos(\theta)$
    • Radius of curvature: $R = \frac{u^2}{g \cos(\theta)}$
  • At highest point:
    • Horizontal velocity $v = u \cos(\theta)$
    • Normal acceleration: $a_\perp = g$
    • Radius of curvature: $R = \frac{(u \cos(\theta))^2}{g} = \frac{u^2 \cos^2(\theta)}{g}$
  • Illustration 9: Parabolic path $y = ax^2$ with constant speed $v$. Find $R$ at $x = 0$
  • Calculus approach: $\frac{dy}{dx} = 2ax \implies \text{at } x=0, \frac{dy}{dx} = 0$; $\frac{d^2y}{dx^2} = 2a$$R = \frac{[1 + 0]^{3/2}}{2a} = \frac{1}{2a}$
  • Kinematic approach: $y = ax^2 \implies v_y = 2ax v_x$. At $x=0$, $v_y = 0 \implies v_x = v$$a_y = \frac{d^2y}{dt^2} = 2a v_x^2 + 2ax a_x$. At $x=0$, $a_y = 2a v^2$       Since $a_y$ is perpendicular to $v_x$, normal acceleration $a_\perp = 2a v^2$$R = \frac{v^2}{a_\perp} = \frac{v^2}{2a v^2} = \frac{1}{2a}$

Dynamics of Circular Motion

• Newton's Second Law Formulation:

  • Circular motion requires a net inward force directed toward the center to supply centripetal acceleration:     $F_c = m a_c = \frac{m v^2}{r} = m \omega^2 r$
  • If speed varies, a net tangential force $F_t$ is required:     $F_t = m a_t = m \frac{dv}{dt}$
  • Total net force vector: $\vec{F}_{net} = \vec{F}_c + \vec{F}_t$
  • If speed increases ($a_t > 0$), $F_t$ acts parallel to velocity. If speed decreases ($a_t < 0$), $F_t$ acts antiparallel to velocity.

• Physical Sources of Centripetal Force:

  • Tension in a whirled string.
  • Gravitational force in planetary and satellite orbits.
  • Electrostatic force in atomic electron orbits.
  • Static friction or normal reaction forces on banked surfaces.

• Conical Pendulum:

  • Particle of mass $m$ tied to string of length $L$ revolving in a horizontal circle of radius $r = L \sin(\theta)$ at angle $\theta$ to vertical.
  • Vertical force balance: $T \cos(\theta) = mg$
  • Horizontal centripetal force: $T \sin(\theta) = \frac{m v^2}{r} = m \omega^2 r$
  • Dividing equations: $\tan(\theta) = \frac{v^2}{rg} \implies v = \sqrt{rg \tan(\theta)}$
  • Angular speed: $\omega = \sqrt{\frac{g \tan(\theta)}{r}} = \sqrt{\frac{g \tan(\theta)}{L \sin(\theta)}} = \sqrt{\frac{g}{L \cos(\theta)}}$
  • Time period of revolution:     $T_{period} = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{L \cos(\theta)}{g}} = 2\pi \sqrt{\frac{r}{g \tan(\theta)}}$

• Illustrations:

  • Illustration 10: Stone $P$ of mass $m$ whirled in free space with string radius $R$
  • Gravity is absent ($g = 0$). Tension provides full centripetal acceleration: $T = \frac{m v^2}{R}$
  • Force exerted by man holding stationary end: Equal in magnitude to tension $T = \frac{m v^2}{R}$, directed away from stone.
  • Illustration 11: Boy inside rotating vertical cylindrical container of radius $R$, back pressed against inner wall, static friction coefficient $\mu_s$. Platform under feet removed. Find minimum angular speed $\omega_{min}$ to avoid falling.

* Horizontal normal force: $N = m \omega^2 R$
* Vertical force balance: $f_s = mg$
* Non-slipping constraint: $f_s \le \mu_s N \implies mg \le \mu_s m \omega^2 R$
* Minimum angular speed:

      $\omega_{min} = \sqrt{\frac{g}{\mu_s R}}$

• Illustration 12: Motorcyclist on horizontal circular track of radius $R$ accelerates tangentially at constant rate $a_t = a$. Static friction coefficient $\mu_s$. Find maximum safe speed $v_{max}$
  • Normal force: $N = mg$
  • Tangential friction force: $F_t = m a$
  • Centripetal friction force: $F_c = \frac{m v^2}{R}$
  • Total friction force $f = \sqrt{F_t^2 + F_c^2} = m \sqrt{a^2 + \frac{v^4}{R^2}}$
  • Setting total friction equal to limiting friction $f_{max} = \mu_s mg$:       $m \sqrt{a^2 + \frac{v^4}{R^2}} = \mu_s mg \implies a^2 + \frac{v^4}{R^2} = \mu_s^2 g^2$$v_{max} = \left[ R^2 (\mu_s^2 g^2 - a^2) \right]^{1/4}$

Circular Turning on Roads

• Centripetal Requirements for Vehicles:

  • When vehicles negotiate horizontal circular turns, centripetal acceleration is supplied via friction, road banking, or both.

• Case 1: By Friction Only (Level Unbanked Road):

  • Centripetal force supplied entirely by static friction $f = \frac{m v^2}{r}$
  • Limiting friction: $f_{max} = \mu_s N = \mu_s mg$
  • Safe condition: $\frac{m v^2}{r} \le \mu_s mg$
  • Maximum safe turning speed:     $v_{max} = \sqrt{\mu_s g r}$
  • Minimum friction coefficient required:     $\mu_s \ge \frac{v^2}{rg}$

• Case 2: By Banking of Roads Only (Frictionless Banked Road):

  • Road surface tilted outward at angle $\theta$ to horizontal.
  • Horizontal component of normal reaction provides centripetal force: $N \sin(\theta) = \frac{m v^2}{r}$
  • Vertical force balance: $N \cos(\theta) = mg$
  • Dividing equations gives design speed:     $\tan(\theta) = \frac{v^2}{rg} \implies v = \sqrt{rg \tan(\theta)}$

• Case 3: By Friction and Banking of Road Both:

• Forces acting: Weight $mg$, normal reaction $N$, friction $f$
• Friction Direction Behavior:
  1. If vehicle is stationary ($v = 0$), friction $f$ acts outward/upward along incline balancing $mg \sin(\theta)$.
  2. If $v < \sqrt{rg \tan(\theta)}$, vehicle tends to slide down incline; friction $f$ acts outward/upward along incline.
  3. If $v = \sqrt{rg \tan(\theta)}$, friction $f = 0$
  4. If $v > \sqrt{rg \tan(\theta)}$, vehicle tends to skid outward; friction $f$ acts inward/downward along incline.
• Maximum Safe Speed ($v_{max}$):
  • Friction $f = \mu_s N$ acts downward along incline:       $N \sin(\theta) + f \cos(\theta) = \frac{m v^2}{r}$$N \cos(\theta) - f \sin(\theta) = mg$
  • Dividing equations yields:       $\frac{\sin(\theta) + \mu_s \cos(\theta)}{\cos(\theta) - \mu_s \sin(\theta)} = \frac{v_{max}^2}{rg}$$v_{max} = \sqrt{rg \left( \frac{\tan(\theta) + \mu_s}{1 - \mu_s \tan(\theta)} \right)}$
• Minimum Safe Speed ($v_{min}$):
  • Friction $f = \mu_s N$ acts upward along incline:       $N \sin(\theta) - f \cos(\theta) = \frac{m v^2}{r}$$N \cos(\theta) + f \sin(\theta) = mg$
  • Dividing equations yields:       $v_{min} = \sqrt{rg \left( \frac{\tan(\theta) - \mu_s}{1 + \mu_s \tan(\theta)} \right)}$

Centrifugal Force

• Definition and Context:

  • Centrifugal force is a pseudo/fictitious inertial force required exclusively when analyzing particle dynamics from a non-inertial reference frame rotating at constant angular velocity $\omega$.

• Magnitude and Direction:

  • Magnitude: $F_{cf} = m r \omega^2 = \frac{m v^2}{r}$
  • Direction: Radially outward from axis of rotation.

• Observer Interpretation Differences:

  • Inertial Observer (Ground): Object moves in a circle due to inward real centripetal force.
  • Rotating Frame Observer: Object is stationary or moving relative to rotating frame under balance between radially outward centrifugal pseudo force and internal system forces.

• Illustrations:

  • Illustration 13: Max speed of car on level curve radius $R = 30\,m$, $\mu = 0.4$, $g = 10\,m/s^2$$v_{max} = \sqrt{\mu g R} = \sqrt{0.4 \times 10 \times 30} = \sqrt{120} \approx 11\,m/s$
  • Illustration 14: Banking angle for traffic at $v = 60\,km/h = \frac{50}{3}\,m/s$, radius $r = 0.1\,km = 100\,m$, $g = 10\,m/s^2$$\tan(\theta) = \frac{v^2}{rg} = \frac{(50/3)^2}{100 \times 10} = \frac{2500/9}{1000} = \frac{5}{18} \implies \theta = \tan^{-1}\left(\frac{5}{18}\right)$
  • Illustration 15: Smooth hemispherical bowl radius $R$ rotates about vertical axis. Small ball rotates at angle $\alpha$ with vertical. Find $\omega$
  • Vertical force balance: $N \cos(\alpha) = mg$
  • Horizontal centripetal force: $N \sin(\alpha) = m (R \sin(\alpha)) \omega^2 \implies N = m R \omega^2$
  • Substituting $N$ into vertical equation:       $(m R \omega^2) \cos(\alpha) = mg \implies \omega = \sqrt{\frac{g}{R \cos(\alpha)}}$

Beginner's Practice Problems

• Beginner's Box 1 (Kinematics):

  1. $\omega = \theta^2 + 2\theta$. Angular acceleration $\alpha$ at $\theta = 1\,rad$:
  • $\alpha = \omega \frac{d\omega}{d\theta} = (\theta^2 + 2\theta)(2\theta + 2)$
  • At $\theta = 1$: $\alpha = (1 + 2)(2 + 2) = 12\,rad/s^2$ (Option C)
  1. Radii ratio $r_1 : r_2 = 1 : 2$. Same centripetal acceleration. Speed ratio $v_1 : v_2$:
  • $a_c = \frac{v^2}{r} \implies v \propto \sqrt{r} \implies \frac{v_1}{v_2} = \frac{1}{\sqrt{2}}$ (Option C)
  1. String length $r = 80\,cm = 0.8\,m$. 14 revolutions in $25\,s$:
  • $\omega = 2\pi \left(\frac{14}{25}\right) = 2 \times \frac{22}{7} \times \frac{14}{25} = 3.52\,rad/s$
  • $a_c = \omega^2 r = (3.52)^2 \times 0.8 \approx 9.9\,m/s^2$ (Option C)
  1. Radius $r = \frac{20}{\pi}\,m$, starts from rest, speed $v = 50\,m/s$ after 2 revolutions ($s = 4\pi r = 80\,m$):
  • $v^2 = 2 a_t s \implies 2500 = 2 a_t (80) \implies a_t = 15.6\,m/s^2$ (Option C)
  1. Particle on turntable with $v_1 = 20\,cm/s$, $a_1 = 20\,cm/s^2$. Radius halved ($r_2 = \frac{r_1}{2}$) at constant $\omega$:
  • $v_2 = \omega r_2 = 10\,cm/s$
  • $a_2 = \omega^2 r_2 = 10\,cm/s^2$ (Option A)
  1. Particle completes first $\frac{1}{3}$ circumference in $2\,s$, next $\frac{1}{3}$ in $1\,s$:
  • Total displacement: $\Delta \theta = \frac{4\pi}{3}\,rad$ in $\Delta t = 3\,s$
  • $\omega_{avg} = \frac{4\pi/3}{3} = \frac{4\pi}{9}\,rad/s$
  1. $\theta = \omega_0 t + \alpha t^2$ with $\omega_0 = 1\,rad/s$, $\alpha = 1.5\,rad/s^2$. Angular velocity at $t = 2\,s$:
  • $\omega = \frac{d\theta}{dt} = \omega_0 + 2\alpha t = 1 + 2(1.5)(2) = 7\,rad/s$
  1. Angular velocity $\omega = 1.5 t^3 - 3t^2 + 2$. Time when $\alpha = 0$:
  • $\alpha = \frac{d\omega}{dt} = 4.5 t^2 - 6t = 0 \implies t = \frac{6}{4.5} = \frac{4}{3}\,s$
  1. Disc starts from rest with $\alpha = 3t - t^2$. Angular velocity after $t = 2\,s$:
  • $\omega = \int_0^2 (3t - t^2) \, dt = \left[ \frac{3t^2}{2} - \frac{t^3}{3} \right]_0^2 = 6 - \frac{8}{3} = \frac{10}{3}\,rad/s$

• Beginner's Box 2 (Relative Angular Speed and Radius of Curvature):

  1. Projectile projected with speed $u$ at angle $\theta$. Radius of curvature near top:
  • $R = \frac{u^2 \cos^2(\theta)}{g}$ (Option B)
  1. Curve $y = 4x^2$. Radius of curvature at origin ($x=0$):
  • $a = 4 \implies R = \frac{1}{2a} = \frac{1}{8}\,m$ (Option C)
  1. Radius of curvature at general point $(x,y)$ for $y = 4x^2$:
  • $\frac{dy}{dx} = 8x$, $\frac{d^2y}{dx^2} = 8 \implies R = \frac{[1 + (8x)^2]^{3/2}}{8}$ (Option A)
  1. Curve path point $P$ with $v = 5\,m/s$, $a = 10\,m/s^2$ at $30^\circ$ to velocity:
  • (i) Rate of change of speed $a_t = 10 \cos(30^\circ) = 5\sqrt{3}\,m/s^2$ (Option A)
  • (ii) Radius of curvature $R = \frac{v^2}{a_n} = \frac{5^2}{10 \sin(30^\circ)} = \frac{25}{5} = 5\,m$ (Option C)
  1. Two points of rod move with $3v$ and $v$ same direction, separation $r$:
  • $\omega = \frac{3v - v}{r} = \frac{2v}{r}$
  1. Two points of rod move with $3v$ and $v$ opposite directions, separation $r$:
  • $\omega = \frac{3v - (-v)}{r} = \frac{4v}{r}$

• Beginner's Box 3 (Dynamics & Banking):

1. Rotating setup on frictionless table with $l_1 = l_2 = l$:
   • $T_2 = m_2 \omega^2 (2l)$
   • $T_1 - T_2 = m_1 \omega^2 l \implies T_1 = (m_1 + 2m_2) \omega^2 l$
   • Ratio $\frac{T_1}{T_2} = \frac{m_1 + 2m_2}{2m_2}$ (Option B)
2. Centrifugal force is pseudo force when observed by:
   • Observer moving with the particle (Option C)
3. Banked curve allows higher speed because:
   • Normal reaction has a horizontal component (Option C)
4. Mass $m = 16\,kg$, string $r = 144\,m$, max tension $T = 16\,N$:
   • $16 = \frac{16 v^2}{144} \implies v_{max} = 12\,m/s$ (Option D)
5. Motorcyclist in well $r = 5\,m$, min speed $v_{min} = 5\sqrt{5}\,m/s$, $g = 10\,m/s^2$:
   • $v_{min}^2 = \frac{gr}{\mu_s} \implies 125 = \frac{50}{\mu_s} \implies \mu_s = 0.4$
6. Car at $v = 30\,km/h = \frac{25}{3}\,m/s$, radius $r = 60\,m$, $g = 10\,m/s^2$:
   • $\mu_s \ge \frac{v^2}{rg} = \frac{(25/3)^2}{600} = \frac{625}{5400} \approx 0.116$
7. Stone $m = 0.25\,kg$, $r = 1.5\,m$, $f = 40\,rpm = \frac{2}{3}\,rev/s$:
   • $\omega = 2\pi \left(\frac{2}{3}\right) = \frac{4\pi}{3}\,rad/s$
   • Tension $T = m \omega^2 r = 0.25 \times \left(\frac{4\pi}{3}\right)^2 \times 1.5 \approx 6.58\,N$
   • Max speed at $T_{max} = 200\,N$:        $200 = \frac{0.25 v_{max}^2}{1.5} \implies v_{max} = \sqrt{1200} = 20\sqrt{3} \approx 34.64\,m/s$