circular motion lec2

Components of Linear Acceleration in Circular Motion

• When a body moves along a circular path, it possesses both linear and angular acceleration.

• Linear acceleration consists of two distinct components:

  • Centripetal Acceleration ($A_c$): This arises due to changes in the direction of the tangential velocity. It is always directed towards the center of the circle. The formula is given by:     $A_c = \frac{V^2}{R}$

  • Tangential Acceleration ($A_t$): This arises from changes in the magnitude (speed) of the tangential velocity. The formulas are:     $A_t = R\alpha$     $A_t = \frac{dV}{dt}$

• The total linear acceleration ($A$) for a body in a circular path is the vector sum of these two perpendicular components:     $A = \sqrt{A_c^2 + A_t^2}$

Numerical Example: Total Acceleration of a Car

• Scenario: A car travels along a circular curve with a radius of $50 \, \text{m}$. Its tangential speed is $16 \, \text{m/s}$ and is increasing uniformly at a rate of $8 \, \text{m/s}^2$.

• Given Data:

  • $R = 50 \, \text{m}$

  • $V = 16 \, \text{m/s}$

  • $A_t = 8 \, \text{m/s}^2$ (indicated by the rate at which speed is increasing).

• Calculation of Centripetal Acceleration ($A_c$):     $A_c = \frac{16^2}{50} = \frac{256}{50} = 5.12 \, \text{m/s}^2$

• Calculation of Total Acceleration ($A$):     $A = \sqrt{(5.12)^2 + (8)^2}$     $A \approx 9.5 \, \text{m/s}^2$

The Concept of Centripetal Force

• According to Newton's Second Law ($F = ma$), any acceleration must be caused by a force.

• Since centripetal acceleration ($A_c$) is always directed towards the center, there must be a resultant force directed towards the center called Centripetal Force ($F_c$).

• Definition: The force responsible for keeping a body in a circular path. It is also known as the "center-seeking" force.

• Formula:     $F_c = m A_c = \frac{mv^2}{R}$

• Direction: The direction of centripetal force is always identical to the direction of centripetal acceleration (towards the center).

• Role in Systems: Centripetal force is not a new type of biological or physical fundamental force; rather, it is a role played by existing forces in a system:

  • Tension in a String: For a stone whirled in a circle.

  • Gravitational Force: For planets orbiting a sun or moons orbiting planets.

  • Coulombic Force (Electrostatic): For electrons revolving around a nucleus.

  • Friction: For a car turning on a flat road.

Rotational Motion and Torque

• Torque (Rotational Analog of Force): Torque is the rotational equivalent of force in the linear domain.

• Key Characteristics of Torque:

  • While a force ($F$) causes linear acceleration ($a$), torque ($\tau$) causes angular acceleration ($\alpha$).

  • If the net unbalanced torque on a body is zero, its angular acceleration is zero, and it rotates with a constant angular velocity ($\omega$).

  • Torque is the "Turning Effect of Force."

• Formulas:

  • Vector Form: $\tau = R \times F$ (It is specifically $R \times F$, not $F \times R$, as order matters in cross products).

  • Scalar Form: $\tau = Fd$, where $d$ is the Moment Arm.

• Moment Arm ($d$): The perpendicular distance between the point of rotation (pivot) and the line of action of the force. If the force is applied directly at the pivot, the moment arm is zero, and no torque is produced ($\tau = 0$).

• Units and Dimensions:

  • Unit: Newton-meter ($N\cdot m$).

  • Dimensionally, torque is identical to Work Done ($W = Fd$), though they represent different physical concepts.

Vector Calculation of Torque (Cross Product Steps)

• To calculate the cross product of two vectors $R$ and $F$ without a $3\times3$ determinant, follow this method:

  1. List the coefficients of the first vector ($R$) twice: $(i_1, j_1, k_1, i_1, j_1, k_1)$.

  2. List the coefficients of the second vector ($F$) twice directly below: $(i_2, j_2, k_2, i_2, j_2, k_2)$.

  3. Discard the first and last columns.

  4. Perform cross-multiplication on the remaining three pairs of columns to find the $i$, $j$, and $k$ components.

  5. Formula reminder: $R = \text{Final Position Vector} - \text{Initial Position Vector}$.

Moment Due to a Couple

• Definition: A couple is formed by two forces that have:

  1. Same magnitude.

  2. Opposite directions.

  3. Different lines of action.

• Torque of a Couple: $\tau = Fd$, where $d$ is the Couple Arm (the perpendicular distance between the two parallel forces).

• Note: If the lines of action are the same, the forces cancel each other out, and no torque/couple is produced.

Angular Momentum and Moment of Inertia

• Angular Momentum ($L$): The quantity of rotational motion contained in a body.

  • Formula: $L = R \times P = RP\sin(\theta)$.

  • Alternative form: $L = I\omega$ (analogous to $P = mv$).

• Moment of Inertia ($I$): The quantitative measure of rotational inertia. It represents the resistance of a body to changes in its rotational state.

  • Formula for a point mass: $I = mr^2$.

  • Units: $kg\cdot m^2$.

  • Dimensions: $[ML^2T^0]$.

• Values of $I$ for Different Geometries:

  • Thin Hoop/Ring/Hollow Cylinder: $I = MR^2$

  • Solid Disk/Solid Cylinder: $I = \frac{1}{2}MR^2$

  • Solid Sphere: $I = \frac{2}{5}MR^2$

  • Hollow Sphere: $I = \frac{2}{3}MR^2$

  • Thin Rod: $I = \frac{1}{12}ML^2$

• Law of Conservation of Angular Momentum: In the absence of an external torque, the total angular momentum of an isolated system remains constant.

Equilibrium in Mechanics

• General Definition: A body is in equilibrium if its total acceleration (linear and angular) is zero.

• Static Equilibrium: The body is at rest ($V = 0$, $\omega = 0$, $A = 0$, $\alpha = 0$).

• Dynamic Equilibrium: The body moves with constant linear velocity ($V$) and/or constant angular velocity ($\omega$).

• Translational Equilibrium (First Condition): The vector sum of all forces is zero ($\sum F = 0$). This implies $A = 0$.

• Rotational Equilibrium (Second Condition): The vector sum of all torques is zero ($\sum \tau = 0$). This implies $\alpha = 0$.

• Complete Equilibrium: A body satisfying both the first and second conditions.

Rotational Kinetic Energy and Rolling Motion

• Translational Kinetic Energy: $K.E.\text{trans} = \frac{1}{2}mv^2$

• Rotational Kinetic Energy: $K.E.\text{rot} = \frac{1}{2}I\omega^2$

• Rolling Motion (Combined Motion): When a body (like a cylinder or sphere) rolls without slipping, it has both translational and rotational kinetic energy.

  • $K.E.\text{total} = K.E.\text{trans} + K.E.\text{rot}$

  • $K.E.\text{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

• Rolling Down an Inclined Plane: If a body starts from rest at height ($h$), potential energy is converted to total kinetic energy:     $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Comparative Speeds of Rolling Objects at the Bottom of an Incline

• By substituting the specific Moment of Inertia ($I$) and the relationship $v = r\omega$ into the conservation of energy equation, the velocities ($v$) at the bottom of a hill for different objects are:

  • Solid Sphere: $v = \sqrt{\frac{10}{7}gh} \approx 1.195\sqrt{gh}$

  • Solid Cylinder/Disk: $v = \sqrt{\frac{4}{3}gh} \approx 1.154\sqrt{gh}$

  • Hollow Sphere: $v = \sqrt{\frac{6}{5}gh} \approx 1.095\sqrt{gh}$

  • Hollow Cylinder/Ring/Hoop: $v = \sqrt{gh} = 1.0\sqrt{gh}$

• Conclusion: The Solid Sphere reaches the bottom with the highest speed, while the Hollow Cylinder reaches with the lowest speed.

Friction, Skidding, and Maximum Safe Speed

• For a car on a flat circular track, friction provides the centripetal force:     $F_{\text{friction}} = F_c = \frac{mv^2}{R}$

• The maximum frictional force a road can apply is $F_{\text{max}} = \mu_s R_{\text{normal}} = \mu_s mg$.

• Maximum Speed Before Skidding ($V_{\text{max}}$):     $\mu_s mg = \frac{m V_{\text{max}}^2}{R}$     $V_{\text{max}} = \sqrt{\mu_s Rg}$

• Implication: If a car exceeds this speed (V > V_{\text{max}}), the road cannot provide enough friction to maintain the circular path, and the car will skid outwards.

• Friction is unreliable because $\mu_s$ changes in wet or oily conditions.

Banking of Roads

• Banking: Tilting the road at an angle (banking angle $\theta$) relative to the horizontal to reduce reliance on friction.

• The normal force ($N$) is resolved into two components:

  • $N\cos(\theta)$: Balances the weight ($mg$).

  • $N\sin(\theta)$: Acts as the necessary centripetal force ($\frac{mv^2}{R}$).

• Banking Angle Formula:     $\tan(\theta) = \frac{v^2}{Rg}$

• This formula allows engineers to design safe turns for specific "rated" speeds where cars can turn even if the road is perfectly frictionless.

Questions & Discussion

• Student Question: How do you determine the initial and final points for the position vector $R$ in torque problems?

• Response: The final point is the point of application of force (where the force acts). The initial point is the point of rotation (around which the body rotates or the point about which torque is being calculated).

• Student Question: Does the total linear acceleration always include tangential acceleration?

• Response: No. If a body moves with a constant speed in a circle, the tangential acceleration ($A_t$) is zero, and the total acceleration equals the centripetal acceleration ($A = A_c$). If the speed is changing, you must include both using the square root formula.

• Student Observation: In the rolling motion problems, the mass ($m$) cancelled out.

• Response: Correct. This shows that the final speed of a rolling object down a frictionless incline depends on its shape (geometry) and height, not its mass.