(Important)Collision and Rotational Dynamics

Collisions: Inelastic vs. Elastic

  • Scenario: Car and Wall Collision

    • Choice 1 (Completely Inelastic): Car gets stuck to the wall (like a soft, sticky material).

    • Choice 2 (Elastic): Car bounces off the wall (like a rubber car).

    • Question: Which scenario causes more damage to the target (the wall or an obstacle)?

    • Initial Intuition: Many people might think getting stuck (inelastic) causes more damage.

    • Correct Answer: Bouncing off (elastic collision) causes more damage to the target.

  • Reasoning Behind Damage Dissipation

    • Energy Loss: The more inelastic a collision, the more kinetic energy is lost or dissipated (converted into heat, sound, deformation). In a completely inelastic collision, the maximum possible kinetic energy is lost while conserving momentum.

    • Damage Received: When energy is dissipated during the interaction, both the colliding object (car/ball) and the target (wall) receive less energy in terms of deformation or damage. Therefore, an elastic collision (where less energy is lost) imparts more energy to the target, causing more damage.

  • Change in Momentum (Impulse of Force)

    • Elastic Collision: If a car of mass m hits a wall with velocity v and bounces off with the same speed in the opposite direction (velocity -v), the change in momentum is ext{delta}p = mv - (-mv) = 2mv .

    • Inelastic Collision: If a car of mass m hits a wall with velocity v and sticks to it (final velocity 0), the change in momentum is ext{delta}p = 0 - mv = -mv (magnitude is mv).

    • Impact: For the same collision time, an elastic collision involves twice the change in momentum compared to a completely inelastic collision. This means the impulse of force on the wall (and the car) is twice as great in an elastic collision.

    • Car Safety Implication: For the occupants of a car, a completely inelastic collision is preferable because they receive less impulse of force. This is why cars are designed to crumple (deform) in an accident, making the collision more inelastic and extending the collision time, thereby reducing the force on the occupants ( F ext{delta}t = ext{delta}p ).

Rotational Dynamics of a Rigid Body

  • Describing Complex Motion

    • Translational motion in two dimensions can be broken into x and y components.

    • Many objects also undergo rotation while moving.

    • Center of Mass (CM): In free flight (no external forces except gravity), a body rotates around its center of mass.

    • Combined Motion: The motion of a rigid body can be expressed as two separate motions:

      1. Translational motion of the center of mass.

      2. Rotational motion around the center of mass (as observed from a frame moving with the CM).

    • Rotation can occur around any axis, not just the center of mass.

  • Polar Coordinates: A Framework for Rotation

    • To describe rotation, polar coordinates (angles) are used, extending concepts from linear (Cartesian) coordinates.

Quantity

Linear (Cartesian)

Rotational (Polar)

Relation

Coordinate

x(t) (meters)

heta(t) (radians) (angle as a function of time)

Velocity

v = rac{dx}{dt} (meters/second)

Angular Velocity: ext{omega} = rac{d heta}{dt} (radians/second)

v = ext{omega} imes r

Acceleration

a = rac{dv}{dt} (meters/second^2)

Angular Acceleration: ext{alpha} = rac{d ext{omega}}{dt} (radians/second^2)

Tangential Acceleration: a_{ ext{tangential}} = ext{alpha} imes r

Cause

Force (F)

Torque ( ext{tau} )

ext{tau} = F imes r (force times lever arm)

Inertia

Mass (m) - A measure of resistance to linear acceleration (Newton's 2nd Law: F = ma )

Moment of Inertia (I) - A measure of resistance to angular acceleration

  • Derivation of Rotational Newton's Second Law

    • Starting from ext{tau} = F imes r and F = ma_{tangential} :

    • ext{tau} = (ma_{tangential})r

    • Substitute a_{tangential} = ext{alpha} r :

    • ext{tau} = (m( ext{alpha} r))r = (mr^2) ext{alpha}

    • This leads to the definition of Moment of Inertia (I) for a point mass: I = mr^2 . Therefore, the rotational equivalent of Newton's 2nd Law is: ext{tau} = I ext{alpha} .

  • Distinction: Angular vs. Centripetal Acceleration

    • Angular acceleration ( ext{alpha} ): Describes how quickly the angular velocity changes (how fast an object spins up or slows down). It's related to tangential acceleration.

    • Centripetal acceleration ( ac ): A linear acceleration always pointing towards the center of rotation, responsible for changing the direction of velocity, keeping an object moving in a circle ( ac = v^2/r or a_c = ext{omega}^2 r ).

  • Parallelism: Momentum and Energy

Quantity

Linear

Rotational

Momentum

p = mv

Angular Momentum: L = I ext{omega}

Kinetic Energy

KE_{linear} = rac{1}{2} mv^2

Rotational Kinetic Energy: KE_{rotational} = rac{1}{2} I ext{omega}^2

  • Rotational Kinematic Equations (for constant angular acceleration)

Linear Equations

Rotational Analogues

x = x0 + v0 t + rac{1}{2} a t^2

heta = heta0 + ext{omega}0 t + rac{1}{2} ext{alpha} t^2

v = v_0 + at

ext{omega} = ext{omega}_0 + ext{alpha} t

a = rac{vf^2 - vi^2}{2 ext{delta}x}

ext{alpha} = rac{ ext{omega}f^2 - ext{omega}i^2}{2 ext{delta} heta}

Moment of Inertia (I)

  • Nature of Moment of Inertia

    • Moment of Inertia is a scalar quantity, much like mass. Its units are ext{kg} imes ext{m}^2 .

    • Superposition Principle: For a system of multiple point masses, their moments of inertia simply add up: I{total} = ext{Sigma} Ii = ext{Sigma} mi ri^2 . This also applies for rigid bodies where individual infinitesimal masses are integrated.

  • Moments of Inertia for Common Shapes (assuming rotation around the center of mass)

    • Point Mass: I = mr^2

    • Hula Hoop / Thin-Walled Cylinder (all mass at radius R): I = MR^2

    • Solid Disc / Solid Cylinder: I = rac{1}{2} MR^2 (smaller because mass is distributed closer to the axis of rotation compared to a hula hoop of the same mass and outer radius).

    • Solid Sphere: I = rac{2}{5} MR^2

    • Hollow Sphere / Thin-Walled Sphere: I = rac{2}{3} MR^2

    • Reference: These formulas are generally found in physics textbooks (e.g., Figure 8.20 in Giancoli).

Parallel Axis Theorem

  • Purpose: This theorem allows calculation of the moment of inertia about any axis, provided you know the moment of inertia about a parallel axis passing through the center of mass.

  • Formula: Iz = I{CM} + Md^2

    • I_z : Moment of inertia about the new axis (z).

    • I_{CM} : Moment of inertia about the parallel axis passing through the center of mass (obtained from tables).

    • M : Total mass of the object.

    • d : Perpendicular distance between the center of mass and the new axis of rotation.

  • Condition: The new axis of rotation (z) must be parallel to the axis for which I_{CM} is known.

  • Example: Hula Hoop Hanging on a Nail

    • If a hula hoop (thin-walled cylinder) is rotating around its center, I_{CM} = Mr^2 .

    • If lifted by an edge and rotating around a nail placed on its rim, the axis of rotation is at a distance d = r from the center of mass.

    • Using the Parallel Axis Theorem: I_z = Mr^2 + M(r)^2 = 2Mr^2 .

Applications: Problem Solving in Rotational Dynamics

  • General Approach: Similar to linear dynamics, rotational problems can often be solved using two main methods:

    1. Newton's Laws Approach (Dynamics): Apply ext{Sigma} ext{tau} = I ext{alpha} to rotational motion, often in conjunction with ext{Sigma}F = ma for translational motion.

    2. Energy Conservation Approach: Utilize KE{rotational} = rac{1}{2} I ext{omega}^2 in the conservation of mechanical energy equation: PE{initial} = KE{translational} + KE{rotational} .

  • Simplifying Assumption: For most introductory problems, angular acceleration ( ext{alpha} ) is assumed to be constant (or zero).

Example 1: Racing Down the Hill (Rolling Race)

  • Scenario: Different rigid bodies (solid cylinder, thin-walled cylinder, solid sphere, thin-walled sphere) of the same mass and radius roll without slipping down a hill of height h.

  • Condition: Rolling without Slipping: This means the relationship between linear and angular velocity holds: v = ext{omega} r .

  • Moment of Inertia Coefficients (X): Each shape has a coefficient X such that I = XMR^2 .

    • Thin-Walled Cylinder: X = 1

    • Solid Cylinder: X = rac{1}{2}

    • Thin-Walled Sphere: X = rac{2}{3}

    • Solid Sphere: X = rac{2}{5}

  • Energy Conservation Method:

    • Initial Potential Energy: PE_i = Mgh

    • Final Kinetic Energy (combined translational and rotational): KE_f = rac{1}{2} Mv^2 + rac{1}{2} I ext{omega}^2

    • Substitute I = XMR^2 and ext{omega} = v/R :
      Mgh = rac{1}{2} Mv^2 + rac{1}{2} (XMR^2)( rac{v}{R})^2
      Mgh = rac{1}{2} Mv^2 + rac{1}{2} XMv^2
      Mgh = rac{1}{2} Mv^2 (1 + X)
      gh = rac{1}{2} v^2 (1 + X)

    • Solving for final velocity (v):
      v = ext{sqrt}{ rac{2gh}{1+X}}

  • Comparison:

    • Slipping Slab (no friction, no rotation): If an object slides down without friction, its velocity would be v_{slipping} = ext{sqrt}{2gh} .

    • Rolling Objects: Rolling objects are slower because some of the potential energy is converted into rotational kinetic energy, not just translational. The (1+X) term in the denominator (since X is always positive) makes v smaller than v_{slipping} .

  • Ranking by Speed (Largest v to Smallest v):

    • The largest velocity corresponds to the smallest coefficient X.

    1. Solid Sphere: X = rac{2}{5} = 0.4 (Fastest) - Its mass is closest to the axis on average, making it easiest to rotate.

    2. Solid Cylinder: X = rac{1}{2} = 0.5

    3. Thin-Walled Sphere: X = rac{2}{3} ext{approx} 0.67

    4. Thin-Walled Cylinder / Hula Hoop: X = 1 (Slowest) - All its mass is at the maximum radius, making it hardest to rotate for its mass and radius.

Example 2: Primitive Yo-Yo (Falling Thin-Walled Cylinder)

  • Scenario: A thin-walled cylinder (like a hula hoop) with string wrapped around its circumference, unwinding as it falls.

  • Moment of Inertia: For a thin-walled cylinder, I = MR^2 .

  • Axis of Rotation: The yo-yo rotates around its center of mass as it falls, not the point where the string leaves.

  • Problem: Find the velocity (v) of the yo-yo's center of mass after it has fallen a distance h.

  • Energy Conservation Method (Re-derivation for this specific case):

    • Mgh = rac{1}{2} Mv^2 + rac{1}{2} I ext{omega}^2

    • Substitute I = MR^2 and ext{omega} = v/R (assuming no slipping of string):
      Mgh = rac{1}{2} Mv^2 + rac{1}{2} (MR^2)( rac{v}{R})^2
      Mgh = rac{1}{2} Mv^2 + rac{1}{2} Mv^2
      Mgh = Mv^2

    • Solving for v: v = ext{sqrt}{gh}

    • Comparison: This is slower than free fall ( ext{sqrt}{2gh} ) by a factor of ext{sqrt}{2} .

  • Dynamics Method (Newton's Laws) for the Yo-Yo:

    • Problem: Find the acceleration of the center of mass (a_{CM}).

    • Free Body Diagram:

      • Force of gravity: Mg (downwards)

      • Tension in string: T (upwards)

    • Translational Motion (along y-axis):
      ext{Sigma}Fy = Mg - T = Ma{CM} (Equation 1)

    • Rotational Motion (around center of mass):

      • The only force causing torque is tension (T).

      • ext{Sigma} ext{tau} = Tr = I ext{alpha}

      • Assuming no slipping: a{CM} = ext{alpha} r ext{Rightarrow} ext{alpha} = a{CM}/r

      • Substitute I = MR^2 and ext{alpha} .
        Tr = (MR^2)( rac{a{CM}}{R}) T = Ma{CM} (Equation 2)

    • Solve the System of Equations:

      • Substitute T = Ma{CM} into Equation 1: Mg - Ma{CM} = Ma{CM} Mg = 2Ma{CM}
        a_{CM} = rac{1}{2} g

    • Conclusion: The yo-yo falls with an acceleration of half that of gravity, explaining its slower descent compared to free fall.

  • Real Yo-Yo Design: Real yo-yos have a small central shaft (where the string wraps) and two larger, massive