Study Notes on Rotational Kinematics and Energy

Overview of rotational motion as integral to everyday life
Importance of specific expressions and quantifications for rotational motion, similar to linear motion
Tutorial's focus:
- Introducing angular or rotational terminology for kinematics
- Discussing rotational kinetic energy and moment of inertia
- Applying conservation of energy laws in rolling motion scenarios

Establishing an angular coordinate system for measurement in rotation
- Linear motion uses coordinate systems based on x or y-directions; rotational motion requires angular terms.
Key angular quantities defined:
- Angular Position (θ):
- Defined as the angle an object is from a reference line, with counterclockwise (θ < 0) considered positive and clockwise (θ > 0) negative.
- Measured in radians, which are dimensionless (arc length s = rθ, one full revolution = 360° = 2π radians).
- Arc length (s) is given by: $s = r heta$
- For a full revolution, the arc length equals the circumference ( $s = 2 ext{πr}$ ) so 1 revolution = 2π radians.
- Angular displacement (Δθ): Change in angular position, defined as: $ext{Δθ} = θf - θi$
- Angular Velocity (ω):
- Defined as angular displacement over time, just like linear velocity, measured in radians per second: $ext{ω}_{avg} = \frac{ ext{Δθ}}{ ext{Δt}}$
- If angular velocity is constant: $ext{ω} = ext{ω}_{avg} = \frac{ ext{Δθ}}{ ext{Δt}}$ where for one revolution, $ext{Δθ} = 2 ext{π}$ and $ext{Δt} = T$ (period). The period is: $T = \frac{2 ext{π}}{ ext{ω}}$
- Angular Acceleration (α):
- Defined as the change in angular velocity over time, measured in radians per second squared: $ext{α}_{avg} = \frac{ ext{Δω}}{ ext{Δt}}$

For constant angular acceleration, similar equations as linear motion can be employed:
- Relation between linear and rotational quantities:

Linear World	Rotational World
Displacement (Δx)	Angular Position (θ)
Velocity (v)	Angular Velocity (ω)
Acceleration (a)	Angular Acceleration (α)
Kinematic Equations Derived:

$v<em>f = v</em>i + a t$ -> $ext{ω}<em>f = ext{ω}</em>i + ext{α} t$
$Δx = v<em>i t + \frac{1}{2} a t^2$ -> $Δθ = ext{ω}</em>i t + \frac{1}{2} ext{α} t^2$
$Δx = v<em>f t - \frac{1}{2} a t^2$ -> $Δθ = ext{ω}</em>f t - \frac{1}{2} ext{α} t^2$
$Δx = \frac{1}{2}(v<em>i + v</em>f)t$ -> $Δθ = \frac{1}{2}( ext{ω}<em>i + ext{ω}</em>f)t$
$v<em>f^2 = v</em>i^2 + 2aΔx$ -> $ext{ω}<em>f^2 = ext{ω}</em>i^2 + 2 ext{α}Δθ$

PRACTICE PROBLEMS

Scenario: Dog fetches a ball thrown with initial angular speed of 36.0 rad/s, caught at 0.595 s with final speed of 34.2 rad/s.
- (a) Calculate the ball’s angular acceleration (α):
- $α = \frac{ωf - ωi}{t} = \frac{34.2 - 36.0}{0.595} = -3.025 ext{ s}^{-2}$
- (b) Calculate revolutions made (03) before being caught:
- Using angular displacement:
 $Δθ = \frac{1}{2}( ext{ω}i + ext{ω}f)t = \frac{1}{2}(36.0 + 34.2)(0.595) = 20.8845 ext{ rad}$
- Convert to revolutions:
 $Δθ = 20.8845 ext{ rad} imes \frac{1 ext{ rev}}{2 ext{π} ext{ rad}} = 3.324 ext{ rev}$

Scenario: A pulley rotates counterclockwise attached to a mass, causing its angular velocity to decrease with a constant angular acceleration of -2.10 rad/s².
- (a) Determine time until the pulley stops:
 Using $ext{ω}f = ext{ω}i + ext{α}t$
 $t = \frac{ ext{ω}f - ext{ω}i}{ ext{α}} = \frac{0 - 5.40}{-2.10} = 2.571 ext{ s}$
- (b) Determine angle turned:
 $Δθ = \frac{ ext{ω}f^2 - ext{ω}i^2}{2 ext{α}} = \frac{0 - (5.40)^2}{2(-2.10)} = 6.943 ext{ rad}$

Discussion on how linear and rotational quantities can coalesce in motion analysis.
Linear speeds and angular speeds relationship: $v_t = rω$
- Tangential speed ( $v_t$ ) related to radius and angular velocity.
Tangential acceleration ( $at$ ): $at = rα$ (analogous relationship)
Centripetal acceleration defined:
$a_{cp} = \frac{v^2}{r} = rω^2$

Definition of rolling motion as a combination of translational and rotational motion.
- Formulation of total kinetic energy for rolling:
 $Ktotal = KEt + KEr$ $KEt = \frac{1}{2} mv^2$ $KE_r = \frac{1}{2} Iω^2$
Moment of Inertia (I) defined as an object’s resistance to angular acceleration.
- $I = mr^2$ where m = mass and r = radius, accounting for distribution of mass.
- For distinct shapes, such as hoops or disks:
- Hoop: $I_{hoop} = MR^2$
- Solid Disk: $I_{disk} = \frac{1}{2} MR^2$

Disk scenario: Given mass of 1.20 kg and radius of 10 cm, find:
- (a) Translational kinetic energy:
 $K{Et} = \frac{1}{2}mv^2 = \frac{1}{2}(1.20)(1.41^2) = 1.193 ext{ J}$
- (b) Rotational kinetic energy needs moment of inertia calculation and angular velocity:
 $K{Er} = \frac{1}{2}Iω^2 = \frac{1}{2} \frac{1}{2} m r^2 \frac{v^2}{r^2} = \frac{1}{4} mv^2 = \frac{1}{4}(1.20)(1.41^2) = 0.596 ext{ J}$
- (c) Total kinetic energy:
 $Ktotal = K{Et} + K{E_r} = 1.193 + 0.596 = 1.789 ext{ J}$

Pulley and block scenario: Calculate height a block rises with conservation of energy when rotation considered:
Set up energy conservation equations to solve for h:
$\frac{1}{2}mvi^2 + \frac{1}{2}Iωi^2 + mgh_i = mgh$

Moment of inertia problem: Given mass of a wheel (0.98 kg) and inertia (0.13 kg·m²), find radius:
I = MR^2
ightarrow R = rac{ ext{I}}{M} = 0.36 m

Two masses (5.0 kg and 3.0 kg) suspended creating an Atwood machine, to find mass of the pulley while considering energy conservation.
Balancing initial and final energies in configurations to find the mass:
$M = 2.148 kg$

Solid sphere descends ramp then launches horizontally. Calculate:
(a) Horizontal distance before landing, (b) revolutions made, and (c) effect of a frictionless ramp on distance.