Comprehensive Physics Study Guide on Rotational Motion and Equilibrium

Angular Quantities and Circular Motion

• Angular Speed and Velocity Definitions:

  • Angular speed ($\omega$) is often measured in radians per second ($\text{rad/s}$) or revolutions per minute ($\text{rpm}$).
  • To convert from $\text{rpm}$ to $\text{rad/s}$, use the conversion factor: \omega = \text{rpm} \times \frac{2\times\text{\pi}}{60}.
  • Example (Flywheel): A flywheel turning at $813.0\,\text{rpm}$ has an angular speed of $85.14\,\text{rad/s}$.
  • Example (Turntable): An old LP record rotating at $33\,1/3\,\text{rpm}$ equates to approximately $3.49\,\text{rad/s}$.
  • Example (Satellite): An artificial satellite circling the Earth every $98\,\text{minutes}$ has an angular speed of $0.0011\,\text{rad/s}$.

• Angular Displacement ($\theta$):

  • Defined as the angle through which an object rotates, measured in radians ($\text{rad}$) or degrees ($^{\circ}$).
  • Example (Turntable): A $33\,\text{rpm}$ turntable rotates through $63^{\circ}$ in $0.32\,\text{s}$.

• Angular Acceleration ($\alpha$):

  • The rate of change of angular velocity, measured in $\text{rad/s}^2$.
  • Magnitude of average angular acceleration is given by $\alpha = \frac{\omega_f - \omega_i}{t}$.
  • Example (Fan): If a fan speed decreases from $10\,\text{rad/s}$ to $6.3\,\text{rad/s}$ in $5.0\,\text{s}$, the average angular acceleration magnitude is $0.74\,\text{rad/s}^2$.

• Angular and Linear Relationships:

  • Linear distance ($s$) moved by a point on a rotating rim: $s = r \times \theta$.
  • Linear velocity ($v$) of a point at radius $r$: $v = r \times \omega$.
  • Linear acceleration components:
    • Tangential acceleration ($a_t$): $a_t = r \times \alpha$.
    • Centripetal/Radial acceleration ($a_r$): $a_r = r \times \omega^2$ or $a_r = \frac{v^2}{r}$.
  • Ahmed and Jacques Case Study: On a merry-go-round, Ahmed (greater distance from center) has a greater tangential speed than Jacques, but both share the same angular speed ($\omega$).
  • Tire Speed Relative to Highway: For a car moving at $65\,\text{mph}$, the highest point of the tire moves at $130\,\text{mph}$, the center at $65\,\text{mph}$, and the lowest point (contact with ground) at $0\,\text{mph}$.

• Angular Measurements in Astronomy:

  • Distance of objects can be determined using the angle subtended ($\theta$) and the diameter ($s$): $s = r \times \theta$.
  • Moon: Diameter is $3.47 \times 10^6\,\text{m}$, subtends $0.00904\,\text{rad}$, distance from Earth is $3.84 \times 10^8\,\text{m}$.
  • Sun: Subtends $0.00928\,\text{rad}$, distance is $1.5 \times 10^{11}\,\text{m}$, diameter is $1.4 \times 10^9\,\text{m}$.

Kinematics of Constant Angular Acceleration

• Equations for Uniform Angular Acceleration:

  1. $\omega_f = \omega_i + \alpha \times t$
  2. $\theta = \omega_i \times t + \frac{1}{2} \times \alpha \times t^2$
  3. $\omega_f^2 = \omega_i^2 + 2 \times \alpha \times \theta$

• Application Examples:

  • Speedup: An object speeding up from $15.0\,\text{rad/s}$ to $33.3\,\text{rad/s}$ with $\alpha = 3.45\,\text{rad/s}^2$ takes $5.30\,\text{s}$.
  • Stopping Ferris Wheel: A wheel rotating at $20\,\text{rad/s}$ slowing at $5.0\,\text{rad/s}^2$ makes $6.4\,\text{revolutions}$ before stopping.
  • Grinding Wheel Sequence: A wheel accelerating from rest for $10\,\text{s}$ to $38\,\text{rad/s}$ has an acceleration of $3.8\,\text{rad/s}^2$.

Moment of Inertia ($I$)

• Definition: The Measure of an object's resistance to rotational acceleration, defined as $I = \sum (m_i \times r_i^2)$.

• Common Geometric Shapes:

  • Solid Sphere (uniform): $I = \frac{2}{5} \times M \times R^2$
  • Solid Cylinder or Disk (uniform): $I = \frac{1}{2} \times M \times R^2$
  • Hollow Pipe/Hoop (thin-walled): $I = M \times R^2$
  • Thin Rod (about center): $I = \frac{1}{12} \times M \times L^2$
  • Thin Rod (about end): $I = \frac{1}{3} \times M \times L^2$

• Systems of Point Masses:

  • Triatomic Molecule: Masses $m$ at origin, $2m$ at $x=a$, and $3m$ at $x=2a$. Moment of inertia about the y-axis is $I = (2m \times a^2) + (3m \times (2a)^2) = 14ma^2$.
  • L-Shaped Object: If masses are distributed on an L-frame, calculate $I_x$ and $I_y$ by summing $m \times r^2$ for each mass relative to the specific axis.

Rotational Kinetic Energy ($K_{rot}$) and Rolling Motion

• Formulas:

  • Rotational Kinetic Energy: $K_{rot} = \frac{1}{2} \times I \times \omega^2$
  • Total Kinetic Energy of a Rolling Object: $K_{total} = K_{trans} + K_{rot} = \frac{1}{2} \times M \times v^2 + \frac{1}{2} \times I \times \omega^2$

• Rolling Without Slipping Concept:

  • Linear and angular speeds are locked: $v = R \times \omega$.
  • For a uniform hoop rolling without slipping, rotational kinetic energy and translational kinetic energy are equal ($K_{trans} = K_{rot}$).
  • For a solid cylinder rolling without slipping, rotational kinetic energy is $1/3$ of its total kinetic energy.

• Competition on an Inclined Plane:

  • When released from rest, the object with the smallest moment of inertia relative to its mass ($\frac{I}{MR^2}$) will have the greatest linear velocity at the bottom and arrive first.
  • Ranking: Sphere ($0.4 \times MR^2$) > Cylinder ($0.5 \times MR^2$) > Hoop/Pipe ($1.0 \times MR^2$).
  • Result: The sphere reaches the bottom first.
  • The linear velocity at the bottom depends on the shape of the object but neither the mass nor the radius themselves ($v = \sqrt{\frac{2gh}{1 + I/MR^2}}$).

Torque ($\tau$) and Rotational Dynamics

• Torque Definition: The measurement of a force's tendency to rotate an object about an axis.

  • $\tau = r \times F \times \sin(\theta)$, where $\theta$ is the angle between the force and the lever arm.
  • Maximum torque occurs when the force is applied perpendicular to the lever arm ($\theta = 90^{\circ}$, $\sin(90^{\circ}) = 1$).
  • Wrench Example: To achieve $90\,\text{N}\cdot\text{m}$ of torque with a $30\,\text{cm}$ wrench, a minimum force of $300\,\text{N}$ is required.

• Newton's Second Law for Rotation:

  • $\sum \tau = I \times \alpha$
  • Application (Bicycle Wheel): A torque of $0.850\,\text{N}\cdot\text{m}$ applied to a hoop-like wheel ($M = 0.750\,\text{kg}$, $R = 0.33\,\text{m}$) results in an angular acceleration $\alpha = 10.4\,\text{rad/s}^2$.

• Work and Power in Rotation:

  • Work done by torque: $W = \tau \times \theta$ (where $\theta$ is in radians).
  • Ice Cream Maker Example: Turning a crank with $4.50\,\text{N}\cdot\text{m}$ for $300\,\text{turns}$ involves software: W = 4.50 \times (300 \times 2\text{\pi}) = 8480\,\text{J}.

Angular Momentum ($L$)

• Definitions and Units:

  • For a rotating rigid body: $L = I \times \omega$.
  • For a point mass: $L = r \times p \times \sin(\theta) = m \times v \times r_{\perp}$.
  • Units: $\text{kg}\cdot\text{m}^2/\text{s}$.

• Conservation of Angular Momentum:

  • $\mathbf{L}$ is constant if the net external torque is zero ($\sum \tau_{ext} = 0$).
  • Ice Skater Application: When a spinner pulls her arms in, her moment of inertia ($I$) decreases. To conserve $L = I \times \omega$, her angular speed ($\omega$) must increase. Her angular momentum remains constant, but her kinetic energy increases (work is done by the muscles).
  • Diego on Merry-Go-Round: Moving to the center decreases the system's moment of inertia and increases angular speed.

• Relationship with Vectors:

  • The direction of the angular velocity vector for a regular bicycle wheel moving forward is to the left (following the right-hand rule).

Static Equilibrium

• Two Degrees of Conditions for Equilibrium:

  1. The net external force must be zero: $\sum \mathbf{F} = 0$.
  2. The net external torque must be zero: $\sum \mathbf{\tau} = 0$.
  • Note: If $\sum F = 0$, $\sum \tau$ is not necessarily zero (e.g., a couple). If both are zero, the object is not necessarily at rest (it could be in uniform motion), but it is in equilibrium.

• Center of Gravity (CG) and Center of Mass (CM):

  • For uniform objects, the CG is at the geometric center.
  • Weighing a Car: The total weight is the sum of the readings under the front and rear wheels.
  • Balance Example (Meter Stick): If a stick balances at the $50.0\,\text{cm}$ mark, and adding a $50.0\,\text{g}$ mass at the $90.0\,\text{cm}$ mark shifts the balance to the $61.3\,\text{cm}$ mark, the stick's mass is $127\,\text{g}$.

• Ladder and Friction Problems:

  • Stability is determined by torque balance around the base of the ladder.
  • Friction against the floor prevents the base from sliding: $f_s \times N_{wall}$.
  • Example (Ladder): A $3.00\,\text{m}$ ladder weighing $200\,\text{N}$ leaning against a smooth wall $1.00\,\text{m}$ from the base with a CG at $1.20\,\text{m}$ needs $28.3\,\text{N}$ of friction to stay stable.

Questions & Discussion

• Q: Does linear velocity of a rolling sphere depend on its mass?
  • A: No, the linear velocity depends on the height and the distribution of mass (shape factor), but not the total mass or radius ($M$ and $R$ cancel out in the conservation of energy equation).
• Q: If units are $\text{kg}\cdot\text{m}^2/\text{s}^2$, what could the quantity be?
  • A: Work, Rotational Kinetic Energy, or Torque (Energy/Torque units are identical).
• Q: Why does a ball roll higher up a hill with friction than without it?
  • A: With friction, the ball can transform both translational and rotational kinetic energy into potential energy. Without friction, the ball cannot stop spinning, so it retains its rotational kinetic energy at the peak, reaching a lower height.
• Q: What is the tension in the tie rod of a stepladder?
  • A: For a person ($800\,\text{N}$) standing $3/5$ of the way up a ladder on a smooth floor with a specific tie rod geometry ($2.5\,\text{m}$ halves, $70\,\text{cm}$ rod), the tension is $140\,\text{N}$.