Physics Exam 3 Study Guide: Rotational Motion, Center of Mass, and Angular Momentum

Center of Mass

Definition and Properties

The center of mass (CoM) is the point at which an object's mass can be considered to be concentrated for the purposes of analyzing translational motion. For objects of uniform density, the CoM coincides with the geometric center.

• Balance Point: The CoM is the balance point of the object.

• Lowest Position: The CoM tends to be at the lowest possible position for stability.

• Immobility Without External Forces: If there are no external forces, the CoM does not move.

Mathematical Definition

• The position of the center of mass along the x-axis is given by: $x_{CM} = \frac{\sum m_i x_i}{\sum m_i}$

• For discrete masses: $x_{CM} = \frac{m_1 x_1 + m_2 x_2 + \ldots}{m_1 + m_2 + \ldots}$

• Example: For two masses, $m_H$ at $x_H$ and $m_N$ at $x_N$: $x_{CM} = \frac{m_H x_H + m_N x_N}{m_H + m_N}$

Rotational Kinematics

Angular Position, Velocity, and Acceleration

Rotational motion describes how objects rotate about an axis. Key quantities include angular position, velocity, and acceleration.

• Angular Position (θ): Measured in radians. $\theta = \frac{s}{r}$, where $s$ is arc length and $r$ is radius.

• Angular Velocity (\omega): Rate of change of angular position. $\omega = \frac{d\theta}{dt}$, units: rad/s.

• Angular Acceleration (\alpha): Rate of change of angular velocity. $\alpha = \frac{d\omega}{dt}$, units: rad/s².

• Linear and Tangential Quantities: Translational (linear) motion refers to $x$, $v$, $a$; tangential velocity $v = r\omega$.

• Direction: Counterclockwise is positive; clockwise is negative.

Equations of Rotational Motion

• $\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$

• $\omega = \omega_0 + \alpha t$

• $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$

Example: Car Wheel

• Given: Car moves at 60 mph ($= 27.3$ m/s), wheel radius $= 0.692$ m.

• Find angular velocity: $v = r\omega \implies \omega = \frac{v}{r} = \frac{27.3}{0.692} = 39.4$ rad/s

• Convert to rpm: $\text{rpm} = \omega \times \frac{60}{2\pi}$

Example: Earth's Surface

• Earth's radius $= 6.37 \times 10^6$ m.

• Find linear speed at equator: $v = \omega r$, $\omega = \frac{2\pi}{\text{day}} = \frac{2\pi}{86400\ \text{s}}$ $v = (7.27 \times 10^{-5}\ \text{rad/s})(6.37 \times 10^6\ \text{m}) = 463$ m/s

Moment of Inertia

Definition and Calculation

The moment of inertia (I) quantifies how mass is distributed relative to an axis of rotation. It determines the resistance to angular acceleration.

• Formula: $I = \sum m_i r_i^2$ for discrete masses; $I = \int r^2 dm$ for continuous bodies.

• Depends on: Mass distribution and axis of rotation.

• Example: Wheel

  • For a solid disk: $I = \frac{1}{2} m r^2$

  • For a ring: $I = m r^2$

  • For a sphere: $I = \frac{2}{5} m r^2$

• Parallel Axis Theorem: If the axis is shifted by distance $d$ from the center of mass axis: $I_{\text{total}} = I_{\text{CM}} + m d^2$

Angular Momentum

Definition and Conservation

Angular momentum (L) is a measure of the rotational motion of an object. It is conserved in the absence of external torques.

• Formula: $L = I\omega$

• Units: kg·m²/s

• Conservation Law: $L_i = L_f$ (if no external torque)

• For a particle: $L = r \times p = mvr$

• Types: Intrinsic (spinning about own axis), Orbital (about another point)

Example: Merry-Go-Round Problem

• A 500 kg merry-go-round (treated as a disk, $I = \frac{1}{2} m r^2$) is initially at rest. A 60 kg person walks to the edge and starts walking at 2 m/s clockwise. Find the new angular velocity.

• Apply conservation of angular momentum: $0 = I_{\text{mgr}} \omega_{\text{mgr}} + I_{\text{person}} \omega_{\text{person}}$ Solve for $\omega_{\text{mgr}}$.

Rotational Kinetic Energy

Energy in Rotational Motion

Rotating objects possess kinetic energy due to their motion about an axis.

• Translational Kinetic Energy: $KE = \frac{1}{2} m v^2$

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

• Units: Joules (J)

• Rotational Work: $W_{\text{rot}} = \Delta KE_{\text{rot}} = \frac{1}{2} I (\omega_f^2 - \omega_i^2)$

Example: Work Done on Merry-Go-Round

• Initial $\omega_i = 1.05$ rad/s, final $\omega_f = 0.984$ rad/s, $I_{\text{mgr}} = 1240$ kg·m², $I_{\text{person}} = 37$ kg·m².

• Calculate initial and final $KE_{\text{rot}}$: $KE_i = \frac{1}{2} (1240)(1.05^2) = 682$ J $KE_f = \frac{1}{2} (1240 + 37)(0.984^2) = 603$ J $W = KE_f - KE_i = 603 - 682 = -79$ J

Summary Table: Key Rotational Quantities

Quantity	Symbol	Formula	Units
Angular Position	$\theta$	$\theta = \frac{s}{r}$	radians (rad)
Angular Velocity	$\omega$	$\omega = \frac{d\theta}{dt}$	rad/s
Angular Acceleration	$\alpha$	$\alpha = \frac{d\omega}{dt}$	rad/s²
Moment of Inertia	$I$	$I = \sum m_i r_i^2$	kg·m²
Angular Momentum	$L$	$L = I\omega$	kg·m²/s
Rotational Kinetic Energy	$KE_{\text{rot}}$	$\frac{1}{2} I \omega^2$	J

Additional info:

• Some notes reference the ability of the center of mass to move for objects that can change shape (e.g., a hollow sphere).

• Intrinsic and orbital angular momentum are both important in physics, especially in atomic and planetary systems.

• Parallel axis theorem is crucial for calculating moments of inertia when the axis does not pass through the center of mass.