Rotation I: Center of Mass and Rotational Kinematics

Center of Mass and Center of Gravity

Definition and Physical Meaning

The center of mass of a system is the point at which the entire mass of the system can be considered to be concentrated for the purpose of analyzing translational motion. The center of gravity is the point where the gravitational force can be considered to act on the body.

• Translational motion refers to the movement of the center of mass of an object.

• Vibrational and rotational motion occur about the center of mass.

Calculating the Center of Mass for Discrete Systems

For a system of particles, the center of mass coordinates $(x_{cm}, y_{cm})$ are given by:

$x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$

$y_{cm} = \frac{\sum m_i y_i}{\sum m_i}$

• Each mass $m_i$ is located at position $(x_i, y_i)$.

• The sum is taken over all masses in the system.

Example: Barbell System

• Two balls: 0.5 kg at (0,0), 2 kg at (0.5,0) m.

• Rod is massless and 0.5 m long.

• Calculation: $x_{cm} = \frac{(0.5)(0) + (2)(0.5)}{2.5} = 0.4$ m, $y_{cm} = 0$ m.

Example: Swapping Masses

• 0.5 kg at (0.5,0), 2 kg at (0,0).

• Calculation: $x_{cm} = \frac{(2)(0) + (0.5)(0.5)}{2.5} = 0.1$ m.

Example: Including Object Size

• 2 kg ball (radius 10 cm) at (0.65,0), 0.5 kg ball (radius 5 cm) at (0,0).

• Rod is 0.5 m long, attached at edges.

• Calculation: $x_{cm} = \frac{(0.5)(0) + (2)(0.65)}{2.5} = 0.52$ m.

Example: Rod with Mass

• Rod mass: 0.1 kg, uniformly distributed, center at (0.3,0).

• Calculation: $x_{cm} = \frac{(0.5)(0) + (0.1)(0.3) + (2)(0.65)}{2.6} = 0.51$ m.

Center of Mass of the Human Body

The center of mass of the human body can be estimated by considering the mass and position of different body segments.

Body Segment	Distance from Floor (% Height)	Center of Mass Location (% Height)
Head & Neck	91.2%	Base of skull
Shoulder Joint	81.2%	Shoulder joint
Hip Joint	61.8%	Hip joint
Knee Joint	28.5%	Knee joint
Ankle Joint	4.1%	Ankle joint

To find the center of mass of the whole body, use the weighted average of the positions and masses of each segment.

Example: Two-Segment Body

• Upper body: 45 kg at (15,50) cm

• Lower body: 30 kg at (30,20) cm

• Calculation: $x_{cm} = \frac{(45)(15) + (30)(30)}{75} = 21$ cm, $y_{cm} = \frac{(45)(50) + (30)(20)}{75} = 38$ cm

Rotational Motion of Rigid Bodies

Basic Concepts

Rotational motion involves the movement of a rigid body about a fixed axis. The key variables are:

• Angular displacement $(\theta)$: The angle through which a point or line has been rotated in a specified sense about a specified axis.

• Angular velocity $(\omega)$: The rate of change of angular displacement.

• Angular acceleration $(\alpha)$: The rate of change of angular velocity.

For all points at radius $r$ from the axis, $\omega$ and $\alpha$ are the same, but the linear velocity $v$ and linear acceleration $a$ depend on $r$.

Relationships Between Linear and Angular Quantities

• Linear velocity: $v = \omega r$

• Linear (tangential) acceleration: $a_t = \alpha r$

• Radial (centripetal) acceleration: $a_r = \omega^2 r$

Equations of Rotational Kinematics (Constant Angular Acceleration)

• Angular displacement: $\Delta \theta = \omega_0 t + \frac{1}{2} \alpha t^2$

• Angular velocity: $\omega = \omega_0 + \alpha t$

• Angular velocity squared: $\omega^2 = \omega_0^2 + 2 \alpha \Delta \theta$

Unit Conversion Example

• Convert 500 rev/min to rad/s: $500 \frac{\text{rev}}{\text{min}} \times \frac{1 \text{ min}}{60 \text{ s}} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} = 52.3 \text{ rad/s}$

Worked Examples in Rotational Motion

Example 1: Tangential Speed on a Rotating Disk

• Disk rotates at 500 rev/min, diameter = 120 mm ($r = 0.06$ m).

• Angular velocity: $\omega = 52.3$ rad/s.

• At edge: $v = \omega r = (52.3)(0.06) = 3.1$ m/s.

• Halfway to edge ($r = 0.03$ m): $v = (52.3)(0.03) = 1.6$ m/s.

Example 2: Rotational Kinematics of a Wheel

• Wheel radius: 0.1 m, $\alpha = 1.0$ rad/s$^2$, $\omega_0 = 2$ rad/s, $t = 5$ s.

• a) Angular velocity: $\omega = \omega_0 + \alpha t = 2 + (1)(5) = 7$ rad/s.

• b) Angular displacement: $\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 = \frac{3\pi}{2} + 2(5) + \frac{1}{2}(1)(5^2) = 27$ rad.

• c) Tangential velocity: $v_t = \omega r = (7)(0.1) = 0.7$ m/s.

Summary Table: Rotational vs. Translational Quantities

Translational	Rotational
Displacement ($x$)	Angular Displacement ($\theta$)
Velocity ($v$)	Angular Velocity ($\omega$)
Acceleration ($a$)	Angular Acceleration ($\alpha$)
Mass ($m$)	Moment of Inertia ($I$) Additional info: Moment of inertia is the rotational analog of mass, quantifying resistance to angular acceleration.

Key Takeaways

• The center of mass is crucial for analyzing both translational and rotational motion.

• Rotational kinematics parallels linear kinematics, with angular variables replacing linear ones.

• Understanding the relationships between angular and linear quantities is essential for solving problems in rotational dynamics.