Rotational Motion, Static Equilibrium, and Elasticity: Study Notes for Physics for Life Sciences I (Lecture 13)

Rotational Motion and Circular Dynamics

Newton’s Second Law for Rotational Motion

Newton’s second law can be extended to rotational motion, describing how a net torque causes angular acceleration in an object.

• Angular acceleration ($\alpha$) is produced when a net torque ($\tau_{\text{net}}$) acts on an object with moment of inertia ($I$):

$\alpha = \frac{\tau_{\text{net}}}{I}$

• This is analogous to Newton’s second law for linear motion: $a = \frac{F_{\text{net}}}{m}$.

• Moment of inertia ($I$) quantifies an object’s resistance to changes in rotational motion, similar to mass in linear motion.

Describing Circular Motion

Special variables are used to describe circular motion, with counterclockwise defined as positive by convention.

• Angular displacement: $\Delta \theta = \theta_f - \theta_i$

• Angular velocity: $\omega = \frac{\Delta \theta}{\Delta t}$

• Angular acceleration: $\alpha = \frac{\Delta \omega}{\Delta t}$

• Angles are measured in radians: $1\ \text{rev} = 360^\circ = 2\pi\ \text{rad}$

• Angular velocity relates to frequency ($f$) and period ($T$): $\omega = \frac{2\pi}{T} = 2\pi f$

Relating Linear and Circular Quantities

Linear and angular motion are closely related for objects moving in circles.

• Linear speed ($v$) and angular speed ($\omega$) are related by: $v = \omega r$

• If the particle’s speed is changing, it has a tangential acceleration ($a_t$) related to angular acceleration ($\alpha$): $a_t = \alpha r$

Torque

Torque is the rotational equivalent of force, causing angular acceleration.

• There are two common interpretations:

  • 1. $\tau = r F_\perp$ (where $F_\perp$ is the component of force perpendicular to the lever arm)

  • 2. $\tau = r_\perp F$ (where $r_\perp$ is the perpendicular distance from the axis to the line of action of the force)

• General formula for torque magnitude: $\tau = r F \sin \phi$ where $\phi$ is the angle between $\vec{r}$ and $\vec{F}$.

Moment of Inertia

The moment of inertia ($I$) is a measure of how mass is distributed relative to the axis of rotation.

• For discrete masses: $I = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 + \ldots = \sum m_i r_i^2$

• For continuous bodies, integration is used (not shown here).

Center of Gravity

The center of gravity is the point where the entire weight of an object can be considered to act.

• For particles at positions $x_1, x_2, \ldots$ with masses $m_1, m_2, \ldots$:

$x_{cg} = \frac{x_1 m_1 + x_2 m_2 + x_3 m_3 + \ldots}{m_1 + m_2 + m_3 + \ldots}$

Rolling Motion and Constraints

Rolling Without Slipping

When an object rolls without slipping, its translational and rotational motions are linked.

• The velocity of the center of mass: $v = \omega R$

• For one full revolution, the center moves forward by the circumference: $\Delta x = 2\pi R$

• The point at the bottom of the wheel is instantaneously at rest relative to the surface.

Constraints Due to Ropes and Pulleys

When a rope does not slip on a pulley, the linear speed and acceleration of the rope match those of the pulley's rim.

• Velocity constraint: $v_{obj} = -\omega R$

• Acceleration constraint: $a_{obj} = -\alpha R$

Static Equilibrium

Conditions for Static Equilibrium

An object is in static equilibrium if it is at rest and remains at rest under the action of applied forces and torques.

• First condition: Net force is zero: $\sum \vec{F} = 0$

• Second condition: Net torque is zero: $\sum \tau = 0$

• Both conditions must be satisfied for extended objects.

Choosing the Pivot Point

The net torque is zero about any point for an object in static equilibrium. Choosing a convenient pivot can simplify calculations, especially where forces are unknown or act at the pivot.

Problem-Solving Approach for Static Equilibrium

1. Strategize: Identify the object and choose a pivot point.

2. Prepare: Draw a diagram, label forces, and measure distances from the pivot.

3. Solve: Write equations for $\sum F_x = 0$, $\sum F_y = 0$, and $\sum \tau = 0$. Solve for unknowns.

4. Assess: Check if the solution is reasonable and answers the question.

Example: Lifting a Load with a Lever

A lever problem involves balancing torques and forces to keep a beam in equilibrium. The force applied by a person and the reaction at the pivot can be found using the above approach.

Example: Ladder Against a Wall

To prevent slipping, the frictional force at the base must be sufficient. The minimum coefficient of static friction ($\mu_s$) can be found by setting up torque and force balance equations.

Stability and Balance

Stability of Extended Objects

An object is stable if its center of gravity is above its base of support. If the center of gravity moves outside the base, the object becomes unstable and may tip over.

• The critical angle for tipping depends on the width of the base ($w$) and the height of the center of gravity ($h$): $\theta_c = \tan^{-1}\left(\frac{w}{2h}\right)$

• Wider bases and lower centers of gravity increase stability.

Elasticity: Springs and Materials

Hooke’s Law for Springs

When a spring is stretched or compressed, it exerts a restoring force proportional to the displacement from equilibrium.

• Hooke’s law: $F_s = -k \Delta x$

• $k$ is the spring constant (N/m), a measure of stiffness.

• The negative sign indicates the force is opposite to the displacement.

Stretching and Compressing Materials

Most solid materials behave elastically for small deformations, obeying a linear relationship between force and extension.

• For a rod of length $L$ and cross-sectional area $A$: $k = \frac{YA}{L}$ where $Y$ is Young’s modulus (a material property).

• Young’s modulus quantifies the stiffness of a material.

Table: Young’s Modulus for Common Materials

Material	Young’s Modulus ($10^{10}$ N/m$^2$)
Cast iron	20
Steel	20
Silicon	13
Copper	11
Aluminum	7
Glass	7
Concrete	3
Wood (Douglas Fir)	1

Stress and Strain

• Stress: $\text{stress} = \frac{F}{A}$ (force per unit area, N/m$^2$)

• Strain: $\text{strain} = \frac{\Delta L}{L}$ (fractional change in length)

• Young’s modulus: $Y = \frac{\text{stress}}{\text{strain}}$

Elastic Limit and Tensile Strength

Materials obey Hooke’s law only up to their elastic limit. Beyond this, permanent deformation or breaking occurs.

• Ultimate stress (tensile strength): maximum stress a material can withstand before breaking.

Table: Young’s Modulus for Biological Materials

Material	Young’s Modulus ($10^{9}$ N/m$^2$)
Tooth enamel	6
Compact bone	1.6
Spongy bone	0.02–0.3
Spider silk	0.2
Tendon	0.15
Cartilage	0.0001
Blood vessel (aorta)	0.00005

Table: Tensile Strengths of Biological Materials

Material	Tensile Strength (N/m$^2$)
Spongy bone	$5 \times 10^6$
Compact bone	$100 \times 10^6$
Tendon	$100 \times 10^6$
Spider silk	$1000 \times 10^6$

Summary of Key Equations

• Angular acceleration: $\alpha = \frac{\tau_{\text{net}}}{I}$

• Linear/angular velocity: $v = \omega r$

• Torque: $\tau = r F \sin \phi$

• Moment of inertia (discrete): $I = \sum m_i r_i^2$

• Hooke’s law: $F_s = -k \Delta x$

• Young’s modulus: $Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L}$

Additional info: Some context and explanations have been expanded for clarity and completeness, including the explicit forms of equations and the inclusion of tables for material properties.