Uniform Circular Motion and Centripetal Forces: Study Notes for College Physics

Uniform Circular Motion

Definition and Characteristics

Uniform circular motion refers to the movement of an object along a circular path at a constant speed. Although the speed remains constant, the direction of the velocity changes continuously, resulting in acceleration.

• Acceleration Direction: The acceleration in uniform circular motion is always directed toward the center of the circle (centripetal acceleration).

• Velocity: The velocity vector is tangent to the circle at every point.

• Key Point: Acceleration is not parallel to velocity; it is perpendicular and points inward.

Example Question

In uniform circular motion, the acceleration:

• Is directed toward the center of the circle.

Period, Frequency, and Speed in Circular Motion

Definitions

The period (T) is the time required for one complete revolution around the circle. The frequency (f) is the number of revolutions per second.

• Frequency Formula: $f = \frac{1}{T}$ Frequency is measured in inverse seconds (s-1), also called Hertz (Hz).

• Speed Formula: $v = \frac{2\pi r}{T}$ Alternatively, $v = 2\pi r f$

Centripetal Acceleration

• Formula for Centripetal Acceleration: $a = \frac{v^2}{r}$ Substituting for v gives: $a = (2\pi f)^2 r = \left(\frac{2\pi}{T}\right)^2 r$

Example: Spinning Table Saw Blade

A table saw blade with a diameter of 25 cm spins at 3600 rpm. Calculate the period, speed at the edge, and centripetal acceleration.

• Period: $T = \frac{1}{f}$ $f = 3600 \text{ rpm} = 60 \text{ rps}$ $T = \frac{1}{60} = 0.017 \text{ s}$

• Speed: $v = \frac{2\pi r}{T} = \frac{2\pi (0.125)}{0.017} = 46.2 \text{ m/s}$

• Centripetal Acceleration: $a = \frac{v^2}{r} = \frac{46.2^2}{0.125} = 17075.52 \text{ m/s}^2$

Forces in Uniform Circular Motion

Newton's Second Law Applied

Objects in uniform circular motion experience a net force directed toward the center of the circle, called the centripetal force. According to Newton's second law:

• Net Force Formula: $\vec{F}_{\text{net}} = m\vec{a} = m \frac{v^2}{r}$ The net force is always toward the center of the circle.

Sources of Centripetal Force

• Tension: For an object tied to a string and swung in a circle, the tension in the string provides the centripetal force.

• Friction: For a car turning on a flat road, static friction between the tires and the road provides the centripetal force.

• Normal Force: In cases such as a car moving through a dip, the normal force from the road can contribute to the centripetal force.

Example: Ice Hockey Puck

An ice hockey puck tied to a string and swung in a circle experiences centripetal acceleration due to the tension in the string.

• Correct Force: Tension in the string

Applications: Forces on a Car in Circular Motion

Road Design and Circular Segments

Roads are often designed with curves and dips as segments of circles to manage the forces experienced by vehicles.

• Normal Force in a Dip: At the bottom of a dip, the normal force is greater than the car's weight due to the upward centripetal acceleration.

• Apparent Weight: Drivers feel heavier at the bottom of a dip because the normal force (apparent weight) exceeds the true weight.

• Turning a Corner: The frictional force between the tires and the road provides the necessary centripetal force for turning. On a frictionless (icy) road, a car cannot turn and will slide straight.

• Static vs. Kinetic Friction: Static friction is responsible for turning because the tires do not slide relative to the road. If the tires skid, kinetic friction takes over, and the car cannot turn effectively.

Example: Maximum Speed in a Turn

What is the maximum speed with which a 1500 kg car can make a turn of radius 20 m on a flat road with a coefficient of friction $\mu = 1.0$?

• Maximum Speed Formula: $v = \sqrt{\mu g r}$

Summary Table: Sources of Centripetal Force

Situation	Source of Centripetal Force
Object on a string	Tension
Car turning on a road	Static friction
Rider in a carnival ride	Normal force, tension, or friction (depending on setup)
Planet orbiting a star	Gravitational force

Key Equations Reference

• Frequency: $f = \frac{1}{T}$

• Speed: $v = 2\pi r f$ or $v = \frac{2\pi r}{T}$

• Centripetal Acceleration: $a = \frac{v^2}{r}$

• Centripetal Force: $F_{\text{c}} = m \frac{v^2}{r}$

• Maximum Speed (friction): $v = \sqrt{\mu g r}$

Additional info: These notes expand on the provided slides and handwritten content to ensure completeness and clarity for exam preparation.