Uniform Circular Motion Notes

Uniform Circular Motion

• Definition: Uniform circular motion is described as motion in a circle of constant radius at a constant speed.

• Characteristics: - The instantaneous velocity of an object in uniform circular motion is always tangent to the circle.

Analysis of Circular Motion with a Car

• Scenario: Consider a car driving at constant speed around a circular path.

• Velocity Representation: At two specific instants, we can sketch the velocity of the car to visualize its direction.

Forces and Friction in Circular Motion

• Friction's Role: If there is inadequate friction between the tires and the road, we must consider: - The direction in which the car would travel if friction were absent.

  • The actual location of the car on the path.

  • The required direction of the friction force exerted on the tires to maintain circular motion.

Radius of Curvature

• Concept: Uniform circular motion can also be applied to paths that are portions of circles.

• Radius of Curvature: This term refers to the radius that the path would have if completed as a full circle.

Acceleration in Uniform Circular Motion (UCM)

• Centripetal Acceleration: If the magnitude of an object's velocity is known, the centripetal acceleration can be calculated. - The formula for centripetal acceleration, $ac$ ,is given by: $ac=\frac{v}{r}^2$

  where:

  • $v$ = velocity

  • $r$ = radius of circular path.

Comparison of Curves A and B

• Given: A comparison is made to determine which curve requires more acceleration.

• Information: The velocity is held constant at 30 mph (approximately 13 m/s).

• Question: Assess which of the two curves requires more friction between the tires and the road.

• Solution: Since centripetal acceleration is given by $ac=\frac{v^2}{r}$ and the required friction force is $Ff=ma_{c}$, the curve with the smaller radius ($r$) will require a larger centripetal acceleration and consequently more friction to negotiate at a constant speed.

Skidding on Curves

• Scenario Analysis: A 1000 kg car attempts to round a curve on a flat road with a radius of 50.0 m at a speed of 13.9 m/s.

• Friction Forces: - On a dry road, the frictional force is 5880 N.

  • On an icy road, the frictional force is 2450 N.

• Key Questions: - A) Will the car successfully follow the curve on dry pavement?

  • B) Will the car remain on the curve on icy pavement?

• Solution: First, calculate the required centripetal force:

  $F_c = m \frac{v^2}{r} = (1000 \text{ kg}) \frac{(13.9 \text{ m/s})^2}{50.0 \text{ m}} = 1000 \cdot \frac{193.21}{50.0} = 3864.2 \text{ N}$

  • A) On dry pavement, the maximum available friction is 5880 N. Since the required force (3864.2 N) is less than the available friction, the car will successfully follow the curve.

  • B) On icy pavement, the maximum available friction is 2450 N. Since the required force (3864.2 N) is greater than the available friction, the car will not remain on the curve; it will skid off.

Loop-the-Loop Problem

• Historical Context: In 1901, Allo "Dare Devil" Diavolo performed a stunt riding a bicycle in a loop.

• Physics Problem: Assuming the loop has a radius of r = 2.7 m, determine the minimum speed, v, that Diavolo must maintain at the top of the loop to stay in contact with it.

• Solution: At the minimum speed at the top of the loop, the normal force becomes zero, and the centripetal force required is provided solely by gravity. Thus, the gravitational force equals the centripetal force:

  $mg = m \frac{v^2}{r}$

  $g = \frac{v^2}{r}$

  $v = \sqrt{gr}$

  Substituting the values:

  $v = \sqrt{(9.8 \text{ m/s}^2)(2.7 \text{ m})} = \sqrt{26.46} \approx 5.14 \text{ m/s}$

  Diavolo must maintain a minimum speed of approximately $5.14 m/s$ at the top of the loop.

Free Body Diagrams

• Analysis Task: Match a provided free body diagram with the appropriate image.

• Discussion: Invent hypothetical free body diagrams for two additional images related to the topic.

Newton's Law of Universal Gravitation

• Observations: Planets follow curved paths as they revolve around the sun. - The direction of this motion is always towards the sun.

  • There must exist a net force acting on the planets to produce the required centripetal acceleration.

Gravitational Force Formula

• Definition: The gravitational force is the attractive force between two masses, acting along the line joining their centers. - This force adheres to Newton's Third Law, which states that for every action, there is an equal and opposite reaction.

• Formula for Gravitational Force: $F=\frac{G\times m1\times m2}{r^2}$ Where:

  • $F$ = gravitational force

  • $G$ = gravitational constant ($6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2$)

  • m1 and m2 = two masses

  • $r$ = distance between the centers of the two masses.

Gravitational Attraction Between Earth and Jupiter

• Context: Examining the gravitational force exerted by Earth on Jupiter when they are at their closest approach.

• Mass of Jupiter: $m_J = 1.898 \times 10^{27} \text{ kg}$

• Mass of Earth: $m_E = 5.972 \times 10^{24} \text{ kg}$

• Distance: $r = 588 \text{ million km} = 588 \times 10^9 \text{ m}$.

• Solution: Using the gravitational force formula:

  $F = \frac{(6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2) \times (5.972 \times 10^{24} \text{ kg}) \times (1.898 \times 10^{27} \text{ kg})}{(588 \times 10^9 \text{ m})^2}$

  $F \approx \frac{7.545 \times 10^{41}}{3.457 \times 10^{23}} \approx 2.183 \times 10^{18} \text{ N}$

  The gravitational attraction between Earth and Jupiter at their closest approach is approximately $2.183 \times 10^{18} \text{ N}$.

Pancreatic Attraction Example

• Humorous Analogy: “My pancreas attracts every other pancreas in the universe with a force proportional to the product of their masses and inversely proportional to the distance between them”.

• Considerations: - Mass of a pancreas = 80 g.

  • If a classmate is 5 m away, calculate the force attracting the two pancreases using the provided analogy.

• Solution: First, convert the mass to kilograms: $80 \text{ g} = 0.080 \text{ kg}$. Then, apply the gravitational force formula:

• $F=\frac{G\times m1\times m2}{r^2}=\frac{\left(6.674\times10^{-11}N\cdot\frac{m}{\operatorname{kg}^2}^2\right)\times\left(0.080\operatorname{kg}\right)\times\left(0.080\operatorname{kg}\right)}{5m^2}$

  $F = \frac{(6.674 \times 10^{-11}) \times 0.0064}{25}$

  $F \approx \frac{4.271 \times 10^{-13}}{25} \approx 1.71 \times 10^{-14} \text{ N}$

  The force attracting the two pancreases is approximately $1.71 \times 10^{-14} \text{ N}$.

Force Exerted by Earth on Humans

• Gravity Calculation: How strong is Earth's pull on a person weighing 60 kg? - Gravitational force can be calculated using the formula: $F = mg$

  • Where: - $m$ = mass (60 kg)

    • $g$ = acceleration due to gravity ($g = 9.8 \text{ m/s}^2$).

• Solution: Using the formula $F = mg$:

  $F = (60 \text{ kg}) \times (9.8 \text{ m/s}^2) = 588 \text{ N}$

  Earth's pull on a 60 kg person is 588 N.

Kepler's Laws of Planetary Motion

• Overview: These laws describe the motion of planets around the sun.

Law of Orbits

• Statement: All planets move in elliptical orbits around the sun, which is located at one focus of the ellipse.

Law of Areas

• Description: A line connecting a planet to the sun sweeps out equal areas in equal times. - This law arises from the conservation of angular momentum.

Law of Periods

• Formula: The relationship can be mathematically expressed as: $\frac{T^2}{r^3} = \text{constant}$

  • Where $T$ is the orbital period in years and $r$ is the average distance (semimajor axis) from the sun in astronomical units (AU).

• Table of Planets and Periods: Observations on various planets, including Earth and Pluto, presented in a comparison of their orbital periods related to the semimajor axes.

Calculation of Pluto's Average Distance from the Sun

• Example Context: Given Pluto's orbital period of 250 years, determine its average distance from the sun using the Law of Periods.

• Solution: According to Kepler's Third Law, $\frac{T^2}{r^3} = \text{constant}$. For objects in our solar system, if $T$ is in years and $r$ is in AU, the constant is 1 (since for Earth, $T=1$ year and $r=1$ AU). Therefore:

  $r^3 = T^2$

  $r = T^{2/3}$

  For Pluto, $T = 250$ years:

  $r = (250)^{2/3} = (250^2)^{1/3} = (62500)^{1/3} \approx 39.69 \text{ AU}$

  The average distance of Pluto from the sun is approximately $39.69 \text{ AU}$.

Formula Sheet

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

  • $a_c$ = centripetal acceleration

  • $v$ = velocity (speed)

  • $r$ = radius of circular path

• Centripetal Force: $F<em>c = m a</em>c = m \frac{v^2}{r}$

  • $F_c$ = centripetal force

  • $m$ = mass of the object

  • $a_c$ = centripetal acceleration

  • $v$ = velocity (speed)

  • $r$ = radius of circular path

• Minimum Speed for Loop-the-Loop (at the top, where normal force is zero): $v = \sqrt{gr}$

  • $v$ = minimum speed

  • $g$ = acceleration due to gravity ($9.8 \text{ m/s}^2$ on Earth)

  • $r$ = radius of the loop

• Newton's Law of Universal Gravitation: $F = \frac{G \times m<em>1 \times m</em>2}{r^2}$

  • $F$ = gravitational force

  • $G$ = gravitational constant ($6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2$)

  • $m_1$ = mass of the first object

  • $m_2$ = mass of the second object

  • $r$ = distance between the centers of the two masses

• Gravitational Force (Weight): $F = mg$

  • $F$ = gravitational force (weight)

  • $m$ = mass of the object

  • $g$ = acceleration due to gravity ($9.8 \text{ m/s}^2$)

• Kepler's Law of Periods: $\frac{T^2}{r^3} = \text{constant}$ (For solar system with $T$ in years and $r$ in AU, this simplifies to $r^3 = T^2$)

  • $T$ = orbital period

  • $r$ = average distance (semimajor axis)