Centripetal Forces, Universal Gravitation, and Satellite Motion Study Guide

Puck on a Frictionless Table: Top-Down and Side Views

• System Setup: A puck of mass $m$ is tied to a fixed string and moves in a circular path at a constant speed on a frictionless horizontal tabletop.
• Given Parameters:     * Mass ($m$): $2\,kg$     * Speed ($v$): $10\,m/s$     * Radius ($r$): $2\,m$
• Free Body Diagram (Side View): While the setup is often viewed from the top to see the path, the side view is essential for a Free Body Diagram (FBD):     * Normal Force ($n$): Acts vertically upward from the table.     * Weight ($mg$): Acts vertically downward.     * Tension ($T$): Acts horizontally toward the center of the circle.     * Acceleration ($a_c$): Points toward the center of the circular path.
• Applying Newton's Second Law:     * In the horizontal dimension, tension is the only force acting toward the center.     * Equation: $T = m \times a_c$     * Substitute centripetal acceleration ($a_c = \frac{v^2}{r}$): $T = m \times \frac{v^2}{r}$
• Numerical Calculation:     * $T = 2\,kg \times \frac{(10\,m/s)^2}{2\,m}$     * $T = 2 \times \frac{100}{2} = 100\,N$

Car Negotiating a Flat Horizontal Curve

• Scenario: A $1500\,kg$ car travels through a curve on a flat road at a constant speed.
• Given Parameters:     * Mass ($m$): $1500\,kg$     * Radius of Curvature ($r$): $35\,m$     * Coefficient of Static Friction ($\mu_s$): $0.5$
• Goal: Find the maximum speed ($v_{max}$) the car can maintain without skidding.
• Radius of Curvature Defined: If the curve were extended to form a full circle, the radius of that circle would be $35\,m$.
• Friction Analysis:     * Type of Friction: Static friction ($f_s$) is used despite the car moving. This is because there is no relative motion between the point of contact of the tire and the road (unless the car is skidding).     * Maximum Static Friction: The limit right before skidding is $f_{s,max} = \mu_s \times n$.
• Free Body Diagram (Side View):     * Vertical: Normal force ($n$) and weight ($mg$). Since there is no vertical acceleration, $n - mg = 0$, so $n = mg$.     * Horizontal: The only force providing centripetal acceleration is static friction ($f_s$).
• Newton's Second Law Equations:     * $f_s = m \times a_c = m \times \frac{v^2}{r}$     * Substituing $f_{s,max}$: $\mu_s \times mg = m \times \frac{v^2}{r}$     * Mass ($m$) cancels from both sides: $\mu_s \times g = \frac{v^2}{r}$     * Solving for speed: $v = \sqrt{\mu_s \times r \times g}$
• Numerical Calculation:     * $v = \sqrt{0.5 \times 35\,m \times 9.8\,m/s^2}$     * $v \approx 13.1\,m/s$

Banked Curves and Design Speed

• Concept: A banked curve is a road slanted toward the inside of the curve. This design allows cars to negotiate turns even when the road is perfectly icy (frictionless).
• Scenario (Icy Road Logic): An exit ramp is banked at an angle $\theta$ so that a car can maintain its path without relying on friction at a specific speed.
• Given Parameters:     * Designated Speed ($v$): $13.4\,m/s$     * Radius of Curvature ($r$): $50\,m$
• Goal: Find the bank angle ($\theta$).
• Coordinate System Selection: Select axes such that the x-axis points toward the center of the circle (the direction of centripetal acceleration). This prevents the need to break acceleration into components.
• Free Body Diagram (Frictionless Case):     * Weight ($mg$): Points straight down.     * Normal Force ($n$): Points perpendicular to the slanted road surface.
• Geometric Relationship: The angle $\theta$ between the horizontal and the road is the same as the angle between the vertical (y-axis) and the normal force.
• Equations from Newton's Second Law:     * Vertical ($y$): $n \cos(\theta) - mg = 0 \rightarrow n = \frac{mg}{\cos(\theta)}$     * Horizontal ($x$): $n \sin(\theta) = m \times a_c = m \times \frac{v^2}{r}$
• Combining Equations:     * Substitute $n$: $(\frac{mg}{\cos(\theta)}) \times \sin(\theta) = m \times \frac{v^2}{r}$     * Mass ($m$) cancels: $g \times \tan(\theta) = \frac{v^2}{r}$     * Solve for angle: $\theta = \arctan(\frac{v^2}{r \times g})$
• Numerical Calculation:     * $\theta = \arctan(\frac{(13.4\,m/s)^2}{50\,m \times 9.8\,m/s^2})$     * $\theta \approx 20.1^{\circ}$
• Motion Dynamics: If speed is less than $13.4\,m/s$, the car tends to slide down the bank; if speed is greater, it tends to slide up and off the road.

Pilot in a Vertical Loop (Airplane Trick)

• Scenario: An airplane performs a vertical loop of radius $r$ at a constant speed $v$.
• Given Parameters:     * Radius ($r$): $2.7\,km = 2700\,m$     * Speed ($v$): $225\,m/s$
• Goal: Determine the normal force ($n$) exerted by the seat on the pilot at the top and bottom of the loop in terms of pilot weight ($mg$).
• Case 1: Top of the Loop:     * FBD: The plane is upside down. Weight ($mg$) is down. Normal force ($n$) is down. Centripetal acceleration ($a_c$) is down.     * Equation: $n + mg = m \times \frac{v^2}{r}$     * Solve for $n$: $n = m(\frac{v^2}{r} - g)$     * In terms of weight: Factor out $g$ to get $n = mg(\frac{v^2}{rg} - 1)$.     * Calculation: $n = mg(\frac{225^2}{2700 \times 9.8} - 1) \approx 0.913 \times mg$ ($91.3\%$ of pilot's weight).
• Case 2: Bottom of the Loop:     * FBD: The plane is upright. Weight ($mg$) is down. Normal force ($n$) is up. Centripetal acceleration ($a_c$) is up.     * Equation: $n - mg = m \times \frac{v^2}{r}$     * Solve for $n$: $n = m(\frac{v^2}{r} + g)$     * In terms of weight: Factor out $g$ to get $n = mg(\frac{v^2}{rg} + 1)$.     * Calculation: $n = mg(\frac{225^2}{2700 \times 9.8} + 1) \approx 2.91 \times mg$ (nearly $3$ times the pilot's weight).

Universal Gravitation

• Definition: The gravitational force between any two objects in the universe is attractive and depends on their masses and the distance between their centers of gravity.
• Universal Gravitation Formula: $F = G \times \frac{m_1 \times m_2}{r^2}$
• Gravitational Constant ($G$): $G \approx 6.67 \times 10^{-11}\,N\,m^2/kg^2$
• Distance ($r$): Measured from the center of gravity of one object to the center of gravity of the second object.
• Application to Earth's Surface:     * $F = G \times \frac{M_E \times m}{R_E^2}$, where $M_E$ is Earth's mass and $R_E$ is Earth's radius.     * The term $\frac{G \times M_E}{R_E^2}$ simplifies to $g = 9.8\,m/s^2$, allowing for the local weight formula $F = mg$.

Satellite Motion

• Dynamics: A satellite in orbit is kept in a circular path by the gravitational force of the Earth, which acts as the centripetal force.
• Newton's Second Law for Satellites:     * $G \times \frac{M_E \times m_s}{r^2} = m_s \times \frac{v^2}{r}$     * Mass of the satellite ($m_s$) cancels.
• Orbital Speed ($v$):     * $v = \sqrt{\frac{G \times M_E}{r}}$     * Orbit radius $r = R_E + h$, where $h$ is the height above the Earth's surface.
• Substitutions with Period ($T$):     * For a constant speed circular orbit, $v = \frac{2\pi r}{T}$.     * Substitute into the force equation: $G \times \frac{M_E}{r} = v^2 = (\frac{2\pi r}{T})^2$     * Resulting Relation: $G \times \frac{M_E}{r} = \frac{4\pi^2 r^2}{T^2}$
• Kepler's Third Law: The derivation leads to the relationship where the square of the period is proportional to the cube of the radius ($T^2 \propto r^3$). For elliptical orbits, $r$ is replaced by the semi-major axis.