Comprehensive Review of Torque, Uniform Circular Motion, and Orbital Mechanics

Advanced Torque Analysis: Angled Beams and Static Equilibrium

In scenarios involving a beam at an angle, the standard torque equations must be adapted to account for the geometry of the system. In the specific example provided, a beam is held in equilibrium while tilted.

System Parameters

• Beam Length ($L$): $2.0\,m$.

• Beam Mass ($m_{beam}$): $100\,kg$.

• Tension Wire Force ($F$): $2000\,N$.

• Overhanging Mass ($M$): Unknown variable to be solved.

• Tension Angle: The wire makes an angle of $78^\circ$ with the beam at the point of attachment.

• Beam Incline Geometry: The beam is inclined such that the internal angle related to the weights is $60^\circ$. This is derived using supplementary angles; if an external angle is given as $120^\circ$, the required interior angle for calculation is $180 - 120 = 60^\circ$.

Individual Torque Calculations ($\tau$)

Every torque is calculated using the formula $\tau = F \times d \times \sin(\theta)$, where $d$ is the distance from the pivot (tail) and $\theta$ is the angle between the force and the beam.

• Torque due to Tension ($\tau_F$):   $\tau_F = F \times \sin(78^\circ) \times 2$   $\tau_F = 2000 \times \sin(78^\circ) \times 2$

• Torque due to Overhanging Mass ($\tau_M$):   $\tau_M = Mg \times \sin(60^\circ) \times 2$   Where $g = 9.81\,m/s^2$.

• Torque due to Weight of the Beam ($\tau_g$):   The force of gravity acts at the center of mass, which is at the midpoint of the beam ($L/2 = 1.0\,m$).   $\tau_g = m_{beam}g \times \sin(60^\circ) \times 1$   $\tau_g = 100 \times 9.81 \times \sin(60^\circ) \times 1$   $\tau_g = 981 \times \sin(60^\circ)$

Balancing the Torques

To maintain equilibrium, the sum of clockwise torques must equal the sum of counter-clockwise torques. Assuming the pivot is fixed at the base of the beam: $\tau_{balanced} = \tau_M + \tau_g$ $\tau_F = (Mg \times \sin(60^\circ) \times 2) + (981 \times \sin(60^\circ)$

By substituting known values, the only unknown remaining is the mass ($M$), which can then be isolated and solved algebraically.

Kinematics and Dynamics of Uniform Circular Motion (UCM)

Uniform Circular Motion occurs when an object travels in a circular path at a constant speed.

Core Vectors and Directions

• Centripetal Force ($F_c$): Always directed toward the center of the circular path.

• Centripetal Acceleration ($a_c$): Always directed toward the center, in the same direction as the centripetal force.

• Velocity ($v$): Always tangential to the circle and perpendicular ($90^\circ$) to the centripetal force/acceleration.

Vector Magnitudes in UCM

• Under uniform conditions, the speed ($v$) is constant.

• While the magnitude of velocity (speed) is constant, the velocity vector is not constant because its direction changes at every point.

• The magnitude of acceleration ($a_c$) is constant, but the acceleration vector is not constant because it rotates to always point toward the center.

• The magnitude of force ($F_c$) is constant.

Mathematical Relationships

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

• Centripetal Force: $F_c = m a_c = \frac{mv^2}{r}$

• Alternative Acceleration Formula: $a_c = \frac{4\pi^2 r}{T^2}$, where $T$ is the period of revolution.

Free Body Diagrams in Vertical Circular Motion

Analyzing objects in vertical loops (like a car in a loop-the-loop) requires identifying how gravity ($F_g$) and normal force ($F_n$) contribute to the net centripetal force ($F_c$).

Scenario 1: Inside a Loop (Object at the Top)

• Force Directions: Gravity ($F_g$) points down toward the center. The normal force ($F_{n1}$) also points down toward the center.

• Net Centripetal Force Equation: $F_c = F_g + F_{n1}$

• Newton's Second Law Application: $m a_c = mg + F_{n1}$

Scenario 2: Inside a Loop (Object at the Bottom)

• Force Directions: Gravity ($F_g$) points down (away from the center). The normal force ($F_{n2}$) points up (toward the center).

• Net Centripetal Force Equation: $F_c = F_{n2} - F_g$

• Newton's Second Law Application: $m a_c = F_{n2} - mg$

• Note: At the bottom, the normal force must be greater than the force of gravity to provide the upward centripetal acceleration (F_{n2} > F_g).

Scenario 3: Outside a Circle (Object at the Top of a Hill)

• Force Directions: Gravity ($F_g$) points down toward the center. The normal force ($F_n$) points up away from the center.

• Net Centripetal Force Equation: $F_c = F_g - F_n$

• Newton's Second Law Application: $m a_c = mg - F_n$

Orbital Mechanics and Satellite Motion

Satellites in orbit are a specific case of circular motion where the centripetal force is provided entirely by gravity.

Derivation of Orbital Velocity

The force of gravity ($F_g$) between a planet and a satellite is given by Newton's Law of Universal Gravitation: $F_g = \frac{GMm}{r^2}$

Since this gravity acts as the centripetal force: $F_c = F_g$ $\frac{mv^2}{r} = \frac{GMm}{r^2}$

Simplifying the Equation

1. The mass of the satellite ($m$) is on both sides and cancels out. The orbital velocity is independent of the mass of the satellite.

2. One factor of the radius ($r$) cancels on both sides. $\frac{v^2}{1} = \frac{GM}{r}$ $v = \sqrt{\frac{GM}{r}}$

Key Variable Definitions

• $M$: The mass of the central body (e.g., the Earth).

• $G$: The Universal Gravitational Constant ($6.67 \times 10^{-11}\,N\,m^2/kg^2$).

• $r$: The center-to-center distance, which is the sum of the planet's radius ($R_{planet}$) and the altitude ($h$) above the surface.

Work and Energy in Gravitational Fields

To move an object from the surface of a planet to a point in space, work must be done against the gravitational field.

Sample Problem Setup

• Object: A rocket/satellite with a mass ($m$) of $10,000\,kg$.

• Target Altitude: $1,000\,km$ above the surface ($1,000,000\,m$).

• Distance Note: Crucial distinction must be made between "above the surface" and "from the center." When calculating potential energy or work, the total distance ($r$) must include the Earth's radius.

• Objective: Calculate the work required to lift the satellite to Point P. This involves calculating the difference in gravitational potential energy between the surface and the final altitude (to be finalized in subsequent lectures).

Questions & Discussion

Q: How did you get the 60-degree angle for the torque problem?A: Looking at the diagram, the angle provided ($120^\circ$) and the angle we need for the sine component are supplementary. Since they must add up to $180^\circ$, subtracting the given $120^\circ$ gives us the $60^\circ$ angle needed for the torque calculation.

Q: Is the acceleration constant in uniform circular motion?A: In magnitude, yes. Because the speed ($v$) and radius ($r$) are constant, $a_c = v^2/r$ remains a constant number. However, as a vector, it is not constant because it is constantly changing direction to stay pointed toward the center of the circle.

Q: Do you need the mass of the satellite to find its orbital speed?A: No. As shown in the derivation $\frac{mv^2}{r} = \frac{GMm}{r^2}$, the small mass ($m$) representing the satellite cancels out. The orbital speed depends only on the mass of the central planet and the orbital radius.