A light aircraft is 60 km north east from OR Tambo International Airport and it flies north for six hours at a velocity of 96 km/h. Determine its position (displacement) with reference to OR Tambo International Airport in magnitude and direction.

QUESTION 1: GENERAL

1.1 Resultant force

The resultant force is the single force that represents the combined effect of all individual forces acting on an object. It is the vector sum of all forces.

1.2 Elastic limit

The elastic limit is the maximum stress a material can withstand without undergoing permanent deformation. If the stress exceeds this limit, the material will not return to its original shape after the load is removed.

1.3 Strain

Strain is a measure of the deformation of a material, defined as the ratio of the change in dimension to the original dimension. It is a dimensionless quantity.

1.4 Hydrostatic pressure

Hydrostatic pressure is the pressure exerted by a fluid at rest due to gravity. It increases with depth and is calculated as $P = \rho gh$, where $\rho$ is the fluid density, $g$ is the acceleration due to gravity, and $h$ is the depth.

1.5 Second moment of area

The second moment of area (or area moment of inertia) is a geometric property of a cross-section that quantifies its resistance to bending. It is calculated by integrating the square of the distance from an axis to each infinitesimal area element over the entire cross-section.

QUESTION 2: KINEMATICS

2.1

• Given: Initial position $P_0 = 60 \text{ km}$ North East from OR Tambo. Velocity $v = 96 \text{ km/h}$ North for $t = 6 \text{ hours}$.

Step 1: Determine the initial displacement vector from OR Tambo. North East implies an angle of $45^\circ$ from the East (or North). $\vec{d_0} = (60 \cos 45^\circ, 60 \sin 45^\circ) km = (60 \times 0.7071, 60 \times 0.7071) km \approx (42.426, 42.426) km$ Step 2: Determine the displacement vector during the flight. The aircraft flies North, so the displacement is purely in the North direction. $\vec{d_f} = (0, 96 km/h \times 6 h) = (0, 576) km$ Step 3: Calculate the total displacement vector from OR Tambo. $\vec{D} = \vec{d_0} + \vec{d_f} = (42.426 + 0, 42.426 + 576) km = (42.426, 618.426) km$ Step 4: Calculate the magnitude of the total displacement. $|\vec{D}| = \sqrt{(42.426)^2 + (618.426)^2} = \sqrt{1799.96 + 382450.5} = \sqrt{384250.46} \approx 619.88 km$ Step 5: Calculate the direction of the total displacement. The angle $\phi$ from the positive x-axis (East) is: $\phi = \arctan\left(\frac{618.426}{42.426}\right) \approx \arctan(14.576) \approx 86.06^\circ$ The direction can be expressed as $86.06^\circ$ North of East, or as a bearing from North: $90^\circ - 86.06^\circ = 3.94^\circ$ East of North.

The position (displacement) is $\boxed{619.88 km at 86.06^\circ North of East}$.

2.2

• Given: Hoist K descends at $v_K = 6.48 \text{ km/h}$. Hoist L ascends at $v_L = 5.04 \text{ km/h}$.
• Assume upward direction is positive. $v_K = -6.48 km/h$ $v_L = +5.04 km/h$

2.2.1 The velocity of hoist K relative to the velocity of hoist L in magnitude and direction.

Step 1: Calculate the relative velocity $v_{K/L} = v_K - v_L$. $v_{K/L} = (-6.48 km/h) - (5.04 km/h) = -11.52 km/h$

The velocity of hoist K relative to hoist L is $\boxed{11.52 km/h downwards}$.

2.2.2 The velocity of hoist L relative to the velocity of hoist K in magnitude and direction.

Step 1: Calculate the relative velocity $v_{L/K} = v_L - v_K$. $v_{L/K} = (5.04 km/h) - (-6.48 km/h) = 5.04 km/h + 6.48 km/h = 11.52 km/h$

The velocity of hoist L relative to hoist K is $\boxed{11.52 km/h upwards}$.

2.3

• Given: Initial velocity $v_0 = 42 \text{ m/s}$. Angle $\theta = 28^\circ$ to the horizontal. Assume $g = 9.81 m/s^2$.

Step 1: Resolve the initial velocity into horizontal and vertical components. $v_{0x} = v_0 \cos\theta = 42 m/s \times \cos(28^\circ) = 42 \times 0.8829 \approx 37.082 m/s$ $v_{0y} = v_0 \sin\theta = 42 m/s \times \sin(28^\circ) = 42 \times 0.4695 \approx 19.719 m/s$

2.3.1 The maximum height that the stone reaches.

Step 2: Use the kinematic equation $v_y^2 = v_{0y}^2 + 2a_y \Delta y$. At maximum height, $v_y = 0$. $0^2 = (19.719 m/s)^2 + 2(-9.81 m/s^2) H_{max}$ $0 = 388.839 - 19.62 H_{max}$ $H_{max} = \frac{388.839}{19.62} \approx 19.818 m$

The maximum height reached by the stone is $\boxed{19.82 m}$.

2.3.2 The horizontal displacement of the stone.

Step 1: Calculate the time to reach maximum height ($t_{up}$). Using $v_y = v_{0y} + a_y t_{up}$: $0 = 19.719 m/s + (-9.81 m/s^2) t_{up}$ $t_{up} = \frac{19.719}{9.81} \approx 2.010 s$ Step 2: Calculate the total time of flight ($T$). For projectile motion returning to the same height, $T = 2 t_{up}$. $T = 2 \times 2.010 s = 4.020 s$ Step 3: Calculate the horizontal displacement (range). $R = v_{0x} T = 37.082 m/s \times 4.020 s \approx 149.17 m$

The horizontal displacement of the stone is $\boxed{149.17 m}$.

QUESTION 3: ROTATIONAL MOTION

3.1 Define angular velocity.

Angular velocity is the rate at which an object rotates or revolves relative to another point, i.e., the rate of change of angular displacement. It is a vector quantity, typically measured in radians per second (rad/s).

3.2

• Given: Diameter $D = 1.76 \text{ m}$, so radius $r = D/2 = 0.88 \text{ m}$. Linear velocity $v = 396 \text{ km/h}$.

Step 1: Convert linear velocity from km/h to m/s. $v = 396 \frac{km}{h} \times \frac{1000 m}{1 km} \times \frac{1 h}{3600 s} = 110 m/s$

3.2.1 The rotational frequency of the wheel in revolutions per minute.

Step 2: Calculate the angular velocity $\omega$ in rad/s using $v = r\omega$. $\omega = \frac{v}{r} = \frac{110 m/s}{0.88 m} = 125 rad/s$ Step 3: Convert angular velocity from rad/s to r/min (revolutions per minute). $Frequency (r/min) = 125 \frac{rad}{s} \times \frac{1 rev}{2\pi rad} \times \frac{60 s}{1 min}$ $Frequency (r/min) = \frac{125 \times 60}{2\pi} = \frac{7500}{2\pi} \approx 1193.66 r/min$

The rotational frequency of the wheel is $\boxed{1193.66 r/min}$.

3.2.2 The angular velocity of the wheel in rad/s.

The angular velocity was calculated in Step 2 of 3.2.1. $\omega = 125 rad/s$

The angular velocity of the wheel is $\boxed{125 rad/s}$.

3.2.3 The number of revolutions made by the wheel during 36 minutes.

Step 1: Convert time from minutes to seconds. $t = 36 min \times 60 s/min = 2160 s$ Step 2: Calculate the total angular displacement $\theta$ in radians. $\theta = \omega t = 125 rad/s \times 2160 s = 270000 rad$ Step 3: Convert angular displacement from radians to revolutions. $Revolutions = \frac{\theta}{2\pi} = \frac{270000 rad}{2\pi rad/rev} \approx 42971.83 revolutions$

The number of revolutions made by the wheel is $\boxed{42971.83 revolutions}$.