An electric motor has a vee pulley with an effective diameter of 11 cm on its shaft. If the motor, after being switched on, takes 6 seconds to reach maximum speed of 1380 r/min, calculate:

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QUESTION 1: GENERAL

1.1 Define momentum. Momentum is a measure of the mass in motion of an object. It is defined as the product of an object's mass and its velocity.

1.2 Distinguish between angular velocity and angular acceleration.

• Angular velocity ($\omega$) is the rate of change of angular displacement, indicating how fast an object rotates or revolves about an axis. It is measured in radians per second (rad/s).
• Angular acceleration ($\alpha$) is the rate of change of angular velocity, indicating how quickly the angular velocity of an object is changing. It is measured in radians per second squared (rad/s$^2$).

1.3 State Hooke's law. Hooke's Law states that the force ($F$) needed to extend or compress a spring by some distance ($x$) is directly proportional to that distance. Mathematically, $F = kx$, where $k$ is the spring constant.

1.4 Distinguish between elasticity and limit of proportionality.

• Elasticity is the property of a material that allows it to return to its original shape and size after deforming forces are removed.
• The limit of proportionality is the point on a stress-strain curve beyond which stress is no longer directly proportional to strain. Up to this limit, Hooke's Law is obeyed.

QUESTION 2: KINEMATICS

2.1 Calculate the velocity of vehicle A relative to vehicle B. Step 1: Resolve the velocities of vehicle A and vehicle B into their x and y components. Vehicle A: $v_A = 105 \text{ km/h}$ northwest (angle $\theta_A = 135^\circ$ from positive x-axis). $v_{Ax} = 105 \cos(135^\circ) \approx -74.246 \text{ km/h}$ $v_{Ay} = 105 \sin(135^\circ) \approx 74.246 \text{ km/h}$ Vehicle B: $v_B = 115 \text{ km/h}$ directly east (angle $\theta_B = 0^\circ$ from positive x-axis). $v_{Bx} = 115 \cos(0^\circ) = 115 \text{ km/h}$ $v_{By} = 115 \sin(0^\circ) = 0 \text{ km/h}$ Step 2: Calculate the components of the relative velocity $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$. $v_{A/Bx} = v_{Ax} - v_{Bx} = -74.246 - 115 = -189.246 \text{ km/h}$ $v_{A/By} = v_{Ay} - v_{By} = 74.246 - 0 = 74.246 \text{ km/h}$ Step 3: Calculate the magnitude of the relative velocity. $|\vec{v}_{A/B}| = \sqrt{(-189.246)^2 + (74.246)^2} = \sqrt{35893.9 + 5512.4} = \sqrt{41406.3} \approx 203.485 \text{ km/h}$ Step 4: Calculate the direction of the relative velocity. $\theta = \arctan\left(\frac{74.246}{-189.246}\right) \approx -21.42^\circ$ Since the x-component is negative and the y-component is positive, the angle is in the second quadrant. $\theta_{actual} = 180^\circ - 21.42^\circ = 158.58^\circ$ The velocity of vehicle A relative to vehicle B is $\boxed{203.5 km/h at 158.6^\circ}$.

2.2 Calculate the resulting velocity in magnitude and direction. Step 1: Resolve the ship's velocity and current's velocity into their x and y components. Ship velocity: $v_s = 95 \text{ km/h}$ south-east (angle $\theta_s = -45^\circ$ from positive x-axis). $v_{sx} = 95 \cos(-45^\circ) \approx 67.175 \text{ km/h}$ $v_{sy} = 95 \sin(-45^\circ) \approx -67.175 \text{ km/h}$ Current velocity: $v_c = 20 \text{ km/h}$ E 25° N (angle $\theta_c = 25^\circ$ from positive x-axis). $v_{cx} = 20 \cos(25^\circ) \approx 18.126 \text{ km/h}$ $v_{cy} = 20 \sin(25^\circ) \approx 8.452 \text{ km/h}$ Step 2: Calculate the components of the resulting velocity $\vec{v}_R = \vec{v}_s + \vec{v}_c$. $v_{Rx} = v_{sx} + v_{cx} = 67.175 + 18.126 = 85.301 \text{ km/h}$ $v_{Ry} = v_{sy} + v_{cy} = -67.175 + 8.452 = -58.723 \text{ km/h}$ Step 3: Calculate the magnitude of the resulting velocity. $|\vec{v}_R| = \sqrt{(85.301)^2 + (-58.723)^2} = \sqrt{7276.26 + 3448.39} = \sqrt{10724.65} \approx 103.56 \text{ km/h}$ Step 4: Calculate the direction of the resulting velocity. $\theta = \arctan\left(\frac{-58.723}{85.301}\right) \approx -34.54^\circ$ The resulting velocity is $\boxed{103.6 km/h at -34.5^\circ (or E 34.5^\circ S)}$.

2.3 A boy throws a cricket ball at an angle of 20° to the horizontal with an initial velocity of 45 m/s. Step 1: Resolve the initial velocity into horizontal and vertical components. $u_x = 45 \cos(20^\circ) \approx 42.2865 \text{ m/s}$ $u_y = 45 \sin(20^\circ) \approx 15.390 \text{ m/s}$ The acceleration due to gravity is $a_y = -9.81 m/s^2$.

2.3.1 The time it will take to reach maximum height. At maximum height, the vertical velocity $v_y = 0$. $v_y = u_y + a_y t$ $0 = 15.390 m/s + (-9.81 m/s^2) t$ $t = \frac{15.390}{9.81} \approx 1.5688 \text{ s}$ The time to reach maximum height is $\boxed{1.57 s}$.

2.3.2 The maximum height reached by the ball. Using the vertical motion equation: $v_y^2 = u_y^2 + 2 a_y y$ $0^2 = (15.390 m/s)^2 + 2 (-9.81 m/s^2) y$ $0 = 236.8521 - 19.62 y$ $y = \frac{236.8521}{19.62} \approx 12.072 \text{ m}$ The maximum height reached by the ball is $\boxed{12.07 m}$.

2.3.3 The horizontal displacement of the ball. The total time of flight is twice the time to reach maximum height: $T = 2 \times 1.5688 s = 3.1376 \text{ s}$. The horizontal displacement is given by $x = u_x T$ (since there is no horizontal acceleration). $x = 42.2865 m/s \times 3.1376 s \approx 132.61 \text{ m}$ The horizontal displacement of the ball is $\boxed{132.6 m}$.

QUESTION 3: ANGULAR MOTION

3.1 An electric motor has a vee pulley with an effective diameter of 11 cm on its shaft. If the motor, after being switched on, takes 6 seconds to reach maximum speed of 1380 r/min, calculate: Step 1: Convert given values to SI units. Diameter $D = 11 cm = 0.11 \text{ m}$, so radius $R = D/2 = 0.055 \text{ m}$. Time $t = 6 \text{ s}$. Initial angular velocity $\omega_0 = 0 \text{ rad/s}$ (starts from rest). Final angular speed $N = 1380 \text{ r/min}$. Convert to rad/s: $\omega_f = 1380 \frac{rev}{min} \times \frac{2\pi rad}{1 rev} \times \frac{1 min}{60 s} = \frac{1380 \times 2\pi}{60} rad/s = 46\pi rad/s \approx 144.513 \text{ rad/s}$

3.1.1 The angular acceleration of the pulley in rad/s$^2$. Using the angular kinematic equation: $\omega_f = \omega_0 + \alpha t$ $144.513 rad/s = 0 rad/s + \alpha (6 s)$ $\alpha = \frac{144.513 rad/s}{6 s} \approx 24.0855 rad/s^2$ The angular acceleration of the pulley is $\boxed{24.09 rad/s^2}$.

3.1.2 The linear acceleration of the vee belt on the pulley in m/s$^2$. The linear acceleration $a$ is related to angular acceleration $\alpha$ by $a = \alpha R$. $a = 24.0855 rad/s^2 \times 0.055 m \approx 1.3247 m/s^2$ The linear acceleration of the vee belt is $\boxed{1.32 m/s^2}$.

3.1.3 The maximum linear velocity of the vee belt in m/s. The maximum linear velocity $v$ is related to the final angular velocity $\omega_f$ by $v = \omega_f R$. $v = 144.513 rad/s \times 0.055 m \approx 7.9482 \text{ m/s}$ The maximum linear velocity of the vee belt is $\boxed{7.95 m/s}$.