QUESTION 3: ROTATIONAL MOTION A flywheel rotates at 200 r/min and accelerates uniformly to 720 r/min in 8 s.

QUESTION 3: ROTATIONAL MOTION

3.1.1 Calculate the angular acceleration of the flywheel.

Step 1: Convert the initial and final angular velocities from revolutions per minute (r/min) to radians per second (rad/s). The conversion factor is $\frac{2\pi rad}{1 rev} \times \frac{1 min}{60 s}$. Initial angular velocity: $\omega_i = 200 r/min \times \frac{2\pi rad}{1 rev} \times \frac{1 min}{60 s} = \frac{400\pi}{60} rad/s = \frac{20\pi}{3} \text{ rad/s}$ Final angular velocity: $\omega_f = 720 r/min \times \frac{2\pi rad}{1 rev} \times \frac{1 min}{60 s} = \frac{1440\pi}{60} rad/s = 24\pi \text{ rad/s}$

Step 2: Calculate the angular acceleration ($\alpha$) using the formula $\alpha = \frac{\omega_f - \omega_i}{t}$. Given time $t = 8 \text{ s}$. $\alpha = \frac{24\pi rad/s - \frac{20\pi}{3} rad/s}{8 s}$ $\alpha = \frac{\frac{72\pi - 20\pi}{3}}{8} rad/s^2$ $\alpha = \frac{52\pi}{24} rad/s^2$ $\alpha = \frac{13\pi}{6} rad/s^2$ $\alpha \approx 6.80678 rad/s^2$ Rounding to three significant figures: $\boxed{6.81 rad/s^2}$

3.1.2 Determine the number of revolutions made during the 8 s.

Step 1: Calculate the angular displacement ($\Delta \theta$) using the formula $\Delta \theta = \frac{(\omega_i + \omega_f)}{2}t$. $\Delta \theta = \frac{\left(\frac{20\pi}{3} rad/s + 24\pi rad/s\right)}{2} \times 8 \text{ s}$ $\Delta \theta = \frac{\left(\frac{20\pi + 72\pi}{3}\right)}{2} \times 8 \text{ rad}$ $\Delta \theta = \frac{92\pi}{6} \times 8 \text{ rad}$ $\Delta \theta = \frac{46\pi}{3} \times 8 \text{ rad}$ $\Delta \theta = \frac{368\pi}{3} \text{ rad}$

Step 2: Convert the angular displacement from radians to revolutions. $Revolutions = \frac{\Delta \theta}{2\pi}$ $Revolutions = \frac{\frac{368\pi}{3} rad}{2\pi rad/rev}$ $Revolutions = \frac{368}{6} = \frac{184}{3} rev$ $Revolutions \approx 61.333 rev$ Rounding to three significant figures: $\boxed{61.3 rev}$

3.2.1 Calculate the torque required on the drum to lift the load.

Step 1: Calculate the force required to lift the load, which is equal to its weight. Given mass $m = 1250 \text{ kg}$ and assuming $g = 9.81 m/s^2$. $F = mg = 1250 kg \times 9.81 m/s^2 = 12262.5 \text{ N}$

Step 2: Calculate the radius of the drum. Given diameter $D = 380 mm = 0.38 \text{ m}$. $r = \frac{D}{2} = \frac{0.38 m}{2} = 0.19 \text{ m}$

Step 3: Calculate the torque ($\tau$) using the formula $\tau = F \times r$. $\tau = 12262.5 N \times 0.19 m = 2330.075 \text{ Nm}$ Rounding to three significant figures: $\boxed{2330 Nm}$

3.2.2 If the drum rotates at 24 r/min, calculate the power developed.

Step 1: Convert the angular speed from revolutions per minute (r/min) to radians per second (rad/s). Given angular speed $N = 24 \text{ r/min}$. $\omega = 24 r/min \times \frac{2\pi rad}{1 rev} \times \frac{1 min}{60 s} = \frac{48\pi}{60} rad/s = \frac{4\pi}{5} \text{ rad/s}$ $\omega \approx 2.51327 \text{ rad/s}$

Step 2: Calculate the power ($P$) developed using the formula $P = \tau \omega$. Using the torque calculated in 3.2.1, $\tau = 2330.075 \text{ Nm}$. $P = 2330.075 Nm \times \frac{4\pi}{5} \text{ rad/s}$ $P \approx 5857.8 \text{ W}$ Rounding to three significant figures: $\boxed{5860 W}$

QUESTION 4: DYNAMICS

4.1.1 Calculate the kinetic energy of the bakkie.

Step 1: Convert the velocity from kilometers per hour (km/h) to meters per second (m/s). Given velocity $v = 90 \text{ km/h}$. $v = 90 km/h \times \frac{1000 m}{1 km} \times \frac{1 h}{3600 s} = 25 \text{ m/s}$

Step 2: Calculate the kinetic energy ($KE$) using the formula $KE = \frac{1}{2}mv^2$. Given mass $m = 1800 \text{ kg}$. $KE = \frac{1}{2} \times 1800 kg \times (25 m/s)^2$ $KE = 900 kg \times 625 m^2/s^2$ $KE = 562500 \text{ J}$ $\boxed{562500 J}$

4.1.2 If the bakkie is brought to rest in 40 m, calculate the average braking force.

Step 1: Apply the work-energy theorem, which states that the work done by the braking force is equal to the change in kinetic energy. Since the bakkie comes to rest, the final kinetic energy is zero. Work done by braking force $W_b = \Delta KE = KE_f - KE_i = 0 - KE_i = -KE_i$. The magnitude of the braking force $F_b$ multiplied by the distance $s$ is equal to the initial kinetic energy. $F_b \times s = KE_i$ Given distance $s = 40 \text{ m}$ and $KE_i = 562500 \text{ J}$ from 4.1.1. $F_b = \frac{KE_i}{s}$ $F_b = \frac{562500 J}{40 m}$ $F_b = 14062.5 \text{ N}$ Rounding to three significant figures: $\boxed{14100 N}$

4.2.1 Calculate the tension in the rope.

Step 1: Apply Newton's second law of motion. The forces acting on the crate are the tension ($T$) upwards and its weight ($mg$) downwards. The net force causes an upward acceleration ($a$). Given mass $m = 50 \text{ kg}$, acceleration $a = 1.2 m/s^2$, and assuming $g = 9.81 m/s^2$. $\sum F = ma$ $T - mg = ma$ $T = mg + ma = m(g+a)$ $T = 50 kg \times (9.81 m/s^2 + 1.2 m/s^2)$ $T = 50 kg \times 11.01 m/s^2$ $T = 550.5 \text{ N}$ Rounding to three significant figures: $\boxed{551 N}$

4.2.2 Calculate the work done in lifting the crate.

Step 1: Calculate the work done ($W$) by the tension force using the formula $W = F \times s \times \cos\theta$. Since the tension force and displacement are in the same direction ($\theta = 0^\circ$, so $\cos\theta = 1$), the formula simplifies to $W = T \times s$. Given distance $s = 30 \text{ m}$ and tension $T = 550.5 \text{ N}$ from 4.2.1. $W = 550.5 N \times 30 \text{ m}$ $W = 16515 \text{ J}$ Rounding to three significant figures: $\boxed{16500 J}$

4.3.1 Calculate the maximum static frictional force acting on the box.

Step 1: Calculate the normal force ($N$) acting on the box. The normal force is perpendicular to the inclined plane. The component of gravity perpendicular to the plane is $mg \cos \theta$. Given mass $m = 25 \text{ kg}$, angle $\theta = 25^\circ$, and assuming $g = 9.81 m/s^2$. $N = mg \cos \theta = 25 kg \times 9.81 m/s^2 \times \cos(25^\circ)$ $N = 245.25 N \times 0.9063077$ $N \approx 222.29 \text{ N}$

Step 2: Calculate the maximum static frictional force ($f_{s,max}$) using the formula $f_{s,max} = \mu_s N$. Given coefficient of static friction $\mu_s = 0.35$. $f_{s,max} = 0.35 \times 222.29 \text{ N}$ $f_{s,max} \approx 77.80 \text{ N}$ Rounding to three significant figures: $\boxed{77.8 N}$

4.3.2 Show that the box is indeed about to slide down the plane.

Step 1: Calculate the component of the gravitational force acting down the inclined plane. $F_{down} = mg \sin \theta$ $F_{down} = 25 kg \times 9.81 m/s^2 \times \sin(25^\circ)$ $F_{down} = 245.25 N \times 0.422618$ $F_{down} \approx 103.6 \text{ N}$

Step 2: Compare the force component down the plane ($F_{down}$) with the maximum static frictional force ($f_{s,max}$) calculated in 4.3.1. We found $F_{down} \approx 103.6 \text{ N}$ and $f_{s,max} \approx 77.8 \text{ N}$. Since $F_{down} (103.6 N) > f_{s,max} (77.8 N)$, the force pulling the box down the plane is greater than the maximum static friction that can oppose it. Therefore, the box will indeed slide down the plane. The condition for sliding is met.

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