Calculate the torque required on the drum to lift the load. If the drum rotates at 24 r/min, calculate the power developed.

Here are the solutions to the questions:

QUESTION 3: ROTATIONAL MOTION

3.1.1 Calculate the angular acceleration of the flywheel.

Step 1: Convert initial and final angular speeds from revolutions per minute (r/min) to radians per second (rad/s). $\omega_i = 200 r/min \times \frac{2\pi rad}{1 rev} \times \frac{1 min}{60 s} = \frac{400\pi}{60} rad/s = \frac{20\pi}{3} rad/s$ $\omega_f = 720 r/min \times \frac{2\pi rad}{1 rev} \times \frac{1 min}{60 s} = \frac{1440\pi}{60} rad/s = 24\pi rad/s$

Step 2: Calculate the angular acceleration $\alpha$ using the formula $\alpha = \frac{\omega_f - \omega_i}{t}$. $\alpha = \frac{24\pi rad/s - \frac{20\pi}{3} rad/s}{8 s}$ $\alpha = \frac{\frac{72\pi - 20\pi}{3}}{8} rad/s^2 = \frac{52\pi}{24} rad/s^2 = \frac{13\pi}{6} rad/s^2$ $\alpha \approx 6.81 rad/s^2$ The angular acceleration is $\boxed{\frac{13\pi}{6} rad/s^2 or 6.81 rad/s^2}$.

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

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

Step 2: Convert the angular displacement from radians to revolutions. $Number of revolutions = \frac{\Delta\theta}{2\pi rad/rev} = \frac{\frac{368\pi}{3} rad}{2\pi rad/rev} = \frac{368}{6} rev = \frac{184}{3} rev$ $Number of revolutions \approx 61.33 rev$ The number of revolutions made is $\boxed{\frac{184}{3} revolutions or 61.33 revolutions}$.

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

Step 1: Calculate the radius of the drum and the force required to lift the load. Diameter $D = 380 mm = 0.380 \text{ m}$ Radius $r = \frac{D}{2} = \frac{0.380 m}{2} = 0.190 \text{ m}$ Mass $m = 1250 \text{ kg}$ Force (weight) $F = mg = 1250 kg \times 9.8 m/s^2 = 12250 \text{ N}$

Step 2: Calculate the torque $\tau$ using the formula $\tau = Fr$. $\tau = 12250 N \times 0.190 m = 2327.5 N\cdotm$ The torque required is $\boxed{2327.5 N\cdotm}$.

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

Step 1: Convert the angular speed from r/min to rad/s. $\omega = 24 r/min \times \frac{2\pi rad}{1 rev} \times \frac{1 min}{60 s} = \frac{48\pi}{60} rad/s = \frac{4\pi}{5} rad/s$

Step 2: Calculate the power developed $P$ using the formula $P = \tau \omega$. $P = 2327.5 N\cdotm \times \frac{4\pi}{5} rad/s$ $P = 1862\pi W \approx 5850.7 W$ The power developed is $\boxed{1862\pi W or 5850.7 W}$.

QUESTION 4: DYNAMICS

4.1.1 Calculate the kinetic energy of the bakkie.

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

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

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

Step 1: Use the work-energy theorem, $W = \Delta KE$. The initial kinetic energy $KE_i = 562500 \text{ J}$ (from 4.1.1). The final kinetic energy $KE_f = 0 \text{ J}$ (since the bakkie comes to rest). $W = KE_f - KE_i = 0 - 562500 J = -562500 J$

Step 2: Calculate the average braking force $F_b$ using $W = F_b d \cos\theta$. The braking force acts opposite to the direction of motion, so $\theta = 180^\circ$ and $\cos(180^\circ) = -1$. $-562500 J = F_b \times 40 m \times (-1)$ $F_b = \frac{562500 J}{40 m} = 14062.5 N$ The average braking force is $\boxed{14062.5 N}$.

4.2.1 Calculate the tension in the rope.

Step 1: Identify the forces acting on the crate and apply Newton's second law. Mass $m = 50 \text{ kg}$ Acceleration $a = 1.2 m/s^2$ (upwards) Gravitational acceleration $g = 9.8 m/s^2$ The forces are tension $T$ (upwards) and weight $mg$ (downwards). $\sum F_y = ma_y \implies T - mg = ma$ $T = m(g+a)$

Step 2: Substitute the values and calculate the tension. $T = 50 kg \times (9.8 m/s^2 + 1.2 m/s^2)$ $T = 50 kg \times 11.0 m/s^2 = 550 N$ The tension in the rope is $\boxed{550 N}$.

4.2.2 Calculate the work done in lifting the crate.

Step 1: Use the formula for work done by a constant force, $W = Fd\cos\theta$. The force doing the work is the tension $T = 550 \text{ N}$ (from 4.2.1). The distance $d = 30 \text{ m}$. The force and displacement are in the same direction, so $\theta = 0^\circ$ and $\cos(0^\circ) = 1$. $W = Td$

Step 2: Substitute the values and calculate the work done. $W = 550 N \times 30 m = 16500 J$ The work done in lifting the crate is $\boxed{16500 J}$.

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

Step 1: Calculate the normal force $N$ acting on the box. Mass $m = 25 \text{ kg}$ Angle of inclination $\theta = 25^\circ$ Normal force $N = mg\cos\theta$ $N = 25 kg \times 9.8 m/s^2 \times \cos(25^\circ)$ $N = 245 N \times 0.9063 \approx 222.04 N$

Step 2: Calculate the maximum static frictional force $f_{s,max}$ using the formula $f_{s,max} = \mu_s N$. Coefficient of static friction $\mu_s = 0.35$ $f_{s,max} = 0.35 \times 222.04 N \approx 77.71 N$ The maximum static frictional force is $\boxed{77.71 N}$.

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

Step 1: Calculate the component of the gravitational force parallel to the incline, which tends to pull the box down. $F_{parallel} = mg\sin\theta$ $F_{parallel} = 25 kg \times 9.8 m/s^2 \times \sin(25^\circ)$ $F_{parallel} = 245 N \times 0.4226 \approx 103.54 N$

Step 2: Compare the parallel component of gravity with the maximum static frictional force. From 4.3.1, the maximum static frictional force $f_{s,max} = 77.71 \text{ N}$. The component of gravity pulling the box down the incline is $F_{parallel} = 103.54 \text{ N}$. Since $F_{parallel} (103.54 N) > f_{s,max} (77.71 N)$, the force pulling the box down the incline is greater than the maximum static friction that can oppose it. This means the box will overcome static friction and slide down the plane, confirming that it is indeed in a state where it will slide.

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