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

Physics

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QUESTION 3: ROTATIONAL MOTION

3.1.1 Calculate the angular acceleration of the flywheel.

  1. Convert initial and final angular speeds from revolutions per minute (r/min) to radians per second (rad/s).

    ωi=200 r/min×2π rad1 rev×1 min60 s=400π60 rad/s=20π3 rad/s\omega_i = 200 \text{ r/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{400\pi}{60} \text{ rad/s} = \frac{20\pi}{3} \text{ rad/s} ωf=720 r/min×2π rad1 rev×1 min60 s=1440π60 rad/s=24π rad/s\omega_f = 720 \text{ r/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{1440\pi}{60} \text{ rad/s} = 24\pi \text{ rad/s}

  2. Calculate the angular acceleration α\alpha using the formula α=ωf−ωit\alpha = \frac{\omega_f - \omega_i}{t}.

    α=24π rad/s−20π3 rad/s8 s\alpha = \frac{24\pi \text{ rad/s} - \frac{20\pi}{3} \text{ rad/s}}{8 \text{ s}} α=72π−20π38 rad/s2=52π24 rad/s2=13π6 rad/s2\alpha = \frac{\frac{72\pi - 20\pi}{3}}{8} \text{ rad/s}^2 = \frac{52\pi}{24} \text{ rad/s}^2 = \frac{13\pi}{6} \text{ rad/s}^2 α≈6.81 rad/s2\alpha \approx 6.81 \text{ rad/s}^2

    The angular acceleration is 13π6 rad/s2 or 6.81 rad/s2\boxed{\frac{13\pi}{6} \text{ rad/s}^2 \text{ or } 6.81 \text{ rad/s}^2}.

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

  3. Calculate the angular displacement Δθ\Delta\theta using the formula Δθ=(ωi+ωf)2t\Delta\theta = \frac{(\omega_i + \omega_f)}{2} t.

    Δθ=(20π3 rad/s+24π rad/s)2×8 s\Delta\theta = \frac{\left(\frac{20\pi}{3} \text{ rad/s} + 24\pi \text{ rad/s}\right)}{2} \times 8 \text{ s} Δθ=(20π+72π3)2×8 rad=92π6×8 rad=46π3×8 rad=368π3 rad\Delta\theta = \frac{\left(\frac{20\pi + 72\pi}{3}\right)}{2} \times 8 \text{ rad} = \frac{92\pi}{6} \times 8 \text{ rad} = \frac{46\pi}{3} \times 8 \text{ rad} = \frac{368\pi}{3} \text{ rad}

  4. Convert the angular displacement from radians to revolutions.

    Number of revolutions=Δθ2π rad/rev=368π3 rad2π rad/rev=3686 rev=1843 rev\text{Number of revolutions} = \frac{\Delta\theta}{2\pi \text{ rad/rev}} = \frac{\frac{368\pi}{3} \text{ rad}}{2\pi \text{ rad/rev}} = \frac{368}{6} \text{ rev} = \frac{184}{3} \text{ rev} Number of revolutions≈61.33 rev\text{Number of revolutions} \approx 61.33 \text{ rev}

    The number of revolutions made is 1843 revolutions or 61.33 revolutions\boxed{\frac{184}{3} \text{ revolutions or } 61.33 \text{ revolutions}}.

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

  5. Calculate the radius of the drum and the force required to lift the load.

    Diameter D=380 mm=0.380 mD = 380 \text{ mm} = 0.380 \text{ m}
    Radius r=D2=0.380 m2=0.190 mr = \frac{D}{2} = \frac{0.380 \text{ m}}{2} = 0.190 \text{ m}
    Mass m=1250 kgm = 1250 \text{ kg}
    Force (weight) F=mg=1250 kg×9.8 m/s2=12250 NF = mg = 1250 \text{ kg} \times 9.8 \text{ m/s}^2 = 12250 \text{ N}

  6. Calculate the torque τ\tau using the formula τ=Fr\tau = Fr.

    τ=12250 N×0.190 m=2327.5 N⋅m\tau = 12250 \text{ N} \times 0.190 \text{ m} = 2327.5 \text{ N}\cdot\text{m}

    The torque required is 2327.5 N⋅m\boxed{2327.5 \text{ N}\cdot\text{m}}.

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

  7. Convert the angular speed from r/min to rad/s.

    ω=24 r/min×2π rad1 rev×1 min60 s=48π60 rad/s=4π5 rad/s\omega = 24 \text{ r/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{48\pi}{60} \text{ rad/s} = \frac{4\pi}{5} \text{ rad/s}

  8. Calculate the power developed PP using the formula P=τωP = \tau \omega.

    P=2327.5 N⋅m×4π5 rad/sP = 2327.5 \text{ N}\cdot\text{m} \times \frac{4\pi}{5} \text{ rad/s} P=1862π W≈5850.7 WP = 1862\pi \text{ W} \approx 5850.7 \text{ W}

    The power developed is 1862π W or 5850.7 W\boxed{1862\pi \text{ W or } 5850.7 \text{ W}}.

    QUESTION 4: DYNAMICS

    4.1.1 Calculate the kinetic energy of the bakkie.

  9. Convert the speed from kilometers per hour (km/h) to meters per second (m/s).

    v=90 km/h×1000 m1 km×1 h3600 s=25 m/sv = 90 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 25 \text{ m/s}

  10. Calculate the kinetic energy KEKE using the formula KE=12mv2KE = \frac{1}{2}mv^2.

    KE=12×1800 kg×(25 m/s)2KE = \frac{1}{2} \times 1800 \text{ kg} \times (25 \text{ m/s})^2 KE=900 kg×625 m2/s2=562500 JKE = 900 \text{ kg} \times 625 \text{ m}^2\text{/s}^2 = 562500 \text{ J}

    The kinetic energy is 562500 J\boxed{562500 \text{ J}}.

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

  11. Use the work-energy theorem, W=ΔKEW = \Delta KE.

    The initial kinetic energy KEi=562500 JKE_i = 562500 \text{ J} (from 4.1.1).
    The final kinetic energy KEf=0 JKE_f = 0 \text{ J} (since the bakkie comes to rest).

    W=KEf−KEi=0−562500 J=−562500 JW = KE_f - KE_i = 0 - 562500 \text{ J} = -562500 \text{ J}

  12. Calculate the average braking force FbF_b using W=Fbdcos⁡θW = F_b d \cos\theta.

    The braking force acts opposite to the direction of motion, so θ=180∘\theta = 180^\circ and cos⁡(180∘)=−1\cos(180^\circ) = -1.

    −562500 J=Fb×40 m×(−1)-562500 \text{ J} = F_b \times 40 \text{ m} \times (-1) Fb=562500 J40 m=14062.5 NF_b = \frac{562500 \text{ J}}{40 \text{ m}} = 14062.5 \text{ N}

    The average braking force is 14062.5 N\boxed{14062.5 \text{ N}}.

    4.2.1 Calculate the tension in the rope.

  13. Identify the forces acting on the crate and apply Newton's second law.

    Mass m=50 kgm = 50 \text{ kg}
    Acceleration a=1.2 m/s2a = 1.2 \text{ m/s}^2 (upwards)
    Gravitational acceleration g=9.8 m/s2g = 9.8 \text{ m/s}^2
    The forces are tension TT (upwards) and weight mgmg (downwards).

    ∑Fy=may  ⟹  T−mg=ma\sum F_y = ma_y \implies T - mg = ma T=m(g+a)T = m(g+a)

  14. Substitute the values and calculate the tension.

    T=50 kg×(9.8 m/s2+1.2 m/s2)T = 50 \text{ kg} \times (9.8 \text{ m/s}^2 + 1.2 \text{ m/s}^2) T=50 kg×11.0 m/s2=550 NT = 50 \text{ kg} \times 11.0 \text{ m/s}^2 = 550 \text{ N}

    The tension in the rope is 550 N\boxed{550 \text{ N}}.

    4.2.2 Calculate the work done in lifting the crate.

  15. Use the formula for work done by a constant force, W=Fdcos⁡θW = Fd\cos\theta.

    The force doing the work is the tension T=550 NT = 550 \text{ N} (from 4.2.1).
    The distance d=30 md = 30 \text{ m}.
    The force and displacement are in the same direction, so θ=0∘\theta = 0^\circ and cos⁡(0∘)=1\cos(0^\circ) = 1.

    W=TdW = Td

  16. Substitute the values and calculate the work done.

    W=550 N×30 m=16500 JW = 550 \text{ N} \times 30 \text{ m} = 16500 \text{ J}

    The work done in lifting the crate is 16500 J\boxed{16500 \text{ J}}.

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

  17. Calculate the normal force NN acting on the box.

    Mass m=25 kgm = 25 \text{ kg}
    Angle of inclination θ=25∘\theta = 25^\circ
    Normal force N=mgcos⁡θN = mg\cos\theta

    N=25 kg×9.8 m/s2×cos⁡(25∘)N = 25 \text{ kg} \times 9.8 \text{ m/s}^2 \times \cos(25^\circ) N=245 N×0.9063≈222.04 NN = 245 \text{ N} \times 0.9063 \approx 222.04 \text{ N}

  18. Calculate the maximum static frictional force fs,maxf_{s,max} using the formula fs,max=μsNf_{s,max} = \mu_s N.

    Coefficient of static friction μs=0.35\mu_s = 0.35

    fs,max=0.35×222.04 N≈77.71 Nf_{s,max} = 0.35 \times 222.04 \text{ N} \approx 77.71 \text{ N}

    The maximum static frictional force is 77.71 N\boxed{77.71 \text{ N}}.

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

  19. Calculate the component of the gravitational force parallel to the incline, which tends to pull the box down.

    Fparallel=mgsin⁡θF_{parallel} = mg\sin\theta Fparallel=25 kg×9.8 m/s2×sin⁡(25∘)F_{parallel} = 25 \text{ kg} \times 9.8 \text{ m/s}^2 \times \sin(25^\circ) Fparallel=245 N×0.4226≈103.54 NF_{parallel} = 245 \text{ N} \times 0.4226 \approx 103.54 \text{ N}

  20. Compare the parallel component of gravity with the maximum static frictional force.

    From 4.3.1, the maximum static frictional force fs,max=77.71 Nf_{s,max} = 77.71 \text{ N}.
    The component of gravity pulling the box down the incline is Fparallel=103.54 NF_{parallel} = 103.54 \text{ N}.
    Since Fparallel(103.54 N)>fs,max(77.71 N)F_{parallel} (103.54 \text{ N}) > f_{s,max} (77.71 \text{ 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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