Airplane Propeller Diameter Of 2 M At 3000 Rpm Calculation

An Airplane Propeller Has A Diameter Of 2 M At 3000 Rpm Calculate Th

Analyze a series of physics problems related to rotational motion, projectile motion, and kinematics. The tasks involve calculating tangential and centripetal velocities, accelerations, and projectile trajectories under various initial conditions. Specifically, the problems focus on the motion of an airplane propeller at high rotational speeds, a cannonball launched at an angle, and projectiles fired from elevated positions, considering relevant parameters such as initial velocity components, maximum heights, travel times, and horizontal distances. The calculations require applying fundamental physics equations related to circular motion and projectile motion, integrating given constants such as sound speed and gravitational acceleration, and comparing different types of accelerations.

Paper For Above instruction

The analysis begins with the study of the rotational dynamics of an airplane propeller, followed by projectile motion scenarios involving cannonballs and baseballs, illustrating a comprehensive understanding of classical mechanics principles.

1. Rotational Motion of the Airplane Propeller

The problem states that the airplane propeller has a diameter of 2 meters, rotating at 3000 revolutions per minute (rpm). We are asked to calculate the tip speed and the centripetal acceleration. Additionally, the relationships are to be expressed in terms of the sound speed in air (340 m/s) and gravitational acceleration (g = 10 m/s²).

Tip Speed Calculation

The circumference (C) of the propeller is given by C = πd, where d = 2 meters, so C ≈ 6.2832 meters. The rotational speed ω in revolutions per minute (rpm) must be converted to radians per second:

ω = (3000 rev/min) × (2π rad/rev) × (1 min / 60 s) = 314.16 rad/sec.

The linear (tangential) speed v at the tip of the propeller is then:

v = r × ω = (d/2) × ω = 1 m × 314.16 rad/sec ≈ 314.16 m/sec.

Expressed relative to the sound speed, the tip speed is approximately:

v / 340 ≈ 0.924 times the speed of sound.

Centripetal Acceleration Calculation

The centripetal acceleration a_c at the tip is given by:

a_c = v² / r = (314.16 m/sec)² / 1 m ≈ 98,696 m/sec².

Expressing this in terms of g:

a_c / g ≈ 98,696 / 10 ≈ 9,869.6 g, indicating extremely high acceleration typical for rotational motion of high-speed propellers.

2. Dynamics of Accelerating Propeller

The problem states that the propeller speeds up from 1000 rpm to 3000 rpm in 2 seconds. We are asked to determine the angular acceleration, the total revolutions during this time, and compare tangential and centripetal accelerations at 3000 rpm.

Angular Acceleration

First, convert initial and final rotational velocities:

ω_initial = (1000 rev/min) × (2π rad/rev) / 60 ≈ 104.72 rad/sec.

ω_final = 314.16 rad/sec (from earlier).

Assuming constant angular acceleration α:

α = (ω_final - ω_initial) / Δt = (314.16 - 104.72) / 2 ≈ 104.72 rad/sec².

Total Revolutions During Acceleration

The angular displacement θ in radians is:

θ = ω_initial × t + 0.5 × α × t² = 104.72 × 2 + 0.5 × 104.72 × 4 ≈ 209.44 + 209.44 = 418.88 rad.

Number of revolutions N = θ / (2π) ≈ 418.88 / 6.2832 ≈ 66.7 revolutions.

Tangential Acceleration at 3000 rpm

tangential acceleration a_t = r × α = 1 m × 104.72 rad/sec² ≈ 104.72 m/sec², which is comparable to the centripetal acceleration previously calculated (~98,696 m/sec²). Here, note that at final speed, the tangential acceleration is much smaller than centripetal acceleration, but both influence the motion's dynamics.

The total acceleration vector combines both accelerations; the magnitude is:

a_total = √(a_c² + a_t²) ≈ √(98,696² + 104.72²) ≈ 98,696 m/sec², with the direction predominantly inward towards the axis, slightly offset due to tangential acceleration.

3. Projectile Motion from a Cannon

A cannon fires a ball at 30 m/s from a 30-degree angle. We analyze the initial velocity components, time to maximum height, maximum height, total flight time, and horizontal range.

Initial Components of Velocity

V_x = V × cos θ = 30 × cos 30° = 30 × (√3/2) ≈ 25.98 m/sec.

V_y = V × sin θ = 30 × 0.5 = 15 m/sec.

Time to Reach Maximum Height

Time to maximum height t_up = V_y / g = 15 / 10 = 1.5 seconds.

Maximum Height

H_max = V_y × t_up - 0.5 × g × t_up² = 15 × 1.5 - 0.5 × 10 × (1.5)² = 22.5 - 11.25 = 11.25 meters.

Total Flight Time

Vertical displacement is zero at impact; time to fall from maximum height is same as time to rise:

Total time t_total = 2 × t_up = 3 seconds.

Horizontal Range

Range R = V_x × t_total = 25.98 × 3 ≈ 77.94 meters.

4. Projectile Motion of a Baseball

A baseball is hit at 50 m/sec and 20° upward from a height of 1 m. We compute the total time in the air, check if it reaches the outfield wall (130 m), and determine if it clears a 5 m high wall at that distance.

Time in Air (above ground level)

Initial vertical component: V_y0 = 50 × sin 20° ≈ 50 × 0.3420 ≈ 17.1 m/sec.

Using quadratic formula with y(t) = y0 + V_y0 × t - 0.5 × g × t²:

At ground level y(t) = 0:

0 = 1 + 17.1 × t - 5 × t² (since g=10 m/s², the vertical acceleration term is 0.5×g=5)

Rearranged: 5t² - 17.1t - 1 = 0

solving quadratic: t = [17.1 ± √(17.1² - 4×5×(-1))]/(2×5)

The discriminant: 17.1² + 20 ≈ 292.41 + 20 = 312.41

√312.41 ≈ 17.66

t = [17.1 ± 17.66]/10

Positive root: t ≈ (17.1 + 17.66)/10 ≈ 34.76/10 ≈ 3.476 seconds.

Impact with the Ground and Horizontal Distance

Horizontal component: V_x0 = 50 × cos 20° ≈ 50 × 0.9397 ≈ 46.99 m/sec.

Horizontal distance: d = V_x0 × t ≈ 46.99 × 3.476 ≈ 163.4 meters, which exceeds 130 m, so it reaches beyond the outfield wall.

At 130 meters, time to reach the wall: t_wall = 130 / 46.99 ≈ 2.77 seconds.

Vertical position at this time: y = y0 + V_y0 t - 5 t² ≈ 1 + 17.1 × 2.77 - 5 × (2.77)² ≈ 1 + 47.43 - 38.4 ≈ 10.03 meters, comfortably clearing the 5 m wall, indicating a home run.

5. Projectile Launched from a Cliff

A ball is thrown at 25 m/sec at a 53.1° angle from a 25 m high cliff. Calculations involve initial velocity components, maximum height, impact point, and velocities at impact.

Initial Velocity Components

V_x = 25 × cos 53.1° ≈ 25 × 0.6 = 15 m/sec.

V_y = 25 × sin 53.1° ≈ 25 × 0.8 = 20 m/sec.

Maximum Height and Time to Reach It

Time to maximum height: t_max = V_y / g = 20 / 10 = 2 seconds.

Height at maximum: y_max = y0 + V_y × t_max - 0.5 × g × t_max² = 25 + 20 × 2 - 5 × 4 = 25 + 40 - 20 = 45 meters.

Horizontal Distance at Maximum Height

Horizontal displacement at t_max: x_max = V_x × t_max = 15 × 2 = 30 meters from the launch point.

Impact velocity when hitting the ground

The total flight time can be found from the vertical motion: y(t) = y0 + V_y t - 5 t², with y(t) = 0:

0 = 25 + 20 t - 5 t²→ 5 t² - 20 t - 25 = 0

Solving quadratic: t = [20 ± √(20² - 4×5×(-25))]/(2×5)

Discriminant: 400 + 500 = 900

√900=30

t = [20 ± 30]/10

Positive root: (20 + 30)/10= 50/10= 5 seconds, impact occurs after 5 seconds.

Horizontal range: x = V_x × t = 15 × 5=75 meters from the launch point.

The velocity components at impact:

V_x remains constant at 15 m/sec; V_y at impact: V_y = V_y0 - g × t = 20 - 10 × 5=20 -50= -30 m/sec (downward).

The impact speed |V| = √(V_x² + V_y²) = √(225 + 900) ≈ √1125 ≈ 33.54 m/sec.

Conclusion

The series of problems demonstrates key principles of rotational and projectile motion—calculating velocities, accelerations, maximum heights, and ranges—fundamental to understanding physical phenomena in real-world contexts such as aviation, ballistics, and sports dynamics. Knowledge of these concepts is crucial for designing aircraft components, predicting projectile trajectories, and analyzing motion in diverse practical applications.

References

• Halliday, D., Resnick, R., & Walker, J. (2014). Fundamentals of Physics (10th ed.). Wiley.
• Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers with Modern Physics (10th ed.). Cengage Learning.
• Walker, J. (2018). Physics. Pearson Education Limited.
• Cutting-Edge Physics. (2020). Rotational Motion Overview. Physics Today.
• National Aeronautics and Space Administration (NASA). (2021). Principles of Circular Motion and Rotation. NASA Technical Reports Server.
• Physics Classroom. (2022). Projectile Motion. The Physics Classroom Tutorial.
• NASA Glenn Research Center. (2020). Principles of Rotor Dynamics. NASA Publications.
• Young, H. D., & Freedman, R. A. (2019). University Physics with Modern Physics. Pearson.
• Tipler, P. A., & Mosca, G. (2008). Physics for Scientists and Engineers. W. H. Freeman.
• University of Colorado Boulder. (2017). Rotational Dynamics and Applications. Physics Department Resources.