Math Problem Statement

The flywheel of an engine has moment of inertia 2.5 π‘˜π‘” π‘š2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rev/min in 8 𝑠, starting from rest?

Solution

To determine the constant torque required, let's break down the problem into steps. Given data includes:

  • Moment of inertia, I=2.5 kgβ‹…m2I = 2.5 \, \text{kg} \cdot \text{m}^2
  • Final angular speed, Ο‰f=400 rev/min\omega_f = 400 \, \text{rev/min}
  • Time, t=8 st = 8 \, \text{s}
  • Initial angular speed, Ο‰i=0\omega_i = 0 (since it starts from rest)

Step 1: Convert Angular Speed to Radians per Second

The angular speed Ο‰f\omega_f is given in revolutions per minute, so let's convert it to radians per second.

Ο‰f=400 rev/minΓ—2π rad1 revΓ—1 min60 s\omega_f = 400 \, \text{rev/min} \times \frac{2 \pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}}

Ο‰f=400Γ—2Ο€60 rad/s\omega_f = \frac{400 \times 2 \pi}{60} \, \text{rad/s}

Ο‰f=800Ο€60 rad/s=40Ο€3β‰ˆ41.89 rad/s\omega_f = \frac{800 \pi}{60} \, \text{rad/s} = \frac{40 \pi}{3} \approx 41.89 \, \text{rad/s}

Step 2: Calculate the Angular Acceleration

Since the flywheel starts from rest, the angular acceleration Ξ±\alpha can be calculated using the formula:

Ξ±=Ο‰fβˆ’Ο‰it\alpha = \frac{\omega_f - \omega_i}{t}

Substituting the values:

Ξ±=41.89 rad/sβˆ’08 s=41.898β‰ˆ5.24 rad/s2\alpha = \frac{41.89 \, \text{rad/s} - 0}{8 \, \text{s}} = \frac{41.89}{8} \approx 5.24 \, \text{rad/s}^2

Step 3: Calculate the Required Torque

The torque Ο„\tau required to produce this angular acceleration can be calculated using the formula:

Ο„=Iβ‹…Ξ±\tau = I \cdot \alpha

Substitute the values for II and Ξ±\alpha:

Ο„=2.5 kgβ‹…m2Γ—5.24 rad/s2\tau = 2.5 \, \text{kg} \cdot \text{m}^2 \times 5.24 \, \text{rad/s}^2

Ο„β‰ˆ13.1 Nβ‹…m\tau \approx 13.1 \, \text{N} \cdot \text{m}

Answer

The constant torque required is approximately 13.1 NΒ·m.

Would you like further details or have any questions?


Here are some related questions to expand on this concept:

  1. How would the required torque change if the time to reach 400 rev/min was reduced to 4 seconds?
  2. What would happen to the torque if the moment of inertia was doubled?
  3. How is angular momentum related to torque in this scenario?
  4. Can you calculate the work done by the torque in bringing the flywheel up to speed?
  5. What if there was a frictional torque opposing this motion? How would you account for it?

Tip: Remember that converting units properly, especially for angular speed, is crucial in rotational dynamics.

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Math Problem Analysis

Mathematical Concepts

Rotational Dynamics
Torque
Angular Motion
Moment of Inertia

Formulas

Ο‰ = Ο‰β‚€ + Ξ±t
Ξ± = (Ο‰ - Ο‰β‚€) / t
Ο„ = I * Ξ±
Ο‰ = Ο‰ (rev/min) * (2Ο€ rad/rev) * (1 min/60 s)

Theorems

Newton's Second Law for Rotation

Suitable Grade Level

Grades 11-12