The flywheel of a steam engine runs with a constant angular velocity of . When steam is shut off, the friction of the bearings and of the air stops the wheel in 2.2h.

(a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown?

(b) How many revolutions does the wheel make before stopping?

(c) At the instant the flywheel is turning at75revmin , what is the tangential component of the linear acceleration of a flywheel particle that is50cm from the axis of rotation?

(d) What is the magnitude of the net linear acceleration of the particle in (c)?

Short Answer

Expert verified
  1. The constant angular acceleration, in revolution per minute squared, of the wheel during the slowdown αis-1.14revmin2
  2. The number of revolutions the wheel makes before stopping θis 9.9×103rev.
  3. The tangential component of the linear acceleration of a flywheel particle that isfrom the axis of rotationatis-0.99mms2.
  4. The magnitude of the net linear acceleration of the particle in (c) a is 31ms2.

Step by step solution

01

Understanding the given information

  1. The constant angular velocity of flywheel of steam engine ω is, 150revmin.
  2. The air stops the wheel when t is2.2h.
  3. At the instant, the flywheel is rotating at angular speedωis75revmin.
  4. Distance from the axis of rotation r is, 0.50m.
02

Concept and Formula used for the question

By using formulas for the angular acceleration α, the kinematic equation θ, the tangential acceleration at,and radial acceleration ar, we can find theconstant angular acceleration, in revolution per minute squared, of the wheel during the slowdown, the number of revolution the wheel makes before stopping, the tangential component of the linear acceleration of a flywheel particle that is 50 cmfrom the axis of rotation, and themagnitude of the net linear acceleration of the particle in (c) respectively.

  1. The angular acceleration is α=ΔωΔt
  1. The kinematic equation is θ=ω0t+12αt2
  2. The tangential acceleration is at=αr
  3. The radial acceleration is ar=ω2r
03

(a) Calculation for the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown

Angular acceleration is given by

α=ΔωΔt

Substitute all the value in the above equation.

role="math" localid="1660906926211" α=0-150revmin2.2×60minα=-1.14revmin2

Hence the value of angular acceleration is, -1.14revmin2.

04

(b) Calculation for the number of revolutions the wheel makes before stopping

Using kinematic equation with t=2.2×60min=132min, we get

θ=ω0t+12αt2

Substitute all the value in the above equation.

θ=150revmin×132min+12×-1.14revmin2×132min2=9.9×103rev

Hence the number of revolution is, 9.9×103rev.

Step 3: (c) Calculation for the tangential component of the linear acceleration of a flywheel particle

With r=500mm, the tangential acceleration is

at=αr

Substitute all the value in the above equation.

role="math" localid="1660907996924" at=-1.14revmin2×2πrad1rev×1min60s2×500mm=-0.99mm/s2

Hence the magnitude of tangential linear acceleration is, =-0.99mm/s2.

05

(d) Calculation for the magnitude of net linear acceleration of the particle in (c)

The angular speed of the flywheel is

ω=75revmin×2πradrev×1min60sω=7.85rads

With r=0.50m, the radial acceleration is given by

ar=ω2r=7.85rads2×0.50m31ms2

which is much bigger than at.

Consequently, the magnitude of acceleration is given by

a=ar2+at2

Substitute all the value in the above equation.

a=312+-0.992=31ms2ar

Hence the magnitude of linear acceleration is, 31ms2.

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