Chapter 10, Problem 26 The flywheel of a steam engine runs with a constant angul

Question

Chapter 10, Problem 26

The flywheel of a steam engine runs with a constant angular velocity of 100 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 1.5 hrs. (a) What is the constant angular acceleration, in revolutions pr minutsquared, of the wheel during slow down? (b) How many revolutions does the wheel make before stopping? (c) At the instant the flywheel is turningat 50 rev/min, what is the tangent component of the linear acceleration ofa particle that is 47 cm from the axis of rotation? (d) What is the magnitude of the linear acceleration of the particle in (c)?

(a) answer is -1.11111 rev/min^2

(b) answer is 4504.5 revs

(c) ???????

(d) ???????

Explanation / Answer

Given

Initial angular velocity , o = 100rev/min

time , t = 1.5 hrs = 1.5*60min

    t = 90 min

a)we find angular acceleration using   = -0/t

= (0-100rev/min)/90 min  

=   -1.11 rev/min^2

b)^2 = o^2 - 2

(100 rev/min)^2 = 2(1.11 rev/min^2)

     = 4500 rev

c)Tangent component of linear acceleration is

at = r

      = 0.47 m *(1.11 rev/min^2)

      = 0.52 m/s^2

d)Radial acceleration ar = ^2r

   ar = (50*(2/60) rad/s)^2 * 0.47 m

   ar = 12.87 m/s^2

Magnitude of linear acceleration is

a = (0.52 m/s^2)^2 +(12.87 m/s^)^2

a = 12.88 m/s^2

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