The flywheel of a steam engine runs with a constant angular velocity of 210 rev/

Question

The flywheel of a steam engine runs with a constant angular velocity of 210 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 1.4 h. (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 at 105 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 27 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

Explanation / Answer

(a) ? = ?o + at 0 = 150 + a * 132 a = -150/132 ˜ -1.136 rev/s² (b) ? = ?ot + ½at² = 150 *132 - ½*1.136*132² ˜ 9903 revolutions (nearest rev) (c) See (b) (d) 75 rpm = 75*2p rad/min a = r?² = 50 *(75*2p)² ˜ 11103300 cm/min² = 111033 m/min² = 111033/3600 m/s² ˜ 30.84 m/s² (e) ???? I do not understand what it is you are seeking.

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