the flywheel of a steam engine runs with a constant angular velcoity of 15o rpm.
ID: 1985788 • Letter: T
Question
the flywheel of a steam engine runs with a constant angular velcoity of 15o rpm. when the steam engine is shut off, the constant frictional forces acting at the bearings stop the wheel in 2.2 hours...what is the angular acceleration of the flywheel in revs/min^2 during the slow down...how many revolutions does the wheel make before stopping....at the instant the flywheel is turning at 75 rpm what is the tangential acceleration in m/s^2 of a point on the wheel that is 50cm from the axis of rotation...and what is the magintude of the net acceleration of that point
Explanation / Answer
(a) = o + t
0 = 150 + * 132
= -150/132
-1.136 rev/s²
(b) = ot + ½t²
= 150 *132 - ½*1.136*132²
9903 revolutions (nearest rev)
(c) See (b)
(d) 75 rpm = 75*2 rad/min
= r²
= 50 *(75*2)²
11103300 cm/min²
= 111033 m/min²
= 111033/3600 m/s²
30.84 m/s²
The net linear acceleration is the vector sum of radial and tangential acceleration [aT^2 + aR^2]
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