The flywheel of a steam engine runs with a constant angular speedof 350 rev/min.

Question

The flywheel of a steam engine runs with a constant angular speedof 350 rev/min. When steam is shut off,the friction of the bearings stops the wheel in 2.7 h. (a) What is the constant angular acceleration,in revolutions per minute-squared, of the wheel during theslowdown?
rev/min2

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

(c) At the instant the flywheel is turning at 75 rev/min, what isthe tangential component of the linear acceleration of a flywheelparticle that is 50 cm from the axis of rotation?
mm/s2

(d) What is the magnitude of the net linear acceleration of theparticle in (c)?
m/s2 (a) What is the constant angular acceleration,in revolutions per minute-squared, of the wheel during theslowdown?
rev/min2

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

(c) At the instant the flywheel is turning at 75 rev/min, what isthe tangential component of the linear acceleration of a flywheelparticle that is 50 cm from the axis of rotation?
mm/s2

(d) What is the magnitude of the net linear acceleration of theparticle in (c)?
m/s2

Explanation / Answer

= / t = 350 / (2.7 * 60) = 2.16 rev/ min2 = 0 t - 1/2 t2 = 350* 2.7 * 60 - (2.16 / 2) * (2.7 * 60)2 = 2.84 * 10E4rev aT = R = (2.16 * 2 / 3600) * 500 =1.88 mm / sec2 aR = 2 R = (75 * 2 /60)2 * 500 = 3.08 * 10E4 mm / sec2 Add the vectors aT and aR in the usualway (aT will be insignificant in the result soaR to the significant figures given) Add the vectors aT and aR in the usualway (aT will be insignificant in the result soaR to the significant figures given)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---