A gyroscope flywheel of radius 3.23 cm is accelerated fromrest at 17.3 rad/s 2 u

Question

A gyroscope flywheel of radius 3.23 cm is accelerated fromrest at 17.3 rad/s2 until its angular speed is 2880rev/min.
a) What is the tangential acceleration of a pointon the rim of the flywheel during this spin-up process? b) What is the radial acceleration of thispoint when the flywheel is spinning at full speed? c) Through what distance does a point on therim move during the spin-up? A gyroscope flywheel of radius 3.23 cm is accelerated fromrest at 17.3 rad/s2 until its angular speed is 2880rev/min.
a) What is the tangential acceleration of a pointon the rim of the flywheel during this spin-up process? b) What is the radial acceleration of thispoint when the flywheel is spinning at full speed? c) Through what distance does a point on therim move during the spin-up?

Explanation / Answer

Given: R = 3.23 cm = 0.0323 m = 17.3 rad/s2 = 0 rad / s ' = 2880 rev / min        = 2880 * 2 rad / 60s        = 301.44 rad / s a). Tangential accleration a t =R                                              =0.55879 m / s 2 b) . radial acceleration of this point when the flywheelis spinning at full speed: a r = R '2        = 2934.97 m /s2 c) from the realtion ' 2 - 2 = 2 => =[   ' 2 - 2 ]/ 2         = 2626.187 rad we know distancemoved = * R                                         = 84.82 m b) . radial acceleration of this point when the flywheelis spinning at full speed: a r = R '2        = 2934.97 m /s2 c) from the realtion ' 2 - 2 = 2 => =[   ' 2 - 2 ]/ 2         = 2626.187 rad we know distancemoved = * R                                         = 84.82 m we know distancemoved = * R                                         = 84.82 m

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