A gyroscope flywheel of radius 3.53 cmis accelerated from rest at 14.4rad/s 2 un

Question

A gyroscope flywheel of radius 3.53 cmis accelerated from rest at 14.4rad/s2 until its angular speed is 2760 rev/min. (a) What is the tangential acceleration of apoint on the rim of the flywheel during this spin-up process?
1 m/s2
(b) What is the radial acceleration of this point when the flywheelis spinning at full speed?
2 m/s2
(c) Through what distance does a point on the rim move during thespin-up?
3 m (a) What is the tangential acceleration of apoint on the rim of the flywheel during this spin-up process?
1 m/s2
(b) What is the radial acceleration of this point when the flywheelis spinning at full speed?
2 m/s2
(c) Through what distance does a point on the rim move during thespin-up?
3 m

Explanation / Answer

Radius of the fly wheel R = 3.53 cm = 0.0353 m Angular accleration = 14.4 rad/s2 Initial angular speed = 0 rad / s Final angular speed ' = 2760 rev / min                                   = 2760 * 2 rad / 60 s                                   = 289.03 rad / s (a). Tangential accleration a t =R                                             
(b) . radial acceleration of this point when the flywheel isspinning at full speed a r = R '2                                                                                                                            
(c). from the realtion ' 2 - 2 = 2 we get angulardisplacement = [   ' 2 - 2 ] / 2                                               = ____ we know distancemoved = * R                                         = _____ we know distancemoved = * R                                         = _____

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