A gyroscope flywheel of radius 2.44 cm is accelerated from rest at 12.3 rad/s2 u

Question

A gyroscope flywheel of radius 2.44 cm is accelerated from rest at 12.3 rad/s2 until its angular speed is 2520 rev/min. (a) What is the tangential acceleration of a point on the rim of the flywheel during this spin-up process? (b) What is the radial acceleration of this point when the flywheel is spinning at full speed? (c) Through what distance does a point on the rim move during the spin-up?

Note: I found (a) and (b) which were .300 m/s^2 and 1700 m/s^2 respectively (with significant digits limited to three). I'm having trouble with (c).

Explanation / Answer

Iniial angular speed w = 0 radius r = 2.44 cm = 0.0244 m angular accleration a = 12.3 rad / s^ 2 final angular speed w ' = 2520 rev / m in = 2520 * 2*pi rad / 60 s = 263.89 rad /s from the relation w'^ 2- w^ 2 = 2a Q Angular displacment Q = [ w'^ 2- w^ 2] / 2a = 2830.89 rad distance on the rim move during the spin-up is S = Q r = 69 m

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