Two disk are mounted (like a merry go round) on low friction bearings on the sam

Question

Two disk are mounted (like a merry go round) on low friction bearings on the same axle and can be brought together so that they couple and rotate as one unit. The first disk, with rotational inertia 3.60 kg*mg^2 about its central axis, is spinning counterclockwise at 450 revs/min. The second disk, with rotational inertia 6.50 k*mg^2 is spinning clockwise at 900 revs/min. They then couple together

a) What is their resulting angular speed in radians per sec? Hint: the net external torque is zero.


b) What is their resulting kinetic energy in joules?

Explanation / Answer

Sign convention: Clockwise: - ve Counter clockwise: +ve Data: Rotational inertia of the first disk, I1 = 3.6 kg m^2 Rotational inertia of the second disk, I2 = 6.5 kg.m^2 Angular speed of the first disk, 1 = 450 rev/min                                                            = 450 * ( 2 / 60 ) rad/s                                                            = 47.12 rad/s Angular speed of the second disk, 2 = - 900 rev/min                                                                 = - 900 * ( 2 / 60 )                                                                 = - 94.25 rad/s Solution: (a) From the law of conservation of angular momentum, = [ I1 1 + I2 2 ] / ( I1 + I2 )     = [ 3.6 * 47.12 - 6.5 * 94.25 ] / ( 3.6 + 6.5 )     = - 43.86 rad/s Ans: Resulting angular speed = 43.86 rad/s ( clockwise ) (b) Resulting kinetic energy: KE = (1/2) ( I1 + I2 ) ^2       = 0.5 * ( 3.6 + 6.5 ) * ( 43.86 )^2       = 9715 J Ans: Kinetic energy, KE = 9715 J

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