5- A uniform disk with mass 2 kg and radius 0.50 m can rotate freely in a horizo

Question

5- A uniform disk with mass 2 kg and radius 0.50 m can rotate freely in a horizontal plane about its center. A smaller uniform disk, with a mass of 1 kg and a radius of 0.25 m, lies on top of this disk and is concentric with it. The two are initially rotating together with an angular velocity of 5 rad/s. What is the angular momentum of this system? The smaller disk then slips, and quickly slides to a new position, where its edge is now touching the edge of the larger disk. At this point it stops slipping and comes to rest relative to the larger disk. What is the angular speed of the rotating system now? How much mechanical energy has been lost in the process?

Explanation / Answer

5. I= M R^2 / 2 (for solid disk)

I1 = (2)(0.50^2)/2 = 0.25 kg m^2

I2 = (1)(0.25^2)/2 = 0.03125 kg m^2

L = (I1 + I2) w = (0.25 + 0.03125) (5)

L = 1.41 kg m^2 /s .......Ans

Applying angular momentum conservation,

1.41 = I1 w^2 /2 + I2' w^2 /2

I2' = I2 + (M2 0.25^2) = 0.09375 kg m^2

2 x 1.41 = (0.25) w^2 + (0.09375) w^2

w = 2.86 rad/s ............Ans

Ki = (0.25 + 0.03125) (5^2)/2 = 3.52 J  

Kf = (0.25 + 0.09375)(2.86^2)/2 = 1.41 J  

energy lost = Ki - Kf = 2.11 J ...Ans

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